math book
TRANSCRIPT
.
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Col
l ege
Al g
ebra
Selected Chapters
M. Affouf
Kean University
September 6, 2011 Version
Contents
Preface 6
1 Rational Expressions 71.1 Basic Definitions and Principles of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1.1 Equivalent Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.1.2 Basic Principle of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.1.3 Signs of a fraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.1.4 Simplifying Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.1.5 The opposite of a quantity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.1.6 Homework: Simplifying Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . 15
1.2 Multiplication and Division of Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . 161.2.1 Multiplication of Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.2 Division of Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.3 Homework: Multiplication and Division of Rational Expressions . . . . . . . . . . . . 20
1.3 Addition of Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.3.1 Addition and subtraction with like denominators . . . . . . . . . . . . . . . . . . . . . . 211.3.2 Addition and subtraction with unlike denominators . . . . . . . . . . . . . . . . . . . . 221.3.3 The least common denominator (LCD) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.3.4 Homework: Adding and Subtracting Rational Expressions . . . . . . . . . . . . . . . . 26
1.4 Complex Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.4.1 Complex Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.4.2 Homework: Complex Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.5 Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321.5.1 Ratio and Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.5.2 Homework: Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401.7 Review Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2 Radicals and Exponents 442.1 Integer Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.1.1 Product Rule for Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.1.2 Division Rule for Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.1.3 Power Rule for Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.1.4 Power Rule for Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.1.5 Power Rule for Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502.1.6 Zero Exponent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.1.7 Negative exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.1.8 Homework: Integer Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
2.2 Roots and Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572.2.1 Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 572.2.2 Cube Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592.2.3 nth Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592.2.4 Properties of radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602.2.5 Radical Simplifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632.2.6 Homework: Roots and Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
2.3 Addition and Subtraction of Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 672.3.1 Homework: Addition and Subtraction of Radicals . . . . . . . . . . . . . . . . . . . . . 69
2.4 Multiplication of Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.4.1 Homework: Multiplication of Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
2.5 Division of Radicals and Rationalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
2
3
2.5.1 Rationalizing Denominators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 732.5.2 Homework: Division of Radicals and Rationalization . . . . . . . . . . . . . . . . . . . 77
2.6 Equations involving radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 782.6.1 Method of solving equations with radicals . . . . . . . . . . . . . . . . . . . . . . . . . . 782.6.2 Homework: Equations in Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
2.7 Rational Exponents and Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 832.7.1 Homework: Rational Exponents and Radicals . . . . . . . . . . . . . . . . . . . . . . . . 86
2.8 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 872.8.1 Homework: Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
2.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 892.10 Review Radicals and Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
3 Quadratic Equations and Inequalities 923.1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
3.1.1 Fundamental Operations with Complex Numbers . . . . . . . . . . . . . . . . . . . . . . 943.1.2 Homework: Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
3.2 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1003.2.1 Solving equations by factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1003.2.2 The square root property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1023.2.3 Homework: Factoring and Square Root Method . . . . . . . . . . . . . . . . . . . . . . . 104
3.3 Completing the square method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1053.3.1 Homework: Completing the Square Method . . . . . . . . . . . . . . . . . . . . . . . . . 110
3.4 The Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1113.4.1 Homework: Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
3.5 Equations in Quadratic Form and the Discriminant . . . . . . . . . . . . . . . . . . . . . . . . 1173.5.1 The Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1183.5.2 Homework: Equations in Quadratic Form . . . . . . . . . . . . . . . . . . . . . . . . . . 119
3.6 Quadratic and Rational Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1203.6.1 Quadratic Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1213.6.2 Rational Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.6.3 Homework: Quadratic and Rational Inequalities . . . . . . . . . . . . . . . . . . . . . . 129
3.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1303.8 Review Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
4 The Distance Formula and Circles 1324.1 The Distance Formula and Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
4.1.1 Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1324.1.2 The Distance Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.1.3 The Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1364.1.4 Homework: The Distance Formula and Circles . . . . . . . . . . . . . . . . . . . . . . . 139
5 Functions 1405.1 Introduction to Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
5.1.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.1.2 Representations of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1435.1.3 Evaluation and equality of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1455.1.4 Domain of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1475.1.5 Graphical Representation of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1485.1.6 Homework: Introduction to Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
5.2 Quadratic Functions: Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1555.2.1 Graphs of f (x) = ax2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1575.2.2 Graphs of f (x) = ax2 +k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1585.2.3 Graphs of f (x) = a(x −h)2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1595.2.4 Graphs of f (x) = a(x −h)2 +k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1595.2.5 Homework: Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
5.3 Quadratic Functions: General form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1625.3.1 Finding the x- and y- intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
4
5.3.2 Homework: Quadratic Functions: General Form . . . . . . . . . . . . . . . . . . . . . . 1655.4 Function Transformations and their Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
5.4.1 Graphs of basic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1665.4.2 Transformations of Functions and their Graphs . . . . . . . . . . . . . . . . . . . . . . 1675.4.3 Homework: Function Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
5.5 The Algebra of Functions and Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1745.5.1 The Algebra of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1745.5.2 Difference Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1755.5.3 Composition of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1765.5.4 Homework: Algebra of Functions and Compositions . . . . . . . . . . . . . . . . . . . . 181
5.6 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1825.6.1 One-to-one Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1825.6.2 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1835.6.3 Homework: Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
5.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
6 System of Linear Equations 1906.1 The Graphing method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1906.2 The Substitution Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1946.3 The Addition Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
6.3.1 Homework: System of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
A Answers and Hints 199Answers and Hints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
Preface
This text is a one semester course in college algebra. These were my lecture notes for teaching Math1000 at Kean University since 2008. The prerequisite to this course is a basic algebra course whichcovers properties of real numbers, linear equations and inequalities and their graphs, and polynomials.
I would like to acknowledge T. Fung for teaching the notes and finding many typos and errors. Iwould also like to thank my students for spotting typos.
Finally, I would appreciate any comments, suggestions, and corrections which can be addressed tothe email below.
M. Affouf
5
1 Rational Expressions
Fractions occur frequently in algebra as a ratio of two algebraic expressions. In this chapter, weintroduce the fundamental operations on fractional expressions, the rules of simplifying fractions andmethods of solving equations with fractional terms. The first section is devoted to reviewing the basicrules of operations on fractions.
☞ Fractions and their manipulations are the most challenging hurdle for algebra learners. Get ready!
1.1 Basic Definitions and Principles of FractionsOBJECTIVES
• Definitions of Fractions and Rational Expressions
• Basic Principles of Fractions
• Simplifying Rational Expressions
• The sign of Fractions
Fractions are introduced early in school. The following problems from elementary school curriculumillustrate the challenging aspect of understanding fractions.
1.1 Example Answer the following three problems from California public school standards [2006]:
1. Which of the numbers 0,1 or 2 is closest to the sum8
9+
14
15? and why?
2. Mark was on a hiking trail and after walking3
4of a mile, he found that he was only
5
8of the way
to the end of the trail. How long is the trail? Explain.
3. Draw a picture that illustrates each of the following problems and its solution. Explain how yourdrawings illustrate the problems and the solutions.
(a)3
4×
1
2
(b)3
4+
1
2
(c) 2×3
4
Solution:
1. 2 is closest to the sum8
9+
14
15since
8
9+
14
15≃ 1+1
2. Assume the length of trail is L, we translate the information to an algebraic equation:
5
8L =
3
4
solve this equation for L =8
5×
3
4=
6
5= 1.2.
3. Draw a picture that illustrates each of the following problems and its solution.
6
Chapter 1
(a)3
4×
1
2: Take a rectangle with base one and height
1
2. Divide vertically the rectangle into
four equal rectangles each with base1
4and height
1
2. Shade the first three rectangles.
Its area is given by3
4×
1
2which is equal to
3
8.
(b)3
4+
1
2Divide a unit square into four equal squares and shade three of them, divide
second square into four equal squares and shade two of them add the shaded areasin the following way:
3
4+
1
2=
1
4+
1
4+
1
4+
1
4+
1
4=
5
4
(c) 2×3
4: Draw two unit square and shade three quarters of each of them, that is we have
two copies of3
4which is
6
4ä
1 Definition The symbola
b, where b 6= 0, is called a fraction with numerator a and denominator b. It
represents the quotient of a and b; thus
a
b·b = b ·
a
b= a
This means that the product of a fraction by its denominator is equal to the numerator.
The fraction1
bis called a unit fraction. Note that b ×
1
b= 1, also we have the trivial case
a
1=
a
1·1 = 1 ·
a
1= a
1.2 Example Identify the unit fraction and how many of them are in each fraction?
1.7
9
2.5
x
3.3a
2b
Solution:
1.7
9: The unit fraction is
1
9, there are 7 of them.
2.5
x: The unit fraction is
1
x, there are 5 of them.
3.3a
2b: The unit fraction is
1
2b, there are 3a of them. ä
1.1.1 Equivalent Fractions
2 Definition We say two fractionsa
band
c
dare equal or (equivalent)
a
b=
c
dif and only if ad = bc
For example, the fractions2
4and
3
6are equivalent (equal), since 2·6 = 4·3 = 12. The following two fractions
are equivalenta
b=
k a
kb, k 6= 0
That is, if the numerator and denominator of a fraction are multiplied ( or divided) by the same non-zeroexpression, the result is a fraction equal to the original.
7
Basic Definitions and Principles of Fractions
1.3 Example Check that the following fractions are equivalent.
3
4=
−6
−8=
3x
4x=
0.3
0.4=
3(x2 +1)
4(x2 +1)
1.1.2 Basic Principle of Fractions
1. Building fractions: This process allows us to multiply both numerator and denominator by thesame factor:
a
b=
a
b·
k
k=
a ·k
b ·k, k 6= 0
2. Reducing fractions (Divide out common factors): Often, this rule is called cancellation law
a ·k
b ·k=
a
b, k 6= 0
This cancellation law can be expressed in terms of division:
(ak)÷k
(bk)÷k=
a
b, k 6= 0
1.4 Example Build up the fraction so both sides are equal3
2ab2c=
?
6a2b2c 4
Solution: Observe that the denominator is multiplied by 3ac 3, perform the same with
numerator to get3
2ab2c=
9ac 3
6a2b2c 4 ä
1.5 Example Build up the fraction so both sides are equal3x −1
x2 −1=
?
(x +1)(x −1)2
Solution: Compare the denominators: the first is x2 − 1 = (x + 1)(x − 1) and the second is
(x +1)(x −1)2 so the multiplier is (x −1). Thus we have3x −1
x2 −1=
(3x −1)(x −1)
(x +1)(x −1)2 ä
1.6 Example Build up the fractions.
1.3
y −1=
?
1− y
2.s
t=
?
t 2
3.a +2
a −2=
?
a2 −4
Solution:
1.3
y −1=
−3
1− y
2.s
t=
st
t 2
3.a +2
a −2=
(a +2)2
a2 −4 ä
1.7 Example Use cancellation law to reduce the following fractions.
1.5x
5t
2.s2t 4
st 5
8
Chapter 1
3.18a
20a
4.3a +6ab
3a
5.x2 −1
x −1
Solution:
1.5x
5t=
x
t
2.s2t 4
st 5=
s2t 4 ÷ st 4
st 5 ÷ st 4=
s
t
3.18a
20a=
18a ÷2a
20a ÷2a=
9
10
4.3a +6ab
3a=
3a(1+2b)
3a= 1+2b
5.x2 −1
x −1=
(x +1)(x −1)
x −1= x +1 ä
3 Definition A fraction is said to be simple, reduced form or in the lowest terms if its numerator anddenominator have no common factor except ∓1.
☞ Simplify a fraction, reduce a fraction, and cancel out common factors in a fraction have the samemeaning.
1.8 Example Reduce the fractions to the lowest terms.
1.12
21
2.27x2
36x3
3.28x3 y2z
14x4 y z 2
4.4x2 −4x
16(x −1)
Solution:
1.12
21=
3×4
3×7=
4
7
2.27x2
36x3=
3×3×3× x × x
2×2×3×3× x × x × x=
3
4x, or
27x2
36x3=
9x2 ×3
9x2 ×4x=
3
4x
3.28x3 y2 z
14x4 y z 2=
28x3 y2 z ÷ (14x3 y z)
14x4 y z 2 ÷ (14x3 y z)=
2y
xz
4.4x2 −4x
16(x −1)=
4x(x −1)
16(x −1)=
x
4 ä
1.9 Example Simplify the fractions.
1.5(x −1)2
10(x −1)(x +1)
2.x2 −9
(x −3)2
9
Basic Definitions and Principles of Fractions
Solution:
1.5(x −1)2
10(x −1)(x +1)=
x −1
2(x +1)
2.x2 −9
(x −3)2=
(x +3)(x −3)
(x −3)(x −3)=
x +3
x −3ä
Common Error
3x +5
2y +56=
3x
2y
Note that only same factors can be divided out. To see the mistake compare the two fractions:
7+5
1+56=
7
1= 7
the correct answer is 2, but7×5
1×5=
7
1= 7
1.1.3 Signs of a fraction
The following fractions are equivalent:a
b=
−a
−b
−a
b=
−a
b=
a
−b
−(−a
b) =−
−a
b=
a
b
☞ If the signs of the numerator and denominator in a fraction are changed, then the sign before the
fraction does not change. If, however, the sign of only the numerator or denominator is changed, then thesign before the fraction must be changed.
1.10 Example Change the sign of only numerator or denominator, so the fraction stays the same.
1.3
10
2.x +3
2− x
Solution:
1.3
10=−
−3
10=−
3
−10
2.x +3
2− x=−
x +3
−(2− x)=−
x +3
x −2=−
−(x +3)
2− x=−
−x −3
2− xä
1.11 Example Change both signs of the numerator and denominator, so the fraction stays the same.
1.3
10
2.x +3
2− x
3.b −4
b −5
Solution:
1.3
10=
−3
−10
2.x +3
2− x=
−x −3
−2+ x=
−x −3
x −2=−
x +3
x −2
3.b −4
b −5=
−b +4
−b +5=
4−b
5−bä
10
Chapter 1
1.1.4 Simplifying Rational Expressions
4 Definition Rational expressions are the quotient (or ratio) of two polynomials, provided the divisor isnot identical to zero. For example,
−8y3z 5
6y4 z 3,
3x
x −6,
3x3 −5
x2 +1
are rational expressions. However, the fractions
x +7
x − x,
x2 +p
x
x2 +1
are not rational expressions, why?.
☞ Rational expressions are meaningful, if denominator is not zero. Thus the rational expressions3x
x −7makes sense for all real numbers except x = 7.
In many algebra problems, there is a question about simplifying the rational expressions. Although,what one considers simple is a matter of personal preference. In general, this means to cancel commonfactors. That is to apply the famous cancellation law:
a ·k
b ·k=
a
b, k 6= 0
Simplifying rational expressions involves two technical steps:
1. Completely factor the numerator and the denominator.
2. Divide out common factors from both the numerator and the denominator.
1.12 Example Simplify the rational expressions.
1.9x3 +18x2 +3x
3x
2.6x2 −6z 2
6
3.28ab3
7a3b3
4.x2 −7x +12
x −4
Solution: Factor numerator and denominator if possible, then cancel common factors
1.9x3 +18x2 +3x
3x=
3x(3x2 +6x +1)
3x= 3x2 +6x +1
2.6x2 −6z 2
6=
6(x2 − z 2)
6= x2 − z 2
3.28ab3
7a3b3=
4
a2
4.x2 −7x +12
x −4=
(x −3)(x −4)
x −4= x −3 ä
1.13 Example Simplify the rational expressions.
1.20x y
15x y2
2.27u3
9u4 −15u2
11
Basic Definitions and Principles of Fractions
3.x −2y
x2 −4y2
4.18a2b5
12a4b3
5.(x −2)2
x2 −4
6.2x3
2x3 −10x2
Solution: Factor and cancel the common factors
1.20x y
15x y2=
5x y(4)
5x y(3y)=
4
3y
2.27u3
9u4 −15u2=
27u3
3u2(3u2 −5)=
9u
3u2 −5
3.x −2y
x2 −4y2=
x −2y
(x −2y)(x +2y)=
1
x +2y
4.18a2b5
12a4b3=
3b2
2a2
5.(x −2)2
x2 −4=
(x −2)2
(x −2)(x +2)=
x −2
x +2
6.2x3
2x3 −10x2=
2x3
2x2(x −5)=
x
x −5 ä
1.1.5 The opposite of a quantity
5 Definition The negative of a quantity is called the opposite of a quantity. For example,
1. The opposite of 6 is −6
2. The opposite of −25 is −(−25)= 25
3. The opposite of 3x is −3x
4. The opposite of x +2 is −(x +2) =−x −2
5. The opposite of x −5 is −(x −5) =−x +5 = 5− x (Positive first!)
The division of any nonzero quantity by its opposite equals to −1, that is
a
−a=
−a
a=−1
For example,−7
7=−1,
x +8
−(x +8)=−1,
x −5
5− x=−1, and
x −4y
4y − x=−1
1.14 Example Simplify the rational expressions.
1.2+ x
x +2
2.2− x
x −2
3.2x − x2
x2 −5x +6
4.(x2 +2x)(x2 +2x −3)
(x2 + x −2)(x2 +3x)
12
Chapter 1
5.9x2 −16
9x2 +24x +16
6.6x −3y +9z
−12x +6y −18z
Solution: Factor and identify the opposite quantities and cancel the common factors
1.2+ x
x +2= 1
2.2− x
x −2=
2− x
−(2− x)=−1
3.2x − x2
x2 −5x +6=
x(2− x)
(x −2)(x −3)=−
x
x −3
4.(x2 +2x)(x2 +2x −3)
(x2 + x −2)(x2 +3x)=
x(x +2)(x +3)(x −1)
(x +2)(x −1)x(x +3)= 1
5.9x2 −16
9x2 +24x +16=
(3x +4)(3x −4)
(3x +4)(3x +4)=
3x −4
3x +4
6.6x −3y +9z
−12x +6y −18z=
3(2x − y +3z)
−6(2x − y +3z)=
−1
2 ä
1.15 Example Simplify the rational expressions.
1.a3 −b3
a2 −b2
2.(2x − y)(2x2 − x y − y2)
(4x2 − y2)(3y2 −2x y − x2)
3.x3 + x2 + x +1
x3 + x
Solution:
1.a3 −b3
a2 −b2=
(a −b)(a2 +ab +b2)
(a −b)(a +b)=
a2 +ab +b2
a +b
2.(2x − y)(2x2 − x y − y2)
(4x2 − y2)(3y2 −2x y − x2)=
(2x − y)(x − y)(2x + y)
(2x − y)(2x + y)(y − x)(3y + x)=
−1
3y + x
3.x3 + x2 + x +1
x3 + x=
(x +1)(x2 +1)
x(x2 +1)=
x +1
x ä
13
Basic Definitions and Principles of Fractions
1.1.6 Homework: Simplifying Rational Expressions
Reduce the rational numbers.
1.1.1−35
−42
1.1.236
−60
Supply the missing term so the two fractions areequal.
1.1.34
3a=
?
18a3
1.1.4a −b
a +b=
a2 −b2
?
Simplify the rational expressions.
1.1.5−52x2 y 3
39x y 5
1.1.664s2t 5
8s2t
1.1.7x2 −9
x2 +3x
1.1.8x2 +x y
x2 − y 2
1.1.912x −8
8x −4
1.1.1020x4 −10x2
5x3 +5x2
1.1.115n2 −3n −2
3n2 −n −2
1.1.12n2 −5n −24
18+3n −n2
1.1.133a −4
4−3a
Reduce the following fractions to lowest terms.
1.1.14x4 −1
(x2 −1)(x2 +1)
1.1.154x2 −3x −1
8x +2
1.1.16x3 − y 3
x2 − y 2
1.1.17x2 −9
x3 −27
1.1.18(x −4)x −5
(x −5)
1.1.19x −3
(3x −8)x −3
1.1.205x5 −1
2−10x5
14
Chapter 1
1.2 Multiplication and Division of Rational ExpressionsOBJECTIVES
• Multiplying and dividing rational expressions
• Mixed Operations
1.2.1 Multiplication of Rational Expressions
6 Definition The product of two fractionsa
band
c
dis defined by
a
b·
c
d=
ac
bd
That is, the product of their numerators divided by the product of their denominators, provided de-nominators are not 0.
☞ It is advisable to apply cancellation law of all common factors before multiplying fractions.
1.16 Example Perform the multiplications.
1.3
4·
5
7
2.x
4·
x
2(x +1)
3.7
6·
6
7
4.x2
y2·
y2
x2
Solution:
1.3
4·
5
7=
3 ·5
4 ·7=
15
28
2.x
4·
x
2(x +1)=
x2
8(x +1)
3.7
6·
6
7=
7 ·6
6 ·7= 1
4.x2
y2·
y2
x2= 1
ä
1.17 Example Perform the multiplications.
1.3
8·
4
5
2.3
5·
5
6
3.6x2 y
4z 3·
2z 2
9x2 y2
4.2x2 y
z·
x y3
4z 3
Solution:
15
Multiplication and Division of Rational Expressions
1.3
8·
4
5=
3 ·4
8 ·5=
12
40=
3
10
2.3
5·
5
6=
3 ·5
5 ·6=
1
2
3.6x2 y
4z 3·
2z 2
9x2 y2=
12x2 y z 2
36z 3x2 y2=
1
3y z
4.2x2 y
z·
x y3
4z 3=
x3 y4
2z 4 ä
Steps in reducing product of fractions
1. Factor numerators and denominators.
2. Divide out common factors.
3. Multiply remaining fractions into one fraction using the product rule.
1.18 Example Find the product ofx2 +6x +9
xand
x2
x +3.
x2 +6x +9
x·
x2
x +3=
(x2 +6x +9)(x2)
x(x +3)
=(x +3)(x +3)xx
x(x +3)= x(x +3)
1.19 Example Multiply4a2 −1
9a2·
3a
2ab −b
Solution:4a2 −1
9a2·
3a
2ab −b=
(2a −1)(2a +1) · (3a)
9a2 ·b(2a −1)=
2a +1
3ab ä
1.20 Example Multiply and simplify.
1.x2 − x −6
x2 −4·
x2 + x −6
x2 −9
2. (2x − x2) ·x
x2 −5x +6
3.x(x −2)−3
x −2·
x(x −1)−2
x(x +2)+1
Solution:
1.x2 − x −6
x2 −4·
x2 + x −6
x2 −9=
(x −3)(x +2)
(x −2)(x +2)·
(x +3)(x −2)
(x +3)(x −3)= 1
2. (2x − x2) ·x
x2 −5x +6=
x(2− x)
1·
x
(x −3)(x −2)=−
x2
x −3
3.x(x −2)−3
x −2·
x(x −1)−2
x(x +2)+1=
x2 −2x −3
x −2·
x2 − x −2
x2 +2x +1=
(x −3)(x +1)
x −2·
(x −2)(x +1)
(x +1)2= x −3
ä
7 Definition Reciprocal: The reciprocal of a number x 6= 0 is1
xand the reciprocal of
a
bis
b
abecause
a
b×
b
a= 1
1.21 Example Find the reciprocals of the following terms −3, x2 +3,3
x
16
Chapter 1
Solution: The reciprocals are1
−3,
1
x2 +3,
x
3 ä
1.22 Example Solve the linear equation for x: ab x = 2
3
Solution: Multiply both sides by ba
b
a·
a
bx =
b
a·
2
3
x =2b
3a ä
1.2.2 Division of Rational Expressions
8 Definition The division of two fractionsa
bby
c
dis to multiply the first fraction by the reciprocal of the
second fraction (divisor) that isa
b÷
c
d=
a
b·
d
c=
ad
bc
For any real numbers a,b,c ,d , where b,c ,d 6= 0.
1.23 Example Perform the divisions.
1.2
3÷
5
7
2.4p2
q 2÷
6p2
q 3
3.x y
3z÷
x2 y2
6z 2
Solution:
1.2
3÷
5
7=
2
3·
7
5=
14
15
2.4p2
q 2÷
6p2
q 3=
4p2
q 2·
q 3
6p2=
2q
3
3.x y
3z÷
x2 y2
6z 2=
x y
3z·
6z 2
x2 y2=
2z
x y ä
1.24 Example Perform the operations.
1.2x
5÷
1
5
2.x y3
15z÷
x y2
3z
3.1
(a −b)3÷
a +b
(a −b)4
4.2a
b÷ (−2b)
Solution:
1.2x
5÷
1
5=
2x
5·
5
1=
2x ·5
5= 2x
2.x y3
15z÷
x y2
3z=
x y3
15z·
3z
x y2=
x y3 ·3z
15z ·x y2=
y
5
17
Multiplication and Division of Rational Expressions
3.1
(a −b)3÷
a +b
(a −b)4=
1
(a −b)3·
(a −b)4
a +b=
a −b
a +b
4.2a
b÷ (−2b) =
2a
b·
1
(−2b)=
−a
b2 ä
1.25 Example Perform the divisions.
1.3
5÷
2
7
2.x +2
x +1÷
x2 +4x +4
2x2 −2
3.b2 −2b
b −1÷ (b −2)
Solution:
1.3
5÷
2
7=
3
5·
7
2=
21
10
2.x +2
x +1÷
x2 +4x +4
2x2 −2=
x +2
x +1·
2(x +1)(x −1)
(x +2)(x +2)=
2(x −1)
x +2
3.b2 −2b
b −1÷ (b −2) =
b(b −2)
b −1·
1
(b −2)=
b
b −1 ä
Mixed Operations
1.26 Example Simplify.
1. (4x2 −9)÷(2x2 +5x +3)
x +2÷ (2x −3)
2.4t 2 + t −5
t 3 − t 2·
8t 4 +10t 3
16t 2 +40t +25
3.m2 −3m −10
m +2÷ (m −5)
4.a2 −4
a2 +5a +6·
a2 +6a +9
a2 +a −2÷
a +3
a −1
Solution: Follow the order of operations:
1. (4x2 −9)÷(2x2 +5x +3)
x +2÷ (2x −3) =
(2x −3)(2x +3)
1·
x +2
(2x +3)(x +1)·
1
(2x −3)=
x +2
x +1
2.4t 2 + t −5
t 3 − t 2·
8t 4 +10t 3
16t 2 +40t +25=
(4t +5)(t −1)
t 2(t −1)·
2t 3(4t +5)
(4t +5)(4t +5)= 2t
3.m2 −3m −10
m +2÷ (m −5) =
(m −5)(m +2)
m +2·
1
m −5= 1
4.a2 −4
a2 +5a +6·
a2 +6a +9
a2 +a −2÷
a +3
a −1=
(a +2)(a −2)
(a +3)(a +2)·
(a +3)(a +3)
(a −2)(a +1)·
a −1
a +3=
a −1
a +1 ä
18
Chapter 1
1.2.3 Homework: Multiplication and Division of Rational Expressions
Perform the operations and reduce to lowest terms.
1.2.1−4
25·
50
16
1.2.2−4
10·
25
−6
1.2.315a3
8·
16
30a
1.2.44
21÷
12
14
1.2.55x
7÷
7x
5
1.2.62a
3÷
3a
4÷
4a
5
1.2.77x y
6y 2·
12x3 y
21
1.2.83x3 y
8x y 3÷
3x3 y
4x y 2
1.2.92ab4
5c÷
4ac2
15bc3
1.2.10x2 +4x +4
x2 −16·
x +4
x +2
1.2.113a(a +2b)2
5b2×
b(a −2b)
12a2×
20ab
a2 −4b2
1.2.12x +4
x +2·
x2 −4
x2 +8x +16
1.2.13b2 +3b
b2 −36÷
b +3
b +6
1.2.14x2 −4
x +3÷ (3x −6)
1.2.156a5
a −5÷
18a4
4a −20
1.2.166n2 +7n −3
6n2 +11n +3·
3n2 −2n −1
3n2 −4n +1
1.2.176y 2
x2 +6x +9÷
9y
x2 +3x
1.2.182t 2 +9t +9
9t −54÷
2t 2 +5t +3
3t 2 −15t −18
1.2.19x2 +3x +2
x2 −1×
(x +3)2
x2 +3x +2÷
x +3
x +1
1.2.20u2 +u −2
u2×
u5 +3u4
u2 −4÷
[u(u +3)
(u −2)
]2
19
Addition of Rational Expressions
1.3 Addition of Rational ExpressionsOBJECTIVES
• Adding and subtracting rational expressions with like denominators
• Finding the least common denominator
• Adding and subtracting rational expressions with unlike denominators
1.3.1 Addition and subtraction with like denominators
Expressions with like denominator can be added and subtracted using distributive property:
a
b+
c
b= a(
1
b)+c(
1
b) = (a +c)(
1
b) =
a +c
b
a
b−
c
b=
a −c
b
9 Definition The sum and the difference of two fractions with the same denominator are defined asfollows
a
b+
c
b=
a +c
band
a
b−
c
b=
a −c
b
To add (or subtract) fractions with like denominators, we simply add or subtract numerators and dividethe result by their common denominator.
1.27 Example Perform the operations.
1.3
4+
9
4
2.2
5−
6
5+
13
5
3.5
8−
3
8
4.4
3x−
7
3x
5.a2
a2 −1−
a
a2 −1
Solution:
1.3
4+
9
4=
3+9
4=
12
4= 3
2.2
5−
6
5+
13
5=
2−6+13
5=
9
5
3.5
8−
3
8=
5−3
8=
2
8=
1
4
4.4
3x−
7
3x=
4−7
3x=
−3
3x=
−1
x
5.a2
a2 −1−
a
a2 −1=
a2 −a
a2 −1=
a(a −1)
(a −1)(a +1)=
a
a +1 ä
20
Chapter 1
1.3.2 Addition and subtraction with unlike denominators
We have the basic rule of building fractions: For any real numbers a,b,k and b 6= 0 and k 6= 0 we havethe following identity
a
b=
a ·k
b ·k
This rule is essential to add fractions with different denominators.
10 Definition Two fractions with different denominators can be added (or subtracted) by rebuilding bothfractions to equivalent fractions with same denominators as follows
a
b+
c
d=
a
b·
d
d+
c
d·
b
b=
ad
bd+
bc
bd=
ad +bc
bd
1.28 Example Combine into a single fraction and simplify.
1.3
4+
7
10
2.7
8−
5
6
3.2
3−
3
2+
5
6
Solution:
1.3
4+
7
10=
3
4·
5
5+
7
10·
2
2=
15
20+
14
20=
29
20
2.7
8−
5
6=
7
8·
3
3−
5
6·
4
4=
21
24−
20
24=
21−20
24=
1
24
3.2
3−
3
2+
5
6=
2 ·2
3 ·2−
3 ·3
2 ·3+
5
6= 0 ä
☞ As a general rule, The product of all denominators is always a common denominator, but it may notbe the best expressions to work with.
1.29 Example Combine each expressions.
1.2x
5+
1
2
2.3
z−
1
3z +1
3.a
3+
3
a
4. 5x −2
x
Solution:
1.2x
5+
1
2=
2x
5·
2
2+
1
2·
5
5=
4x +5
10
2.3
z−
1
3z +1=
3
z·
3z +1
3z +1−
1
3z +1·
z
z=
9z +3− z
z(3z +1)=
8z +3
3z 2 + z
3.a
3+
3
a=
a
3·
a
a+
3
a·
3
3=
a2 +9
3a
4. 5x −2
x= 5x ·
x
x−
2
x=
5x2 −2
x ä
21
Addition of Rational Expressions
1.30 Example Combinex
x − y+
y
y − x
Solution: Note that the denominators are opposite of each other so we factor −1 from thesecond ( change of sign) and obtain:
x
x − y+
y
y − x=
x
x − y−
y
x − y=
x − y
x − y= 1 ä
1.31 Example Combine into a single fraction and simplify.
1.3
x+
5
2x
2.2x
2x −1−
1
1−2x
3.x2
4+2x2 −
2x2
3
4. 3−7
x −2
5.2x
x +2−
2x
x −2
Solution:
1.3
x+
5
2x=
3 ·2
x ·2+
5
2x=
6+5
2x=
11
2x
2.2x
2x −1−
1
1−2x=
2x
2x −1+
1
2x −1=
2x +1
2x −1
3.x2
4+2x2 −
2x2
3=
x2 ·3
4 ·3+
2x2 ·12
12−
2x2 ·4
3 ·4=
(3+24−8)x2
12=
19x2
12
4. 3−7
x −2=
3(x −2)
(x −2)−
7
x −2=
3x −13
x −2
5.2x
x +2−
2x
x −2=
2x(x −2)
(x +2)(x −2)−
2x(x +2)
(x −2)(x +2)=
−8x
(x +2)(x −2) ä
1.32 Example Combine the rational expressions:1
2x−
2
3x2+
3
4x3
Solution:
1. Method 1: Find a common denominator such as the product of all denominators: 2x ·
3x2 ·4x3 = 24x6 and build up all fractions to this denominator:1
2x
12x5
12x5−
2
3x2
8x4
8x4+
3
4x3
6x3
6x3=
12x5 −16x4 +18x3
24x6=
2x3(6x2 −8x +9)
24x6=
6x2 −8x +9
12x3
2. Method 2: Find the least common denominator that is 12x3 and build up all fractions to
this denominator:1
2x
6x2
6x2−
2
3x2
4x
4x+
3
4x3
3
3=
6x2 −8x +9
12x3
Both approaches lead to the same fraction, but the second method is simpler to implement. ä
1.3.3 The least common denominator (LCD)
The simplest common denominator is called the least common denominator (LCD). Here are thedefinite steps to construct it.
1. Factor each denominator completely.
2. List all different factors of each denominator with their highest exponent.
22
Chapter 1
3. the LCD is the product of different factors with their highest exponent.
1.33 Example Find the LCD of the rational expressions.
1.5a
24b5+
11a
18b4
2.1
x2 −12x +36+
3− x
x2 −6x
Solution:
1.5a
24b5+
11a
18b4. Factor the denominators: 24b5 = 23 ·3 ·b5 and 18b4 = 2 ·32 ·b4 so the LCD is
23 ·32 ·b5 = 72b5
2.1
x2 −12x +36+
3− x
x2 −6x. Factor the denominators: x2 −12x +36 = (x −6)2 and x2 −6x = x(x −6)
so the LCD is x(x −6)2 ä
1.34 Example Find the LCD and combine.
1.5a
24b5+
11a
18b4
2.3
x3 y2 z+
2
x y3 z 2
3.2
x2 − y2−
1
x − y
Solution:
1. The LCD is 72b5. Thus5a
24b5
3
3+
11a
18b4
4b
4b=
15a +44ab
72b5
2. The LCD is x3 y3 z 2. Thus3
x3 y2 z
y z
y z+
2
x y3 z 2
x2
x2=
3y z +2x2
x3 y3z 2
3. The LCD is (x − y)(x + y). Thus2
(x − y)(x + y)−
1
x − y
x + y
x + y=
2− x − y
(x − y)(x + y) ä
1.35 Example Find the LCD and subtractx
x2 −2x +1−
4
x2 −1
Solution: Factor both denominators: x2 −2x +1 = (x −1)2 and x2 −1 = (x −1)(x +1). The LCD isthe product of all different factors with highest power, that is LC D = (x −1)2(x +1). Build up thefractions to have the LCD:
x
x2 −2x +1−
4
x2 −1=
x(x +1)
(x −1)2(x +1)−
4(x −1)
(x −1)(x +1)(x −1)=
x2 + x −4x +4
(x −1)2(x +1)=
x2 −3x +4
(x −1)2(x +1) ä
1.36 Example Simplify into a single rational expression.
1.3
x +1−
2
x −1+
x +3
x2 −1
2.7
9x y3−
4
3x+
5
2y2
3. 5y −3−5y2 −4
y +3
4. 4+3
x−
2
x2
5.5y
x2 − y2+
5
x + y
23
Addition of Rational Expressions
Solution:
1.3
x +1−
2
x −1+
x +3
x2 −1=
3(x −1)
(x +1)(x −1)−
2(x +1)
(x −1)(x +1)+
x +3
x2 −1=
3x −3−2x −2+ x +3
(x +1)(x −1)=
2(x −1)
(x +1)(x −1)=
2
x +1
2.7
9x y3−
4
3x+
5
2y2=
7 ·2
9x y3 ·2−
4 ·6y3
3x ·6y3+
5 ·9x y
2y2 ·9x y=
−24y3 +45x y +14
18x y3
3. 5y −3−5y2 −4
y +3=
(5y −3)(y +3)
y +3−
5y2 −4
y +3=
5y2 +15y −3y −9−5y2 +4
y +3=
12y −5
y +3
4. 4+3
x−
2
x2=
4x2 +3x −2
x2
5.5y
x2 − y2+
5
x + y=
5y
(x + y)(x − y)+
5(x − y)
(x + y)(x − y)=
5x
(x + y)(x − y) ä
1.37 Example Combine the following fractions into a single fraction.
1.3
x−
2
x −2y+
12y2 +2x y
x(x −2y)(x +2y)
2.1
a2 −4a +3−
1
a2 −5a +4
3.m +4
m −4+
m −5
m +5−
2m2 −7m +5
m2 +m −20
Solution:
1. The LCD is x(x −2y)(x +2y)
3
x·
(x −2y)(x +2y)
(x −2y)(x +2y)−
2
x −2y·
x(x +2y)
x(x +2y)+
12y2 +2x y
x(x −2y)(x +2y)=
3x2 −12y2 −2x2 −4x y +12y2 +2x y
x(x −2y)(x +2y)
=x(x −2y)
x(x −2y)(x +2y)
=1
x +2y
2. The LCD is (a −1)(a −3)(a −4)
1
a2 −4a +3−
1
a2 −5a +4=
1
(a −1)(a −3)−
1
(a −1)(a −4)
=(a −4)− (a −3)
(a −1)(a −3)(a −4)
=−1
(a −1)(a −3)(a −4)
3.
m +4
m −4+
m −5
m +5−
2m2 −7m +5
m2 +m −20=
m +4
m −4+
m −5
m +5−
2m2 −7m +5
(m −4)(m +5)
=(m +4)(m +5)
(m −4)(m +5)+
(m −5)(m −4)
(m +5)(m −4)−
2m2 −7m +5
(m −4)(m +5)
=m2 +7m +10+m2 −9m +20−2m2 +7m −5
(m −4)(m +5)
=5m +25
(m −4)(m +5)=
5(m +5)
(m −4)(m +5)
=5
m −4 ä
24
Chapter 1
1.3.4 Homework: Adding and Subtracting Rational Expressions
Combine and simplify.
1.3.11
8−
3
8
1.3.2x
4+
2−x
4
1.3.35s
7t+
2s
7t
1.3.4a2 +6
a2 +1−
5
a2 +1
1.3.53
x2 −4−
1−x
x2 −4
1.3.64z
4z −1+
1
1−4z
1.3.7u2 +1
u2 −1+
2u
1−u2
1.3.8−4
5−
3
10
1.3.91
6+
3
−10
1.3.104
3−
5
6+
1
4
1.3.11x −3
4+
x +4
3
1.3.122x −3
2−
3x −2
5
1.3.137
10x−
9
8x
1.3.143
10x2+
4
15x
1.3.151
2n2−
2
9n+
5
6
1.3.162
3−x−
3
x
1.3.177x
1+7x−1
1.3.18b
2−b−
3b
4−b2
1.3.19x
x2 −2x y + y 2−
x
x − y
1.3.20x
y 2 −x y+
y
x2 −x y
1.3.211
x−
5
2x −3+
3
x(2x −3)
1.3.223
2x2 +5x−
1
x
1.3.232n
n2 −16−
3
5n −20
1.3.245
x2 +1−
5
(x2 +1)(x +1)
1.3.251
x2 −1+
1
x2 +2x +1
1.3.26x −1
x −2−
x −2
x −3+
3
x2 −5x +6
1.3.275x2 −5
5x2 +4x −1−
3x +1
5x −1+3
1.3.283x
x2 −1−
x −1
x(x +1)+
2x2 +1
x3 −x
1.3.29x2 −x −2
x2 +x −2−
x2 +x −2
x2 −x −2
1.3.30a
bc+
b
ca+
c
ab
25
Complex Fractions
1.4 Complex Fractions1.4.1 Complex Fractions
11 Definition A complex fraction is a fraction in which a fraction occurs in the numerator or denomina-tor or both. For example
1
2
3
7
,
2
3
6,
3x2
x +2x
x −3
,
1
2+ x
1
x−3
,
1
x+
1
y
x + y
1.38 Example Simplify the complex fractions.
1.
2
3
4
5
2.
12
11
24
55
Solution: Use the definition of fractional division
1.
2
3
4
5
=2
3÷
4
5=
2
3·
5
4=
5
6
2.
12
11
24
55
=12
11÷
24
55=
12
11·
55
24=
5
2
ä
1.39 Example Simplify
3x2
x +2x
x −3
Solution:
3x2
x +2x
x −3
=3x2
x +2÷
x
x −3=
3x2
x +2·
x −3
x=
3x(x −3)
x +2ä
Strategies to simplify complex fractions
There are two methods to reduce complex fractions to single fractions:
First Method
1. Simplify the numerator and the denominator separately to simple fractions.
2. Apply the division of two simple fractions.
1.40 Example Simplify the complex fractions.
1.
2a
b2
4a
b
26
Chapter 1
2.1+
1
x
1−1
x
Solution:
1.
2a
b2
4a
b
=2a
b2÷
4a
b=
2a
b2·
b
4a=
b
2
2.1+
1
x
1−1
x
=
x +1
xx −1
x
=x +1
x·
x
x −1=
x +1
x −1
ä
1.41 Example Simplify the complex fractions.
1.
3x
2y2
5x2
4y
2.
3
x+
1
2x
1+2
x
3.x −
4
x −3
x −1−8
x −3
Solution:
1.
3x
2y2
5x2
4y
=3x
2y2·
4y
5x2=
6
5x y
2.
3
x+
1
2x
1+2
x
=
3 ·2
x ·2+
1
2xx
x+
2
x
=
7
2xx +2
x
=7
2x·
x
x +2=
7
2(x +2)
3.x −
4
x −3
x −1−8
x −3
=
x(x −3)−4
x −3
(x −1)(x −3)−8
x −3
=x2 −3x −4
x2 −4x −5=
(x −4)(x +1)
(x −5)(x +1)=
x −4
x −5
ä
Second Method
1. Find the LCD of every denominator in the complex fraction.
2. Multiply both the numerator and denominator by this LCD.
3. Distribute to each term. All denominators should be cancelled.
4. Simplify final fraction if possible.
1.42 Example Simplify using the LCD method.
27
Complex Fractions
1.
3
x+
1
2x
1+2
x
2.1+
1
x
1−1
x2
Solution:
1.
2x
(3
x+
1
2x
)
2x
(
1+2
x
) =6+1
2x +4=
7
2x +4
2.
x2
(
1+1
x
)
x2
(
1−1
x2
) =x2 + x
x2 −1=
x(x +1)
(x +1)(x −1)=
x
x −1
ä
1.43 Example Simplify
1
3+
1
2x
1+2
x
using the LCD method.
Solution: The denominators are: 3,1,2x, x. Thus the LCD is 6x. Multiply the numerator anddenominator by 6x:
6x(1
3+
1
2x)
6x(1+2
x)
=2x +3
6x +12
ä
1.44 Example Simplify
1
x+
1
yx + y
x y
using the LCD method.
Solution: Multiply the numerator and denominator by the LCD x y:
x y(1
x+
1
y)
x y(x + y
x y)
=y + x
x + y= 1
ä
1.45 Example Simplifyx −
1
x
1−1
x2
Solution: Multiply the numerator and denominator by x2:
x2(x −1
x)
x2(1−1
x2)
=x3 − x
x2 −1=
x(x2 −1)
x2 −1= x
ä
1.46 Example Simplify3a
2−1
a
−1
28
Chapter 1
Solution: Simplify the first part and then subtract −1:3a
1
2a −1
a
−1 =3a2
2a −1−1 =
3a2 −2a +1
2a −1ä
1.47 Example Simplify1+
1
a−
2
a2
1+3
a+
2
a2
Solution: Multiply the numerator and denominator by a2:
a2(1+1
a−
2
a2)
a2(1+3
a+
2
a2)
=a2 +a −2
a2 +3a +2=
(a +2)(a −1)
(a +2)(a +1)=
a −1
a +1ä
1.48 Example Simplify
1
x −1−
2
x1
x −1−
2
x(x −1)
Solution: Multiply and divide by x(x −1):
x(x −1)(1
x −1−
2
x)
x(x −1)(1
x −1−
2
x(x −1))
=x −2(x −1)
x −2=
−x +2
x −2=−1
ä
1.49 Example Simplify
1
x−
1
h
x −h
Solution: Change the numerator to a simple fraction:1
x−
1
h
x −h=
h − x
hxx −h
1
=h − x
hx·
1
x −h=
−1
hxä
29
Complex Fractions
1.4.2 Homework: Complex Fractions
Simplify the complex fractions.
1.4.1
x4
4
x3
8
1.4.2
28ab3
5c3
7a2b2
10c2
1.4.3
m2
m2 −1m
m2 −1
1.4.4
10x2
x2 −25
25x
x2 +10x +25
1.4.5
1
4−
1
2
5
8−
3
4
1.4.64−
2
5
3−1
7
1.4.72−
1
a2
a
1.4.8
y
3
1−2
3y
1.4.91−
4
x2
1−2
x
1.4.10
2
ab−
3
a2
1
a−
2
b
1.4.113−
1
n −2
5−3
n −2
1.4.12
1
x +2−
5
y
5
y−
3
y (x +2)
1.4.13
1
x−
1
y
1
x+
1
y
1.4.14
1
x−
1
x +1
1−2
x2 +x
1.4.15
1
x +2
x
x2 −4−
2
x2 −4
1.4.16
1
t−
1
h
t −h
1.4.17
1
x2 − y 2
1
x − y
1.4.18
3
x−
5
2x +1
2
x−
3
2x +1
1.4.191−
2
x−
3
x2
1+2
x−
3
x2
1.4.20x −1+
x −1
x +1
x −2
x +1
1.4.216+
6
a
6−6
a
1.4.224+
2
a1
8+
a
4
30
Chapter 1
1.5 Rational EquationsOBJECTIVES
• Solving rational equations.
• Extraneous solutions.
12 Definition Equations containing rational expressions are called rational equations, such as
1
x−
2
5=
1
5x−2,
3
4+
5
x +2= 1,
x +3
x −3=
2
x2 −4
Compare these equations with linear equations in fractional form
2x
3−
5
2=
x +1
6−5
Method of solving rational equations
Solving fractional equations is based on the process of clearing the equation of fractions.
1. Factor all denominators and find the LCD. Exclude numbers that make denominators zero.
2. Multiply both sides (every term) by the LCD provided the LCD 6= 0.
3. Clear out all denominators by dividing common factors.
4. The resulting equation does not have any fraction, solve and check the obtained solutions.
☞ The approach in solving fractional equations is to clear out fractions. Do not combine fractions intoa single fraction. Again, we are not adding fractions, but we are removing denominators.
1.50 Example Solve the fractional equation2x
3−
5
2=
x +1
6−5
Solution: This is a linear equation in fractional form. The LCD is 6.
6(2x
3−
5
2) = 6(
x +1
6−5)
6 ·2x
3−6 ·
5
2= 6 ·
x +1
6−6 ·5
4x −15 = x +1−30
4x − x =−29+15
x =−14
3ä
1.51 Example Solve the fractional equation2
x−
3
5x=
9
20−
1
4x
Solution: Multiplying every member of the equation by the LCD = 20x, where x 6= 0 becausedenominators can not be zero:
20x ·2
x−20x ·
3
5x= 20x ·
9
20−20x ·
1
4x
40−12 = 9x −5
x =33
9=
11
3ä
Extraneous solution. This happens when multiplying both sides by expression that can be zero.Extraneous solutions must be discarded.
31
Rational Equations
1.52 Example Solve2x
x −1= 5+
2
x −1
Solution: Multiplying by the LCD (x −1) provided x 6= 1, yields
(x −1) ·2x
x −1= (x −1) ·5+ (x −1) ·
2
x −1
2x = 5(x −1)+2
2x = 5x −5+2
−3x =−3
x = 1
This solution is excluded since denominator can not be zero. We refer to such solution asextraneous (false) solution. ä
1.53 Example Solve2x +3
x+
x
x −1= 3
Solution: Multiplying by the LCD = x(x −1), where x 6= 0,1, yields
x(x −1) ·2x +3
x+ x(x −1) ·
x
x −1= x(x −1) ·3
(2x +3)(x −1)+ x2 = 3x(x −1)
2x2 + x −3+ x2 = 3x2 −3x
4x −3 = 0
x =3
4ä
1.54 Example Solve4
x −2−
7
3− x=
40
x2 −5x +6
Solution: Since x2−5x+6 = (x−2)(x−3), the LCD = (x−2)(x−3),where x 6= 2,3, we multiply bothsides by the LCD and we factor −1 from the second denominator as follows:
4
x −2−
7
3− x=
40
x2 −5x +6
(x −2)(x −3)
(4
x −2+
7
x −3
)
= (x −2)(x −3)
(40
(x −2)(x −3)
)
(x −2)(x −3) ·4
x −2+ (x −2)(x −3) ·
7
x −3= 40
4(x −3)+7(x −2) = 40
4x −12+7x −14 = 40
11x −26 = 40
11x = 66
x = 6 ä
1.55 Example Solve3
5+
7
x +2= 2
Solution: Multiplying by the LCD 5(x +2) provided x 6= −2, yields
5(x +2) ·3
5+5(x +2) ·
7
x +2= 5(x +2) ·2
3(x +2)+35 = 10(x +2)
3x +6+35 = 10x +20
−7x = 20−41 =−21
x = 3 ä
32
Chapter 1
1.56 Example Solve−x2 +10
x2 −1+
3x
x −1=
2x
x +1.
Solution: The LCD is (x −1)(x +1). Multiply every member by the LCD provided x 6= 1,−1.
(x −1)(x +1) ·−x2 +10
(x −1)(x +1)+ (x −1)(x +1) ·
3x
x −1= (x −1)(x +1) ·
2x
x +1
−x2 +10+ (x +1) ·3x = (x −1) ·2x
−x2 +10+3x2 +3x = 2x2 −2x
5x =−10
x =−2, true solution. ä
1.57 Example Solvex +1
5−2 =−
4
x
Solution: Multiplying by the LCD 5x provided x 6= 0, yields
5x ·x +1
5−5x ·2 =−5x ·
4
x
x(x +1)−10x =−20
x2 + x −10x +20= 0
x2 −9x +20= 0
(x −5)(x −4) = 0
x = 4,5
ä
1.58 Example Solve the fractional equation and check the solutions12x
2x −1+
3
x=
6
2x −1
Solution: Multiplying by the LCD x(2x −1) provided x 6= 0, x 6=1
2
x(2x −1) · [12x
2x −1+
3
x]= x(2x −1) ·
6
2x −1
12x2 +3(2x −1)= 6x
12x2 −3 = 0
4x2 −1 = 0
(2x −1)(2x +1)= 0
x =1
2or x =−
1
2
The solution x =1
2is an extraneous since this number is excluded initially, so there is only one
true solution x =−1
2. ä
1.59 Example Solve the fractional equation and check the solutions3
x2 −25−
4
x2 +4x −5=
6
x2 −6x +5
Solution: Factor the denominators
1. x2 −25 = (x −5)(x +5)
2. x2 +4x −5 = (x +5)(x −1)
3. x2 −6x +5 = (x −5)(x −1)
33
Rational Equations
Multiply by the LCD (x −5)(x +5)(x −1) provided x 6= 5,−5,1
(x −5)(x +5)(x −1) · [3
(x −5)(x +5)−
4
(x +5)(x −1)] = (x −5)(x +5)(x −1) ·
6
(x −5)(x −1)
3(x −1)−4(x −5) = 6(x +5)
3x −3−4x +20 = 6x +30
x =−13
7is a true solution ä
1.60 Example Solvex −4
x −3+
x −2
x −3= x −3
Solution: Multiply by the LCD = (x −3) provided x 6= 3
(x −3) ·x −4
x −3+ (x −3) ·
x −2
x −3= (x −3) · (x −3)
x −4+ x −3 = x2 −6x +9
2x −7 = x2 −6x +9
x2 −8x +15 = 0
(x −3)(x −5) = 0
x = 5 is a true solution
x = 3 is a false solution ä
1.61 Example Solve10
x2+
7
x+1 = 0
Solution: Multiply by the LCD = x2 provided x 6= 0
x2 ·10
x2+ x2 ·
7
x+ x2 ·1 = x2 ·0
10+7x + x2 = 0
(x +2)(x +5) = 0
x =−2,−5 are true solutions ä
1.62 Example Solve2(x +1)
x −5=
x +7
x −5
Solution: Multiply by the LCD = x −5 provided x 6= 5
(x −5) ·2(x +1)
x −5= (x −5) ·
x +7
x −5
2(x +1) = x +7
2x +2 = x +7
2x = x +5
x = 5 A false solution, so this equation does not have any solution. ä
1.63 Example Solve3x −2
x +2+
5
3= 0
Solution: Multiply by the LCD = 3(x +2) provided x 6= −2
3(x +2) ·3x −2
x +2+3(x +2) ·
5
3= 0
3(3x −2)+5(x +2) = 0
9x −6+5x +10 = 0
14x +4 = 0
x =−2/7 ä
34
Chapter 1
1.64 Example Solve3
x +1=
7
x −1+
1
x +1
Solution: Multiply by the LCD = (x +1)(x −1) provided x 6= +1,−1
(x +1)(x −1) ·3
x +1= (x +1)(x −1) ·
7
x −1+ (x +1)(x −1) ·
1
x +1
3(x −1) = 7(x +1)+ (x −1)
3x −3 = 7x +7+ x −1
3x −3 = 8x +6
−5x = 9
x =−9/5 ä
1.65 Example Solve9x
3x +1−
1
x=−
3
3x +1
Solution: Multiply by the LCD = x(3x +1) provided x 6= 0,−1/3
x(3x +1) ·9x
3x +1− x(3x +1) ·
1
x=−x(3x +1) ·
3
3x +1
9x2 − (3x +1) =−3x
9x2 −3x −1 =−3x
9x2 −1 = 0
(3x −1)(3x +1) = 0
x = 1/3
x =−1/3 A false solution. ä
1.5.1 Ratio and Proportion
13 Definition The Ratio of two numbers a and b is defined asa
b. For example, in a classroom with 8
males and 12 females, the ratio of females to males is12
8=
3
2.
14 Definition The Proportion of two ratios is defined as the equality between them:
a
b=
c
d
The proportion has the cross-multiplication property:
ad = bc
1.66 Example Use the cross-multiplication to solve.
1.4
13=
8
x
2.2
x +3=
5
2x −1
3.2x +5
5x +2=
2x −4
5x −2
Solution:
1.
4
13=
8
x
4x = 8 ·13
x = 26
35
Rational Equations
2.
2
x +3=
5
2x −1
2(2x −1) = 5(x +3)
4x −2 = 5x +15
x =−17
3.
2x +5
5x +2=
2x −4
5x −2
(2x +5)(5x −2) = (2x −4)(5x +2)
10x2 −4x +25x −10 = 10x2 +4x −20x −8
21x −10 =−16x −8
x =2
37 ä
1.67 Example Solve the equation 1−x
x +2=
3
x +4
Solution: Note that we can not use the cross-multiplication to this equation. However, aftercombining left side into one fraction, we can use it:
x +2− x
x +2=
3
x +4
2
x +2=
3
x +4
2(x +4) = 3(x +2)
2x +8 = 3x +6
x = 2 ä
1.68 Example How to divide 63 dollars between two students in the ratio2
5
Solution: If one student gets x dollars, then the second gets 63− x. The division must be
in ratio2
5. That is
x
63− x=
2
5. Use the cross multiplication property to get 5x = 2(63− x),⇒ 5x =
126−2x, the solution is x = 18 and 63− x = 45. ä
1.69 Example A sum of $ 1750 is to be divided between two people in the ratio of 3 to 4. How muchdoes each person receive?
Solution:x
1750− x=
3
4; x = 750 ä
1.70 Example The sum of a number and its reciprocal is53
14. Find the number.
Separation of fractionsIn many problems, it is advantages to break a fraction into several fractions using the property
a +b
c=
a
c+
b
c
1.71 Example Solve the equationx +2
x +1−
x +3
x +2=
x +4
x +3−
x +5
x +4
36
Chapter 1
Solution: First we use the separation property to simplify the fractions then we add bothsides separately then we cross multiply the ratios:
x +1+1
x +1−
x +2+1
x +2=
x +3+1
x +3−
x +4+1
x +4
1+1
x +1−1−
1
x +2= 1+
1
x +3−1−
1
x +4
(x +2)− (x +1)
(x +2)(x +1)=
(x +4)− (x +3)
(x +4)(x +3)
1
(x +2)(x +1)=
1
(x +4)(x +3)
x2 +7x +12 = x2 +3x +2
4x =−10
x =−5
2
ä
1.72 Example Solve the equationx +4
x +3−
x +5
x +4=
x +6
x +5−
x +7
x +6
37
Rational Equations
1.5.2 Homework: Rational Equations
Solve the rational equations and check youranswers.
1.5.1x −3
2−
x −4
5= 1
1.5.23x +2
6−
x −3
4= x −
5
3
1.5.37
n+
2
3=
5
n
1.5.43
7x−
3
4=
5
2x
1.5.52
3x −1=
5
2x +1
1.5.63
x +1+
2
x +5=
6
x2 +6x +5
1.5.75x −9
x −3+2 =
2x
x −3
1.5.81
x −2+
1
x +2=
3
x2 −4
1.5.9x
−3=
4
x −8
1.5.10x
x +4−1 =
2
x −3
1.5.11n
n −1−
1
2=
6
n +3
1.5.125
x+
2
x2+2 = 0
1.5.133
x +2−
3
x −2= 4
1.5.142
3−5x−
3
2−3x= 0
1.5.15x +1
x −1−
x2
x2 −1=
3
x +1
1.5.16x +30
x= x
1.5.173
x−
2x
x −3= 0
1.5.181
x2 +3x +2−
2
x +2−
1
2= 0
1.5.19x
x −1−1 =
1
(x +4)(x −1)
1.5.20x
x +2+
4
x +6= 1
1.5.213
1+3x=
1
4
1.5.22 A sum of $350 is to be divided between two stu-dents in the ratio of 3 to 4. Set up an equation and solveit to find their share.
1.5.23 The sum of a number and its reciprocal is34
15.
Find the number.
1.5.24 The sum of two numbers is 60. If the larger is di-vided by the smaller, the quotient is 3 and the remainderis 8. Find the numbers.
38
Chapter 1
1.6 ExercisesReduce the following fractions to lowest terms.
1.70
98
2.81
108
3. −−4
−28
4.6cd
−4c
5.7
5·
15
4
6.2x y
7·
14
6y
7.3
5a÷
2b
3
8.2x
3y·
x
2y z
9.2
3·
4
5·
3
7
10.a
b·
2x
y·
3b
d x
11.30
5a÷6
12. 5a ÷3a
4
13. (−2
3) · (−
2
3) · (−
2
3)
14.
12
7
14
15.
3h
5
15h
6
16.3a −3b
6a −6b
17.4a +5b
20a +25b
18.ax −a y
5x −5y
19.6a −4b
6b −9a
20.2x −5y
10y −4x
21.3x2
12x5
22.18a2b4
12a5b2
23.2a2 +4ab
2a2 −6ab
24.x − y
x2 − y 2
25.9x2 −16
9x2 −24x +16
26.am −an −2bm +2bn
2cm −2cn −d m +d n
27.6x +3y +9z
12x −6y +9z
Change the following fractions into equivalent frac-tion in which the letters occur in alphabetical order.
28.(x −2y )(y −x)
(h −2g )(m +2n)
29.(u −v )(v −2u)
(x +5y )(x −2y )
30.(b −2a)(c −2d )
(r + s)(u +3v )
Perform the indicated operations and simplify.
31.2
3·18
32.7
8÷3
33.4x
7y 2·
14y 2
20x3
34. (p2 −q2) ·2p q
4p −4q
35.3y +6
4x +8· (2x2 +8x +8)
36.m +n
u −v·
u2 −v 2
m3 +n3
39
Exercises
37.3x5
2y 3÷
9x2
16y
38.4a2
b3÷3a4
39. (c2 −9d 2)÷c +3d
c −2d
40.4y 2 −1
y 2 +7y +6÷
4y 2 −4y +1
y 2 −1
41.a2 −9
3x −3y÷
a2 −6a +9
y 2−x2
42.x + y
x − y·
2x + y
2x +2y÷ (4x +2y )
43.4
11+6
44.13
16−
5
8
45.5a
2−
d
5
46.3a −b
9+
4b
3
47. 2−4x − y
3
48.5
12t−
3
2t 2
49.2
3x−
2x −3
6x +6
50.1
3n −3−
n +4
n2 +3n −4
51. 3−4x −2x −4
3+4x
52.2u +v
u −v−
u +v
v −2u
53.3
4+
1
2−
4
5
54.3
x + y+
5
x − y
55.3
x2+
5
2x
56. 7−6
y+
3
y 2
57.5x
x2 − y 2+
3
x + y
58. 4+a
a −3b
59.2a +b
a2 −ab−
2
a −b
60.3
x+
2
y−
1
z
61.5
x2 −x −2+
1
4−x2
62.7x
2x2 −x −6+
2x
2x2 +7x +6
63. 5y −3−5y 2 −4
y +3
64.5x y
x2 −4y 2+
x − y
3x +6y−
y
2x −4y
65.5
2x2 +x −6+
3
16−4x2−
2
2x2 −x −3
66.
2
3−
1
6
5
6+
1
2
67.3−
3
4
3
8+1
68.
2
a−5
5
a+4
69.
2−3x
2x − y
4+2y
2x − y
70.3+
1
x
4−1
x2
71.
1
2x−1
4x2 −1
40
Chapter 1
72.1−
1
a+
2
a2
1+3
a+
2
a2
73.
5
x + y−
2
x − y
3
x + y+
1
x − y
74.
4a
a2 −b2+
2
a −b
3
a +b−
b
a2 −b2
Find the reciprocal of the number.
75. (2
3+
5
7)
76. (3x −a/c)
77. (2
3−5/x)
Solve the equation and check the solution.
78.5
4x−
3
2x=−
1
2
79. 2−5x +7
2x −2= 0
80.2
x +2=
5
2x +2
81.5
2x +3+
1
x −6= 0
82.7
2x−
1−2x
x= 4
83.3x
x +2= 3+
5
x
84.3
x +1=
7
x −1+
2
x +1
85. Solve S =ar n −a
r −1for a
86. Solve1
x−
1
y=
1
zfor y
87. Solvex4
2=
4
x2
1.7 Review Rational Expressions1. Basic Definitions:
A rational expression is the ratio of two polynomials:P(x)
Q(x), Such as
3− x
x2 −5x +6
2. Simplifying Rational Expressions: Use the property
PK
QK=
P
Q
For example,x2 −9
x2 −5x +6=
(x −3)(x +3)
(x −3)(x −2)=
x +3
x −2
3. Multiplying Rational Expressions: Use the property
P
Q·
U
V=
PU
QV
4
x2 −1·
x −1
x +1=
4(x −1)
(x −1)(x +1)(x +1)=
4
(x +1)2
4. Dividing Rational Expressions: Use the property
P
Q÷
U
V=
PV
QU
4
x2 −1÷
x −1
x +1=
4(x +1)
(x −1)(x +1)(x −1)=
4
(x −1)2
41
Review Rational Expressions
5. Basic Rules for Simplifying Rational Expressions:
(a)a
b·b = a
(b)a
b=
c
dif and only if ad = bc
(c)a
1= a and
a
a= 1
(d) Divide out common factorsak
bk=
a
b
(e) −a
b=
−a
b=
a
−b
(f) −(−a
b) =−
−a
b=
a
b
6. Addition of Rational Expressions: For like denominators: Use the property
P
Q+
U
Q=
P +U
Q
4
x2 −1+
x
x2 −1=
4+ x
x2 −1For unlike denominators: Find the LCD and write each fraction with the LCD.
3
x −2−
5
x=
3x
x(x −2)−
5(x −2)
x(x −2)=
−2x +10
x(x −2)
7. Equations with fractions: Multiply both sides by the LCD,
1
x−2 =
4−3x
x
x(1
x−2) = x(
4−3x
x)
1−2x = 4−3x
x = 3
42
2 Radicals and Exponents
In the following sections, we shall introduce the integer exponents notation which include whole positiveexponents and extend their interpretation to zero and negative whole powers, and develop their lawsand rules. Next, we define the radical notation and its connection to rational powers and introduce therules of operating on radicals. Finally, we solve equations involving radicals and their applications.
2.1 Integer ExponentsFor problems where a number is multiplied by itself repeatedly, such as the products:
2×2×2×2×2×2×2×2×2×2 = 1024
5×5×5×5×5×5×5×5×5×5×5 = 48828125
3x ×3x ×3x ×3x ×3x = 243× x × x × x × x × x,
Mathematicians have invented a very powerful notation to represent this type of products. It is calledexponential notation. In this notation, the repeated factor (multiplier) is called the base, and thenumber of factors (multipliers) is called the exponent. So the above products are written as follows:
2×2×2×2×2×2×2×2×2×2 = 210
5×5×5×5×5×5×5×5×5×5×5 = 511
3x ×3x ×3x ×3x ×3x = (3x)5
Thus, we have the notation
21 = 2
22 = 2 ·2
23 = 2 ·2 ·2
24 = 2 ·2 ·2 ·2
and so on!
1 Definition For any real number a and any natural number n, we use the exponential notation an todenote the product of n factors , each equal to a, that is
an =n
︷ ︸︸ ︷a ·a ·a . . . a ·a
here, a is called the base, n is called the exponent or power, an is called exponential number, and weread it as: a to the n-th power or a to exponent n.
2.1 Example Write the expressions using exponential notation
1. 3×3×3×3×3×3×3
2. 4×4×4×4×5×5×5
Solution:
1. 3×3×3×3×3×3×3 = 37
43
Integer Exponents
2. 4×4×4×4×5×5×5 = 44 ×53. Note the different bases. ä
2.2 Example Compute
1. (−10)4
2. 107
3. (−10)9
Solution:
1. (−10)4 = (−10) · (−10) · (−10) · (−10)= 10,000
2. 107 = 10 ·10 ·10 ·10 ·10 ·10 ·10 = 10,000,000
3. (−10)9 =−1,000,000,000 ä
2.3 Example Compute the following the values
1. 53
2. 72
3. (−3)2
4. 24 ×52
5. (−2)2 × (−3)2
6. −32
7. (−5)2
8. −52
9. −(−5)2
10. (−1)90
11. −1235
12. −(−10)3
13. (−2)3 − (−3)2
14. (−5)2 + (−5)3
15. (−1)8 + (−1)9
Solution:
1. 53 = 125
2. 72 = 49
3. (−3)2 = (−3)(−3) = 9
4. 24 ×52 = 16×25 = 400
5. (−2)2 × (−3)2 = 36
6. −32 ==−3×3 =−9
7. (−5)2 = 25
8. −52 =−25
9. −(−5)2 =−25
10. (−1)90 = 1, even power.
11. −1235 =−1, odd power.
12. −(−10)3 =−(−1000) = 1000
13. (−2)3 − (−3)2 =−8−9 =−17
14. (−5)2 + (−5)3 = 25−125 = 100
15. (−1)8 + (−1)9 = 1−1 = 0 ä
2.4 Example Identify the base and the exponent of the following exponential numbers
1. (−a)2
2. −a2
3. 5x3
4. (5x)3
Solution:
1. (−a)2: The base is −a, the exponent is 2.
2. −a2: The base is a, the exponent is 2.
3. 5x3: The base is x, the exponent is 3.
4. (5x)3: The base is 5x, the exponent is 3. ä
As a result of the exponential definition, we are going to introduce the main rules of exponentialnumbers.
44
Chapter 2
2.1.1 Product Rule for Exponents
Based on the definition of the exponential numbers we have
x1 = x
x2 = x ·x
x3 = x ·x ·x
x4 = x ·x ·x ·x
It follows x2 ·x3 = x ·x ·x ·x ·x = x5
x2 ·x4 = x ·x ·x ·x ·x ·x = x6
The last two equalities justify the product rule for any real number a and any natural numbers n,m:
am ·an = am+n
☞ The product of exponential numbers with same base is an exponential number with same base andexponent equals to the sum of their exponents. Keep the base and add the powers.
For example, a3a4 = a3+4 = a7, and x y x2 y2 x3 y5 = x1+2+3 y1+2+5 = x6 y8. The product rule is not applicable tothe product a4b3 because the bases are different!
2.5 Example Simplify
1. 6567
2. x5x−2
3. (7x)3(7x)2
4. (x +2)5(x +2)3
Solution:
1. 6567 = 612
2. x5x−2 = x3
3. (7x)3(7x)2 = (7x)5
4. (x +2)5(x +2)3 = (x +2)8
ä
2.6 Example Simplify
1. x11x5
2. a2b3a3b2
3. −8x4x3x
4. (t +1)5(t +1)4(t +1)
Solution:
1. x11x5 = x11+5 = x16
2. a2b3a3b2 = a2+3+b3+2 = a5b5
3. −8x4x3x =−8x4+3+1 =−8x8
4. (t +1)5(t +1)4(t +1) = (t +1)5+4+1 = (t +1)10
ä
45
Integer Exponents
2.1.2 Division Rule for Exponents
The product 23 ·22 = 25, implies25
23= 2
2, and
25
22= 2
3
We also can use the definition of exponential notation and cancellation principle of fractions to divideexponential numbers with same base as in the examples:
57
53=
5 ·5 ·5 ·5 ·5 ·5 ·5
5 ·5 ·5= 5 ·5 ·5 ·5 ·1 ·1 ·1 = 5
4
63
65=
6 ·6 ·6
6 ·6 ·6 ·6 ·6=
1
1 ·1 ·1 ·6 ·6=
1
62
From these examples, point to the general quotient rule of exponential numbers with same bases a andfor any natural numbers m,n:
1. If m > n, thenam
an= am−n. For example,
x6
x2= x6−2 = x4
2. If m < n, thenam
an=
1
an−m. For example,
y7
y9=
1
y9−7=
1
y2
☞ The division of exponential numbers with same base is an exponential number with same base andexponent equals to the difference of their powers. Usually, we subtract the small power.
2.7 Example Simplify
1.x4
x3
2.t 4
t 7
3.x3 y7
y5
4.4(x +2)5
(x +2)3
Solution:
1.x4
x3= x4−3 = x
2.t 4
t 7=
1
t 7−4=
1
t 3
3.x3 y7
y5= x3 y7−5 = x3 y2
4.4(x +2)5
(x +2)3= 4(x +2)5−3 = 4(x +2)2
ä
2.8 Example Simplify
1.38
36
2.2x5
x11
3.−6a3b6c 2
−2ab4c 2
46
Chapter 2
4.x2 y3
x y4
Solution:
1.38
36= 38−6 = 32 = 9
2.2x5
x11=
2
x11−5=
2
x6
3.−6a3b6c 2
−2ab4c 2= 3a3−1b6−4 = 3a2b2, note c 2/c 2 = 1
4.x2 y3
x y4=
x2−1
y4−3=
x
y ä
2.1.3 Power Rule for Exponents
To evaluate (32)3 we proceed according to the definition as follows
(32
)3 = (3
2)× (3
2)× (3
2) = 3
2+2+2 = 36
Note that 6 = 3×2. Thus raising a power to a power, multiply the powers.
2 Definition For any real number a and any natural numbers m,n, we have
(am)
n =
n︷ ︸︸ ︷
(am) · (am
) · (am) . . . (am
)
= a
n︷ ︸︸ ︷
m +m +m +·+m
= amn
For example,
(x3)
4 = x3·4 = x12
(y2)
4 = y2·4 = y8
(x4x3)
2 = (x4+3)
2 = x7·2 = x14
2.9 Example Simplify
1. (32)3
2. (x2x3)6
3. (x2)4(x3)2
4. (xk )3
5. x(x2)4
Solution:
1. (32)3 = 32·3 = 36
2. (x2x3)6 = (x2+3)6 = (x5)6 = x30
3. (x2)4(x3)2 = x8 x6 = x14
4. (xk )3 = xk ·3 = x3k
5. x(x2)4 = x(x8) = x9 ä
47
Integer Exponents
2.1.4 Power Rule for Products
To evaluate the products (5x)4 and (−2x y2 z 3)3 we proceed according to the definition of exponentials asfollows:
(5x)4 = (5x) · (5x) · (5x) · (5x) = 5
4x4
(−2x y2 z 3)
3 = (−2x y2 z 3) · (−2x y2 z 3
) · (−2x y2 z 3) = (−2)
3x3 y2×3z 3×3 =−8x3 y6 z 9
(24x3
)2 = 2
4·2x3·2 = 28x6 = 256x6
3 Definition For any real numbers a,b and any natural number n, we have
(ab)n =
n︷ ︸︸ ︷
(ab) · (ab) · (ab) . . . (ab)
=n
︷ ︸︸ ︷a ·a ·a . . . a
n︷ ︸︸ ︷
b ·b ·b . . . b
= an bn
Similarly,
(abc)n = an bn c n
☞ Note that it is not true that (a +b)n = an +bn .
For example,
(uv )2 = u2 v 4
(2 ·5)3 = 2
3 ·53 = 8×125 = 1000
(2x y)4 = 2
4x4 y4 = 16x4 y4
(−4x)3 = (−4)
3x3 =−64x3
2.10 Example Simplify
1. (−2x2)3
2. (5x3 y)4
3. (x yk )2
4. (xk y2)k
5. (x +3)2
Solution:
1. (−2x2)3 = (−2)3x2·3 =−8x6
2. (5x3 y)4 = 54x12 y4
3. (x yk )2 = x2 y2k
4. (xk y2)k = xk ·k y2·k = xk2
y2k
5. (x +3)2 = (x +3)(x +3) = x2 +3x +3x +33 = x2 +6x +9ä
48
Chapter 2
2.1.5 Power Rule for Fractions
Exponentiation of fractions is accomplished according to the definition. For example,
(3
5)
3 = (3
5) · (
3
5) · (
3
5) =
33
53
4 Definition For any real numbers a,b 6= 0 and any natural number n, we have
(a
b)
n =an
bn
(a
bc)
n =an
bn c n
☞ Note it is not true that (a
b +c)n =
an
bn +cn. The correct answer is (
a
b +c)n =
an
(b +c)n.
For example,
(2
3
)3
=23
33=
8
27(
x
y
)4
=x4
y4
( x
x2 +1
)3
=x3
(x2 +1)3
2.11 Example Simplify
1. (2
5)3
2. (−3a
b2)2
3. (18a3b
9a2b2)2
4. (5x
10x2)k
5. (5x
x + y)8
Solution:
1. (2
5)3 =
23
53=
8
125
2. (−3a
b2)2 =
9a2
b4
3. (18a3b
9a2b2)2 = (
2a
b)2 =
4a2
b2
4. (5x
10x2)k = (
1
2x)k =
1
2k xk
5. (5x
x + y)8 =
58x8
(x + y)8
ä
49
Integer Exponents
2.1.6 Zero Exponent
The exponential number 35 means 3 ·3 ·3 ·3 ·3, but 30 is meaningless in terms of product of repeatedfactors. However, we can give this a consistent explanation as follows: It is known that for any a 6= 0,a
a= 1, using this fact and the division rule of exponential numbers to simplify 1 =
35
35= 35−5 = 30. This
implies that both sides of the equation are equal so we define 30 to be 1. Similarly, for any base a 6= 0,we define
a0 = 1
For example, 50 = 1, (3x2 +1)0 = 1 and (23
6)0 = 1. Note that a can not be zero because the division by
zero is undefined.
2.12 Example Find the values of the following expressions provided all variables are none zero.
1. −3250
2.3x0
2x0
3. (70 −80)0
4. x0
5. 5x y 0
6. (23x4 +x2 +1)0
7. (5x)0
8. 5x0
9. −5x0 y
10. 00
Solution:
1. −3250 =−9 ·1 =−9
2.3x0
2x0=
3
2
3. (70 −80)0 = (1−1)0 = 00, undefined value.
4. x0 = 1
5. 5x y 0 = 5x
6. (23x4 +x2 +1)0 = 1
7. (5x)0 = 1
8. 5x0 = 5
9. −5x0 y =−5y
10. 00, undefined value. ä
In the following example, the plausibility of power zero is illustrated by a population growth variable.
2.13 Example Derive a population model so the number of rabbits are doubling each six months, thefarm starts with three pairs. How many rabbits will be there in 5 years, 10 years?
Solution: In 5 years we have 10 periods of doubling so we compute the product:
6 ·2 ·2 ·2 ·2 ·2 ·2 ·2 ·2 ·2 ·2 = 6 ·210 = 6144
In general, we can derive a formula for the population growth p = 6·2t where t is the time period.So after one period t = 1, p = 621 = 12. In this example the number of rabbits at starting time canbe written as
p = 6t 0 = 6 ä
2.1.7 Negative exponents
How to interpret exponential numbers with negative exponents?What does 5−2 mean?Assume that 5−2 stands for a certain number. How to find the value of this number?
We start with the equality52
52= 1 multiply both sides by 5−2 and switch the sides
5−2 ·1 = 5
−2 ·52
52=
5−2+2
52=
50
52=
1
52
50
Chapter 2
This means
5−2 =
1
52=
1
25
Similarly, we define 2−3 =1
23=
1
8.
5 Definition We define the negative exponent of any none zero number a:
a−n =1
an
Based on this definition, we can interpret the division of exponential numbers
am
an= am−n =
1
an−m
a−m
a−n= a−m−(−n) = a−m+n =
1
a−n−(−m)=
1
a−n+m
a−n =1
an
1
a−m= am
2.14 Example Rewrite without negative exponents and simplify
1. 2−3 ×3−2
2.3−8
3−5
3.8x4 y−2
4x7 y−2
4. 6a−3b3
5. −2x−5 y−3z−1
Solution:
1. 2−3 ×3−2 =1
23×
1
32=
1
8×
1
9=
1
72
2.3−8
3−5=
1
3−5−(−8)=
1
33=
1
27
3.8x4 y−2
4x7 y−2=
2y−2−(−2)
x7−4=
2
x3
4. 6a−3b3 =6b3
a3
5. −2x−5 y−3 z−1 =−2
x5 y3 Z ä
In the following illustration, we justify the definition of negative exponents by following a patternformation of exponential numbers:
32 16 8 4 2 11
2
1
4
1
8
1
16
25 24 23 22 21 20 2−1 2−2 2−3 2−4
51
Integer Exponents
The pattern in the first row is dividing by 2 left to right, and the pattern in the second row is to subtract1 from the power left to right. The two rows are equal so we have
25 = 32, 2
4 = 16, 23 = 8, 2
2 = 4, 21 = 2, 2
0 = 1
and similarly
2−1 =
1
2, 2
−2 =1
22=
1
4, 2
−3 =1
23=
1
8
So we conclude that for any nonzero number a, the negative exponent is defined as follows:
a−n =1
an
For example,
x−5 =1
x5, 5
−3 =1
53=
1
125
Similarly we define the reciprocal of negative exponents as
1
a−n= an
For example1
x−2= x2
,1
3−2= 3
2 = 9,1
(−2)−3= (−2)
3 =−8
☞ The negative exponent of a fraction has an interesting short cut:
(a
b)−n =
a−n
b−n=
bn
an= (
b
a)
n
This means that instead of distributing the negative power, we flip the fraction and change to positivepower. For example,
(23
5)−2 = (
5
23)
2 =52
26=
25
32
2.15 Example Simplify
(2x2
3y3
)−2
Solution: We can simplify in two ways (they are really equivalent)
1. Apply the exponent rules:
(2x2
3y3
)−2
=2−2(x2)−2
3−2(y3)−2=
32(y3)2
22(x2)2=
9y6
4x4
2. Apply the negative exponent of fractions rule:
(2x2
3y3
)−2
=(
3y3
2x2
)2
=9y6
4x4 ä
2.16 Example Change to positive exponents and simplify
1. −3a−2b−3
2.1
2−3
3. (2−3 −3−2)2
4.a−2
a2
Solution:
1. −3a−2b−3 =−3
a2b3
2.1
2−3= 23 = 8
52
Chapter 2
3. (2−3 −3−2)2 = (1
8−
1
9)2 = (
1
72)2
4.a−2
a2=
1
a2a2=
1
a4 ä
2.17 Example Why (a−2 +b−2) 6=1
a2 +b2?
Solution: Because
a−2 +b−2 =1
a2+
1
b2=
a2 +b2
a2b2 ä
2.18 Example Simplify
1. (a4
b−3)−2
2. 3x ·312
3. (2a +5)(2a +5)3
4.a3b−2
a−2b−4
5. (3x3)2 − (2x2)3
Solution:
1. (a4
b−3)−2 =
a−8
b6=
1
a8b6
2. 3x ·312 = 3x+12
3. (2a +5)(2a +5)3 = (2a +5)4
4.a3b−2
a−2b−4= a3−(−2)b−2−(−4) = a5b2
5. (3x3)2 − (2x2)3 = 9x6 −8x6 =−2x6 ä
2.19 Example Which expressions are the same?
1. xn
2. x−n
3. −xn
4. (−x)n
5.1
x−n
6.1
xn
Solution:
1. xn =1
x−n
2. x−n =1
xn ä
Summary of Basic Rules of Exponents
For any real numbers a and b and any integers numbers m and n, the following basic laws of exponen-tial numbers
1. am an = am+n
2. (ab)n = anbn , and( a
b
)n=
an
bn
3. (am)n = amn
53
Integer Exponents
4.am
an= am−n , a 6= 0
(a) If m > n, thenam
an= am−n
(b) If m < n, thenam
an=
1
an−m
5. Zero exponent: If a 6= 0, then a0 = 1.
6. Negative Exponent: a−n =1
anand
1
a−n= an.
7. Negative exponent of fraction: (a
b)−n = (
b
a)n for nonzero bases.
54
Chapter 2
2.1.8 Homework: Integer Exponents
Perform the indicated operations
Perform the indicated operations
2.1.1 (−2)3(−3)2
2.1.2 55256
2.1.377
75
2.1.4 (3
2)4
2.1.5 (2x y 3)5
2.1.6a4b6c
a3b3c
2.1.724a5b2c3
33a5bc2
2.1.8 (12x4 y 3
4x2 y 2)3
Find the value of the expressions
2.1.9 −11−2
2.1.10 20−2 ·20−2
2.1.11 (2−3 ·32)−1
2.1.1224
2−2
2.1.13 (2
3)−1 − (
1
3)−1
2.1.14 (3−1)2(9−1)−2
Simplify the expressions without negativeexponents
2.1.15 a4 ·a−3 ·a3
2.1.16 (x−4
y−2)−1
2.1.1790x−4 y−5
6x−2 y
2.1.18a−4
a4
2.1.19 (a−2
b5)−2
2.1.20 5a−1 −5
a
2.1.21x
y−1+
y
x−1
2.1.22x−1
y−1+
y
x
2.1.23 (a−3b2c−3
a0b−2c−3)−4
2.1.24 x−2 +2x−3
55
Roots and Radicals
2.2 Roots and RadicalsIn this section, we introduce the definition of roots of numbers as opposed to raising a number tointeger powers. Geometrically, computing the area of a square and the volume of a cube given the sideare the same as raising to power 2 and 3. However, computing the side of a square and the side of acube from the area and the volume are the same as finding the square root and the cube root. Thefollowing table shows the familiar process of raising to powers:
b A = b2 V = b3 W = b4
2 4 8 16
-2 4 -8 16
3 9 27 81
-3 9 -27 81
4 16 64 256
-4 16 -64 256
1.4 1.96 2.744 3.8416
Raising a number to an integer power is a multiplication process. For example, the number 16
in 42 = 16, is the product of 4×4. The reverse process is to find numbers whose square is 16. Thesenumbers are +4 and −4. We call these numbers square roots of 16. Similarly, 3 is the cube root of 27
because 33 = 27. In general we have the definition:
6 Definition A number b is called the nth root of a if
bn = a
Roots of a number a are denoted by radical notation: A radical sign p , a radicand a,p
a, and an
index n specifying the power of the root: np
a. It is standard to omit square root index. For example,p9 = 3, 3
p64 = 4, 5
p32= 2.
2.2.1 Square Roots
Every positive real number p has two square roots: +pp and −pp. The positive square root is theprincipal square root. Thus,
p36 = 6 and −
p36=−6.
☞ For any real positive number p,
1.p
p is the positive principal square root. For example,p
49 = 7 and√
x2 is a positive number whose
square is x2, this number is not x but it is |x|, this absolute value insures that the answer is positive.
2. −pp is the negative square root. For example, −p
49 =−7.
3.p−p is not real number. For example,
p−9 is not real number because the square of any real number
is a positive number.
2.20 Example Evaluate the square roots:
56
Chapter 2
1.p
36
2. −p
121
3.
√4
25
4.p
100
5.p
0.04
6. −√
1
81
7.p
0.0
8.p−16
Solution:
1.p
36 = 6
2. −p
121 =−11
3.
√4
25=
2
5
4.p
100 = 10
5.p
0.04 = 0.2
6. −√
1
81=−
1
9
7.p
0.0 = 0
8.p−16, not real number.
ä
Every number a ≥ 0 has a real square root, but only few numbers are perfect squares. For example,p5 is a number that it impossible to express it in decimal or fractional form, that is irrational, but we
can approximate itp
5 ≈ 02.236.
☞ Thus, in all your algebra work, you should never substitute a radical number with its approximation,unless you are asked to do that explicitly.
In the following list, we identify all perfect square numbers less than 1000:
1,4,9,16,25,36,49,64,81,100,121,144,
169,196,225,256,289,324,361,400,441,
484,529,576,625,676,729,784,841,900,961
Square roots of expressions
Definition 1 For any real number x:p
x2 =| x |. Absolute value is needed to insure that the selectedroot is positive number. 2
2.21 Example Simplify the square roots:
1.√
25x2
2. −√
64u2
3.√
x4
4.√
9y 4
5.√
a2 +2a +1
6.√
x2 −10x +25
Solution:
1.√
25x2 = 5|x|
2. −√
64u2 =−8|u|
3.√
x4 = x2
4.√
9y 4 = 3y 2
5.√
a2 +2a +1 =√
(a +1)2 = |a +1|
6.√
x2 −10x +25 =√
(x −5)2 = |x −5| ä
2.22 Example The area of a square given by the formula A = s2, where s is the length of its side. In thefollowing, find the side of a square whose area is
1. A = 144
2. A = 9a2b2
3. A = 100x2
Solution:
1. s =p
144 = 12
2. s =p
9a2b2 = 3ab, provided a and b are positive.
3. s =p
100x2 = 10x, provided x > 0 ä
57
Roots and Radicals
2.2.2 Cube Roots
We say 2 is a cube root of 8 because 23 = 8 and -3 is a cube root of -27 because (−3)3 =−27.
7 Definition The cube root of a number a is denoted by 3p
a = b so that b3 = a. This definition implies theidentity
3√
a3 = a, for any real number a
2.23 Example Simplify the cube roots:
1. 3p
64
2. 3p−64
3. − 3p−27
4. 3
√
−y 3
5.3√
8a3b3c3
6. 3
√
−125(x − y )3
Solution:
1. 3p
64 = 4
2. 3p−64 =−4
3. − 3p−27 = 3
4. 3
√
−y 3 =−y
5.3√
8a3b3c3 = 2abc
6. 3
√
−125(x − y )3 =−5(x − y ) ä
2.24 Example The volume of a cube is given by the formula V = s3, where s is the side of the cube. Inthe following, find the side of a cube whose volume is
1. V = 1000
2. V = 64x3 y3
3. V = 8a3
Solution:
1. s = 3p
1000= 10
2. s = 3√
64x3 y3 = 4x y
3. s = 3p
8a3 = 2a ä
2.2.3 nth Roots
8 Definition Similarly, we can define roots of any index such as, the 4th, 5th, 6th,..., roots:4p
, 5p
, 6p . In general, the nth root of a number a, denoted by n
pa is a number b:
np
a = b if bn = a
With restrictions on even indices: The radicand must be positive and the positive principal root mustbe assigned.
2.25 Example Simplify the radical expressions:
1.p
4
2. 3p
8
3. 4p
16
4. 5p−32
5. 4p−16
6. 7
√1
128
7.4√
81x4
8.5√
−32x5
9.6√
(2x −1)6
58
Chapter 2
Solution:
1.p
4 = 2
2. 3p
8 = 2
3. 4p
16 = 2
4. 5p−32 =−2
5. 4p−16 is not a real number.
6. 7
√1
128=
1
2
7.4√
81x4 = 3|x|
8.5√
−32x5 =−2x
9.6√
(2x −1)6 = |2x −1| ä
☞1. The even roots are defined for positive numbers only, and the even roots are always positive.
2. The even roots of negative numbers are not real numbers.
3. The odd roots are defined for positive and negative numbers.
2.2.4 Properties of radicals
True or false?
1.p
a2 = a
2. np
an = a
3.p
a +b =p
a +p
b
4.p
2x +p
3x =p
5x
5.p
8 = 2p
2
The answers to these questions will be clear by the end of this section.
Fundamental Rules of Radicals
1. Perfect nth power Rule:
np
an = a if n is oddnp
an = |a| if n is even
For example,
(a)√
(3.5x)2 = 3.5|x|(b) 3
√
(6x + y)3 = 6x + y
(c) 4√
(−5)4 = |−5| = 5
(d) 5√
(22a)5 = 22a
(e) 6√
(3y)6 = 3|y |
2. Product Rule of Radicals:
np
ab = np
anp
b Radicals are distributed to each factor in the radicand.
np
anp
b = np
ab Radicals with same index can be multiplied.
provided the roots are real numbers. For example,
(a)p
4 ·5 =p
4 ·p
5 = 2p
5
(b) 3p
8 ·27 = 3p
8 · 3p
27 = 2 ·3 = 6
(c) 4√
81x8 y = 4p
81 · 4p
x8 · 4p
y = 3x2 4p
y
(d) 3p
9x · 3p
3x2 = 3p
9x ·3x2 = 3p
27x3 = 3x
59
Roots and Radicals
☞ This rule is valid for products only. It is not true to distribute radicals if an addition or asubtraction is in the radicand. For example,
p4+9 6=
p4+
p9.
3. Division Rule of Radicals:
n
√a
b=
np
anp
bRadicals are distributed to numerator and denominator in the the radicand.
np
anp
b= n
√a
bRadicals with same index can be divided.
provided the roots are real numbers and b 6= 0. For example,
(a)
√5
36=
p5
p36
=p
5
6
(b) 3
√
x3
125y6=
3p
x3
3√
125y6=
x
5y2
(c)
√
48x y4
p3x
=
√
48x y4
3x=
√
16y4 = 4y2
More Examples of Radical Rules
1. np
an = a or = |a|? In generalnp
amn = am ornp
amn = |a|m
depending on odd or even index. This property is used very often for complete extraction of perfectroots!
2.26 Example Simplify the radical expressions using the perfect nth power rule:
(a) 3p
8
(b) 3p−8
(c) 3
√
y 3
(d)7√
(1−2x)7
(e)5√
(−13)5
(f)p
25
(g)√
y 2
(h)4√
(1−2x)4
(i)4√
(−15)4
(j) 6√
(−15)6
(k)√
y 4
(l)√
y 6
(m) 3
√
y 6
Solution:
(a) 3p
8 = 3√
23 = 2
(b) 3p−8 = 3
√
(−2)3 =−2
(c) 3
√
y 3 = y
(d)7√
(1−2x)7 = 1−2x
(e)5√
(−13)5 =−13
(f)p
25 =√
52 = 5
(g)√
y 2 = |y |
(h)4√
(1−2x)4 = |1−2x|
(i)4√
(−15)4 = 15
(j)6√
(−15)6 = 15
(k)√
y 4 =√
(y 2)2 = |y |2 = y 2
(l)√
y 6 =√
(y 3)2 = |y |3
(m) 3
√
y 6 = 3
√
(y 2)3 = y 2 ä
2. np
ab = np
anp
b This property is applied to factor terms from under the radical signs! ( Partialextraction)
2.27 Example Simplify the radical expressions using the product rule of radicals:
60
Chapter 2
(a)p
48
(b)p
8
(c) 3
√
16y 3
(d)√
25x4 y 10
(e)√
y 2(1− y )
(f)p
(−4)(−4) =p
16 = 4 6=p−4 ·
p−4, explain why?
Solution:
(a)p
48 =p
16 ·3 =p
16 ·p
3 = 4p
3
(b)p
8 =p
4 ·2 =p
4 ·p
2 = 2p
2
(c) 3
√
16y 3 = 3
√
8 ·2 · y 3 = 3
√
8y 3 · 3p
2 = 2y3p
2
(d)√
25x4 y 10 =p
25√
x4
√
y 10 = 5x2 y 5
(e)√
y 2(1− y ) = |y |√
1− y
(f)p
(−4)(−4) =p
16 = 4 6=p−4 ·
p−4, because the
even roots are defined for positive radicandsonly. ä
3. n
√a
b=
np
anp
b. This property is similar to the product property!
2.28 Example Simplify the radical expressions using the division rule of radicals:
(a)
√25
36
(b)
√
4x8
9y 4
(c)3
√
16x4
27
(d) 3
√−27
x3
Solution:
(a)
√25
36=
p25
p36
=5
6
(b)
√
4x8
9y 4=
√
4x8
√
9y 4
=2x8
3y 2
(c)3
√
16x4
27=
3√
8 ·2x3 ·x3p
27=
2x3p
2x
3
(d) 3
√−27
x3=−
3
xä
We mention the following important rule for completeness,
Radical of a Radical Rule:
n√
mp
a = mnp
a,
multiply the indices, provided the roots are real numbers. For example,
1.√p
x = 4p
x
2. 3√
3p
a = 9p
a
3.3√p
7 = 6p
7
2.29 Example Simplify
1.√p
5
2.√
3p
5
3.5√
3p
8x
Solution:
1.√p
5 = 2·2p5 = 4
p5
2.√
3p
5 = 2·3p5 = 6
p5
3.5√
3p
8x = 5·3p8x = 15
p8x ä
☞We will assume that all variables are positive in the following sections. This is important to avoid
repeated distinction between odd and even indices and selecting the principal roots.
61
Roots and Radicals
2.2.5 Radical Simplifications
A radical is in a simple form if the radical satisfies the following conditions:
1. No radical appears in the denominator of a fraction. For example5p
xp
4a4is not a simple radical,
but it can be simplified to5p
x
2a2.
2. The radicand does not have any fractions. For example√
3x4
is not a simple radical, but it can
be simplified to
p3x
2
3. The radicand does not have any negative factors. For example 3p−5 is not a simple radical, but
it can be simplified to − 3p
5
4. The power of each factor in the radicand is less than the index of the radical. For example,√
4x4 y,3p
8x4 are not simple fractions, but they can be simplified to 2x2py, 2x 3p
x.
Strategy to Simplify Radicals
How to simplifyp
18x5?
1. Rewrite the coefficient as product of two numbers one of them is a perfect number and rewrite theexponential number as product of two exponentials one of them is raised to a multiple power ofthe radical index, if possible:p
18x5 =p
9 ·2 ·x4 x1
2. Apply the product rule of radicals. That is, factor out all perfect multiples outside the radical signas follows:p
18x5 =p
9 ·2 ·x4 x1 = 3x2p
2x
3. Rationalize denominators: This will be discussed after addition and multiplication sections.
☞ Perfect nth powers
• Perfect square numbers are: 1,4,9,16,25,36, · · ·
• Perfect square exponentials have multiple of two powers: a2, a4, a6, · · ·
• Perfect cube numbers are: 1,8,27,64,125, · · ·
• Perfect cube exponentials have multiple of three powers: a3, a6, a9, · · ·
• Perfect fourth orders numbers are: 1,16,81,256,625, · · ·
• Perfect fourth order exponentials have multiple of four powers: a4, a8, a12, · · ·
2.30 Example Simplify the following radicals using the fundamental rules of radicals
1.√
32x3 y z 2
2. 3√
−16x2 y5
3.p
50x ·p
2x
4.4p
400x6
5.
√8
4x4
6. 3√
5000x4 y2 · 3√
x2 y2
7.3
√
32x3
27
62
Chapter 2
Solution:
1.√
32x3 y z 2 =√
16 ·2x2x y z 2 = 4xz√
2x y
2. 3√
−16x2 y5 = 3√
−8 ·2x2 y3 · y2 =−2y 3√
2x2 y2
3.p
50x ·p
2x =p
50x ·2x =p
100x2 = 10x
4.4p
400x6 = 4p
16 ·25 ·x4 x2 = 2x4p
25x2 = 2xp
5x
5.
p8
p2x4
=√
8
2x4=
√4
x4=
2
x2
6. 3√
5000x4 y2 · 3√
x2 y2 = 3√
5000x4 y2x2 y2 = 3√
1000 ·5 ·x6 y3 y1 = 10x2 y 3√
5y
7.3
√
32x3
27=
3p
8 ·4x3
3p
27=
2x3p
4
3 ä
2.31 Example Simplify the following radicals using the fundamental rules of radicals
1.p
12
2. 3p
54
3.3p
24ab4c 3
4. 3p
3x3p
9x2
5.
p8x3
p2x
6.5p
a9b5
7.
√15
49x2
8.
√
45x y2
p5x
9. 3p
100x3p
10x
10.p
32 +42
Solution:
1.p
12 =p
4 ·3 = 2p
3
2. 3p
54= 3p
27 ·2 = 33p
2
3.3p
24ab4c 3 = 3p
8 ·3ab3b1c 3 = 2bc3p
3ab
4. 3p
3x3p
9x2 = 3p
3x ·9x2 = 3p
27x3 = 3x
5.
p8x3
p2x
=
√
8x3
2x=p
4x2 = 2x
6.5p
a9b5 = 5p
a5a4b5 = ab5p
a4
7.
√15
49x2=
p15
p49x2
=p
15
7x
8.
√
45x y2
p5x
=
√
45x y2
5x=
√
9y2 = 3y
9. 3p
100x3p
10x = 3p
1000x2 = 103p
x2
10.p
32 +42 =p
25 = 5 ä
2.32 Example Simplify the radicals
63
Roots and Radicals
1.p
4x3 +12x4
2.√
40x5 y4 +80x6 y6
3.3
√
40x5 y4 +80x6 y6
27
4. 4
√
40x8 y7 +80x9 y9
40+80x y2
Solution:
1.p
4x3 +12x4 = sqr t4x2(x +3x2) = 2x√
x +3x2)
2.√
40x5 y4 +80x6 y6 =√
40x5 y4(1+2x2 y2) =√
4 ·10 ·x4 x1 y4(1+2x y2) = 2x2 y2√
10x(1+2x y2)
3.3
√
40x5 y4 +80x6 y6
27= 3
√
40x5 y4(1+2x2 y2)
27= 3
√
8 ·5 ·x3 x2 y3 y1(1+2x2 y2)
27=
2x y
3
3√
5x2 y(1+2x y2)
4. 4
√
40x8 y7 +80x9 y9
40+80x y2= 4
√
40x8 y7(1+2x2 y2)
40(1+2x y2)= 4
√
x8 y7 = 4√
x8 y4 y3 = x2 y 4√
y3
ä
64
Chapter 2
2.2.6 Homework: Roots and Radicals
Evaluate the expressions
2.2.1 −p
81
2.2.2 − 4p
81
2.2.34√
814
2.2.43
5
p50
2.2.5
√22
16
2.2.6
√3
4
2.2.7
p15
p5
2.2.814
p21
p7
2.2.9−6
p20
4p
45
2.2.10 10p
32
Simplify the radical expressions. Assume allvariables are positive.
2.2.11√
9x4 y 6
2.2.123√
8a6b3
2.2.13
√
25x8
16y 4z0
2.2.14p
75
2.2.15√
108x4
2.2.16√
28a8b12
2.2.17 3
√
−16x3 y 6
2.2.18√
x3 y 5z
2.2.19 3
√
x3 y 5z8
2.2.20 4
√
48x2 y 4z6
2.2.21
√
5x5
9y 4
2.2.22p
5xp
20x
2.2.23p
6ap
24a
2.2.24p
3ab√
15ab3
2.2.253√
4x2 3p
2x
2.2.263√
100x2 3√
20x2
2.2.27 3
√
18x2 y 3 3
√
12x y 4
2.2.28√
28x y 2
65
Addition and Subtraction of Radicals
2.3 Addition and Subtraction of RadicalsThe addition and subtraction of radicals are possible for similar radicals.
9 Definition Radical expressions are called like or similar if they have the same index and the sameradicand. For example
1. The radicals 2p
6 and 5p
6 are similar, but not 2p
6 and 53p
6 .
2. The radicalsp
3x and 7p
3x are similar, but notp
3x and 7p
2x.
3. The radicals 3x√
2x y and 8x√
2y x are similar, but not 3p
x y and 8x√
2x y.
Like radicals can be added and subtracted by combining their coefficients using the distribution law
ba +c a = (b +c)a
2.33 Example Combine the following
1. 3x +2y −5x2 + y2 +7x −2y +6x2
2. 2p
5+7p
5
3. 9p
2a +p
2a −5p
2a +10p
a
4. 3 3p
x +2p
x −6 3p
x −3p
x
Solution:
1. 3x +2y −5x2 + y2 +7x −2y +6x2 = 10x + x2 + y2
2. 2p
5+7p
5 = 9p
5
3. 9p
2a +p
2a −5p
2a +10p
a = 5p
2a +10p
a
4. 3 3p
x +2p
x −6 3p
x −3p
x =−3 3p
x −p
x ä
2.34 Example True or False?
1.p
2 ·p
8 = 4
2.p
4+p
16 =p
4+16
3.p
xp
x = x
4.p
xp
y = x y
5. 3p
2x +5p
2x = 8p
2x
6.p
x +py =p
x + y
7. 3x2p
x + yp
x = (3x2 + y)p
x
Solution:
1. True:p
2 ·p
8 =p
2 ·8 =p
16= 4
2. False:p
4+p
16 6=p
4+16, 6 6=p
20
3. True:p
xp
x = x
4. False:p
xp
y =px y
5. True: 3p
2x +5p
2x = (3+5)p
2x = 8p
2x
6. False:p
x +py 6= p
x + y
7. True: 3x2p
x + yp
x = (3x2 + y)p
x ä
66
Chapter 2
2.35 Example Simplify and find the sum of the radicals
1.p
27−p
12
2. 2p
3−5p
48+8p
75
3. 83p
2−53p
2−33p
2
4. 4xp
y +√
x2 y −2xp
y
5.3p
16x4 + 3p
54x4 − 3p−128x4
6. xp
32x +p
50x3
7. x3p
3x − 3p
3x4
8. 5p
a +6 5
√a
32
9.p
4x +8+p
16x +32
10. 5y2√
3x5 y6 −4x√
12x3 y10
Solution:
1.p
27−p
12=p
9 ·3−p
4 ·3 = 3p
3−2p
3 =p
3
2. 2p
3−5p
48+8p
75 = 2p
3−5p
16 ·3+8p
25 ·3 = 2p
3−20p
3+40p
3 = 22p
3
3. 83p
2−53p
2−33p
2 = 0
4. 4xp
y +√
x2 y −2xp
y = 4xp
y + xp
y −2xp
y = 3xp
y
5.3p
16x4+ 3p
54x4− 3p−128x4 = 3
p8 ·2x3x1+ 3
p27 ·2x3 x1− 3
p−64 ·2x3x1 = 2x
3p
2x+3x3p
2x+4x3p
2x9x3p
2x
6. xp
32x +p
50x3 = xp
16 ·2x +p
25 ·2x2x1 = 4xp
2x +5xp
2x = 9xp
2x
7. x3p
3x − 3p
3x4 = x3p
3x − 3p
3x3x = x3p
3x − x3p
3x = 0
8. 5p
a +6 5
√a
32= 5p
a +6
2
5p
a = 4 5p
a
9.p
4x +8+p
16x +32=p
4(x +2)+p
16(x +2) = 2p
x +2+4p
x +2 = 6p
x +2
10. 5y2√
3x5 y6−4x√
12x3 y10 = 5y2√
3x4x y6−4x√
4 ·3x2 x y10 = 5y2x2 y3p
3x−4x·2x y5p
3x = 5x2 y5p
3x−4x2 y5
p3x = x2 y5
p3x ä
67
Addition and Subtraction of Radicals
2.3.1 Homework: Addition and Subtraction of Radicals
Simplify the Radical Expressions. Assume allvariables are positive.
2.3.1 5p
27−2p
48
2.3.2 −5p
24+3p
54−p
6
2.3.3 −53p
16+23p
2− 3p
54
2.3.4 −103p
81+203p
3−43p
24
2.3.5√
54a5b8
2.3.6√
80x y 7
2.3.7
√5
4−
√5
16
2.3.83√
32x2
2.3.9 5
√
9ab2 −2
√
16a2b −√
ab2 −√
25a2b
2.3.10√
24x3 −2xp
54x
2.3.11 3
√
18x5 y 3−2x y
√
12x3 y +10x2√
3x y 3
2.3.12 3
√
16x5 y 3 −2x y3
√
54x3 y +5y3√
2x5
2.3.13 53√
8x3 −2x3√
16x3
2.3.14 3
√
40x3 −2x3p
40x +2
√
10x3 +4x3p
5x
2.3.15√
8x3 +12x4
2.3.16 5
√
40x5 −2
√
90x5 +3
√
10x5 +√
20x5
2.3.17 63√
24x3 −2x3p
75−43√
81x3
2.3.18 4p
20a −5p
45a +p
5a
2.3.19 5 3p
a −3x3p−27a
2.3.20p
6a +3p
24a −4p
54a
68
Chapter 2
2.4 Multiplication of RadicalsMultiplication of radicals is based on the law
np
a · np
b = np
ab
That is, radicals with same index can be multiplied into a single radical. For example,p
5 ·p
3 =p
5 ·3 =p
15
In general, to perform the multiplication of radicals expressions:
1. Multiply their coefficients
2. Multiply the radicands under the radical sign with same index
3. Simplify the result
2.36 Example Multiply the following radicals:
1. (−5p
3)(
2p
6)
2.p
20 ·p
10
3. 3p
2 ·4p
6
4. 3p
2 ·23p
4
5. (2p
5)2
6. 3p
2(
4p
2−2)
7. 3p
2(
4p
8−3p
3)
8.(
3p
2+p
5)(p
2−p
5)
Solution:
1. (−5p
3)(
2p
6)
= (−5 ·2)(p
3 ·6) =−10p
2 ·9 =−30p
2
2.p
20 ·p
10 =p
200 = 10p
2
3. 3p
2 ·4p
6 = 12p
2 ·6 = 12p
12 = 12p
3 ·4 = 24p
3
4. 3p
2 ·23p
4 = 23p
8 = 4
5. (2p
5)2 = 4p
5 ·5 = 20
6. 3p
2(
4p
2−2)
= 12p
2 ·2−6p
2 = 24−6p
2
7. Distribute: 3p
2(
4p
8−3p
3)
= 12p
16−9p
6= 48−9p
6
8. Distribute:(
3p
2+p
5)(p
2−p
5)
= 3p
4−3p
10+p
10−p
25 = 6−2p
10−5 = 1−2p
10 ä
2.37 Example Multiply the radicals and simplify
1.p
5a ·p
5a
2. (2p
5)2
3. (3p
8)(2p
2)
4. (p
7−p
2)(p
7−3p
2)
5. (p
x +5)(p
x −3)
6. (2−p
x)2
7. (p
3+2)(p
3−2)
69
Multiplication of Radicals
8. (p
a −p
b)(p
a +p
b)
9. (p
5−p
2)2
10. (3+p
9− x2)2
Solution:
1.p
5a ·p
5a = 5a
2. (2p
5)2 = 4 ·5 = 20
3. (3p
8)(2p
2) = 6p
16 = 24
4. (p
7−p
2)(p
7−3p
2) =p
7(p
7−3p
2)−p
2(p
7−3p
2) = 7−3p
14−p
14+3 ·2 = 13−4p
14
5. (p
x +5)(p
x −3) = x −3p
x +5p
x −15 = x +2p
x −15
6. (2−p
x)2 = 4−4p
x + x
7. (p
3+2)(p
3−2) = 3−4 =−1
8. (p
a −p
b)(p
a +p
b) = a −b
9. (p
5−p
2)2 = (p
5)2 −2p
5p
2+ (p
2)2 = 5−2p
10+2 = 7−2p
10
10. (3+p
9− x2)2 = 9+6p
9− x2 +9− x2 = 18− x2 +6p
9− x2 ä
70
Chapter 2
2.4.1 Homework: Multiplication of Radicals
Multiply and simplify the radical expressions.
2.4.1 (5p
6)(−2p
3)
2.4.2 (4p
5)(3p
5)
2.4.3p
3(p
2+p
7)
2.4.4p
3(4p
6−p
12)
2.4.5 5p
2(2p
12−4p
27)
2.4.6 (2p
6−5p
5)(2p
6+p
5)
2.4.7 (p
6−p
3)(p
6+p
3)
2.4.8 (p
3x −√
7y)(p
3x +√
7y )
2.4.9 (53p
3)(23p
9+ 3p
18)
2.4.10p
5a(p
5a −p
5b)
2.4.11p
2ab(p
5bc )(−p
10ac )
2.4.12 (1+p
6x )(1−p
6x )
2.4.13 (p
x −5)2
2.4.14 (p
3x +2)(p
3x −3)
2.4.15 (p
x +2y )(p
x −2y )
2.4.16 (p
3a +p
3b)(p
3a −p
3b)
2.4.17 (p
x −1)(p
x +1)
2.4.18p
a(p
2a −p
a)
2.4.19 (p
3x +3)2
2.4.20 (p
5x −√
5y )2
2.4.21 (1+p
7)2
2.4.22 (p
4−x −p
x)2
71
Division of Radicals and Rationalization
2.5 Division of Radicals and RationalizationDivision of radicals and simplifying radical denominators are based on the following facts:
1. The law of radical division isnp
anp
b= n
√a
b
For example,p
27a7b5
p9a5b9
=
√
27a7b5
9a5b9=
√
3a2
b4=
ap
3
b2
2. The complement of the radical root np
am where m < n is the radical np
an−m. The multiplication ofthe radical root and its complement is given by
np
an−m · np
am = a
• The complement ofp
a isp
a. Their product isp
a ·p
a = a
• The complement of 3p
a is3p
a2. Their product is 3p
a · 3p
a2 = a
• The complement of7p
a2 is7p
a5. Their product is7p
a2 · 7p
a5 = a
3. The binomials a −b and a +b are called conjugate binomials. They have the interesting property
(a −b)(a +b) = a2 −b2
For example (2x −3)(2x +3) = 4x2 −9.The product of conjugate binomials with square root radicals is an expression without any radical.For example,
(a) (2−p
3)(2+p
3) = 22 − (p
3)2 = 4−3 = 1
(b) (p
5−1)(p
5+1) = 5−1 = 4
(c) (p
x +1)(p
x −1) = x −1
(d) (p
14+p
6)(p
14−p
6) = 14−6 = 8
(e) (3p
2+p
5)(3p
2−p
5) = 18−5 = 13
In the the following subsection, we will simplify fractions containing radicals in the denominators.
2.5.1 Rationalizing Denominators
1. Denominators consisting of radical monomials such as
2p
5,
p3
p2x
,1
3p
25
are simplified by multiplying and dividing the fraction by the complement of the radical. Forexample, we simplify the above radical fractions as follows
(a)2p
5=
2p
5·p
5p
5=
2p
5p
25=
2p
5
5
(b)
p3
p2x
=p
3p
2x·p
2xp
2x=
p6x
2x
(c)1
3p
25=
1
3p
25·
3p
5
3p
5=
3p
5
3p
125=
3p
5
5
2.38 Example Rationalize the following radical expressions and simplify
(a)2p
2
72
Chapter 2
(b)1p
x
(c)1
px −5
(d)
√a
b
Solution:
(a)2p
2·p
2p
2=
2p
2
2=p
2
(b)1p
x·p
xp
x=
px
x
(c)1
px −5
·p
x −5p
x −5=
px −5
x −5
(d)
√a
b=
pa
pb·p
bp
b=
pab
bä
2. Denominators consisting of binomial with square root radicals such as
3
2−p
2,
2p
5−1,
1p
5x +5,
p3x
p3x −
p2x
are simplified by multiplying and dividing the fraction by the conjugate of the denominator. Forexample, we simplify the above radical fractions as follows
(a)3
2−p
2=
3
2−p
2·
2+p
2
2+p
2=
6+3p
2
(2)2 − (p
2)2=
6+3p
2
4−2=
6+3p
2
2
(b)2
p5−1
=2
p5−1
·p
5+1p
5+1=
2(p
5+1)
5−1=
p5+1
2
(c)1
p5x +5
=1
p5x +5
·p
5x −5p
5x −5=
p5x −5
5x −25
(d)
p3x
p3x −
p2x
=p
3xp
3x −p
2x·p
3x +p
2xp
3x +p
2x=
3x +p
6x2
3x −2x=
3x + xp
6
x= 3+
p6
2.39 Example Rationalize the following radical expressions and simplify
1.1
p2−1
2.1
p5+
p3
3.1− x
1+p
x
Solution:
1.1
p2−1
=1
p2−1
·p
2+1p
2+1=
p2+1
2−1=p
2+1
2.1
p5+
p3=
1p
5+p
3
p5−
p3
p5−
p3=
p5−
p3
5−3=
p5−
p3
2
3.1− x
1+p
x=
1− x
1+p
x
1−p
x
1−p
x=
(1− x)(1−p
x)
1− x= 1−
px
ä
2.40 Example Rationalize and simplify the radicals
73
Division of Radicals and Rationalization
1.
p70
p3
2. 5p
3−6p
3
3.3p
3
4.p
x y +x yp
x y
5.
p3
p4x
6.
√11
20a5
7.3
3p
9x
8.p
a −1+1
pa −1
Solution:
1.
p70
p3
=p
70p
3·p
3p
3=
p210
3
2. 5p
3−6p
3= 5
p3−
6p
3·p
3p
3= 5
p3−
6p
3
3= 5
p3−2
p3 = 3
p3
3.3p
3=
3p
3
p3
p3=
3p
3
3=p
3
4.p
x y +x yp
x y=p
x y +x yp
x y
px y
px y
=px y +
x yp
x y
x y= 2
px y
5.
p3
p4x
=p
3
2p
x·p
xp
x=
p3x
2x
6.
√11
20a5=
p11
2a2p
5a·p
5ap
5a=
p55a
10a3
7.3
3p
9x=
3
3p
9x·
3p
3x2
3p
3x2=
33p
3x2
3p
27x3=
3p
3x2
x
8.p
a −1+1
pa −1
=p
a −1
1·p
a −1p
a −1+
1p
a −1=
a −1+1p
a −1=
ap
a −1·p
a −1p
a −1=
ap
a −1
a −1 ä
2.41 Example Rationalize and simplify the radicals
1.1
p2+1
2.
px +
p2
px −
p2
3.12
p6−2
−6p
6
4.3+
p6
3−p
6
5.
pm −
pn
pm +
pn
74
Chapter 2
Solution:
1.1
p2+1
=1
p2+1
·p
2−1p
2−1=
p2−1
2−1=p
2−1
2.
px +
p2
px −
p2=
px +
p2
px −
p2·p
x +p
2p
x +p
2=
x +2p
2x +2
x −2
3.12
p6−2
−6p
6=
12p
6−2·p
6+2p
6+2−
6p
6·p
6p
6=
12(p
6+2)
6−4−
6p
6
6= 6(
p6+2)−
p6 = 5
p6+12
4.3+
p6
3−p
6=
3+p
6
3−p
6·
3+p
6
3+p
6=
9+6p
6+6
9−6=
3(5+2p
6)
3= 5+2
p6
5.
pm −
pn
pm +
pn
=p
m −p
np
m +p
n·p
m −p
np
m −p
n=
m −2p
mn +n
m −n ä
2.42 Example Simplify (p
x −2)2 − (p
x −2)2
Solution: (p
x −2)2 − (p
x −2)2 = x −2p
x +4− (x −2) = 6−2p
x ä
75
Division of Radicals and Rationalization
2.5.2 Homework: Division of Radicals and Rationalization
Rationalize the denominator and simplify
2.5.11p
2
2.5.23p
3
2.5.32xp
6x
2.5.4x + y
2p
x + y
2.5.5a +4
3p
a +4−p
a +4
2.5.61
1+p
2
2.5.71
p3−2
2.5.81
p5−2
2.5.91
p3+
p2
2.5.102−
p3
2+p
3
2.5.112(p
15+p
3)p
5−p
3
2.5.12x − y
px +p
y
2.5.132a −1p
2a −1
2.5.145
p2x −5
2.5.151
3p
4
2.5.161
3p
25t
2.5.173p
25
3√
100x2
2.5.18x −4p
x −4
2.5.19
p10
p10+3
2.5.201
px −2
−1
px +2
76
Chapter 2
2.6 Equations involving radicals10 Definition Equations involving at least one radical expression are referred as equations with radicals,such as p
15−2x = x,p
4x − x +3 = 0,p
2x −1−p
3x −2 = 4,3√
x2 +2 = x +2
Finding the solutions of equations with radicals is based on the power rule of equality
If a = b, then an = bn
This means that if two expressions are equal, then their squares, cubes, and nth power are equal.However, the converse is not true. For example, squaring the trivial equation x = 5 is the equationx2 = 25 which can be solved by factoring x2−25 = (x−5)(x+5) = 0 which has two solutions x = 5 and x =−5.The first solution x = 5 satisfies the original equation, but the second solution x = −5 does not satisfythe original equation and it is called extraneous solution.
2.43 Example Find the power of the radicals
1. (p
5)2=
2. (p
3x)2=
3. (p
x +7)2=
4. (3p
x +7)3=
In this section, we will restrict ourselves to equations with square radicals.
2.6.1 Method of solving equations with radicals
1. Isolate one of the radicals on one side of the equation.
2. Square both sides.
3. Repeat first step if needed.
4. Solve the resulting equation
5. Check the solution in the original equation
2.44 Example Solve the equationp
4x −3 = 5
Solution: Square both sides of the equation
(p
4x −3)2 = 5
2
4x −3 = 25
4x = 28
x = 7
Check√
4(7)−3=p
28−3X= 5
ä
2.45 Example Solve the equationp
2x +3 = 1
Solution: Square both sides of the equation
(p
2x +3)2 = 1
2
2x +3 = 1
x =−1
Check√
2(−1)+3=p
1X= 1 ä
77
Equations involving radicals
2.46 Example Solve the equationp
2x +3 = 7
Solution: Subtract 3 and square both sides of the equation
(p
2x)2 = 4
2
2x = 16
x = 8
Check√
2(8)+3 =p
16+3X= 7 ä
2.47 Example Solve the equation x =p
15−2x
Solution:
(x)2 = (
p15−2x)
2
x2 = 15−2x
x2 +2x −15 = 0
(x +5)(x −3) = 0
x = 3, x =−5
Check x = 3 : 3 =p
15−6X= 3
Check x =−5 : −5 =p
15+10= 5 Extraneous solution.
ä
2.48 Example Solve the equationp
4a +5−p
2a +13 = 0
Solution: Isolate one of the radicals and square both sides
p4a +5−
p2a +13 = 0p
4a +5 =p
2a +13
4a +5 = 2a +13
2a =+8
a = 4 ä
2.49 Example Solve the equationp
2x +1−p
x = 1
Solution: Isolate one of the radicals and square both sides
p2x +1 = 1+
px
(p
2x +1)2 = (1+
px)
2
2x +1 = 1+ x +2p
x
x = 2p
x
x2 = 4x
x2 −4x = 0
x(x −4) = 0
x = 0, x = 4 Both are true solutions
ä
2.50 Example Solve the equationp
5x2 −1 = 2x
78
Chapter 2
Solution:
(
√
5x2 −1)2 = (2x)
2
5x2 −1 = 4x2
x2 −1 = 0
(x +1)(x −1) = 0
x = 1, x =−1
Check x = 1 :p
5−1X= 2
Check x =−1 :p
5−1 = 2 6= −2 Extraneous solution.
ä
2.51 Example Solve the equation 2x −1 =p
x +7
Solution:
(2x −1)2 = (
px +7)
2
4x2 −4x +1 = x +7
4x2 −5x −6 = 0
(4x +3)(x −2) = 0
x = 2, x =−3/4
Check x = 2 : 4−1 =p
2+7X= 3
Check x =−3/4 : −2 ·3
4−1 =
√
−3
4+7 Left side is negative. Extraneous solution.
ä
2.52 Example Solve the following equations
1.p
2a −3−4 = 1
2.p
5x −10 =p
x +2
3.p
4x − x +3 = 0
4.p
3x +1 = 3+p
x
5.p
x +p
x +3 = 3
Solution:
1.p
2a −3−4 = 1 Isolate the radical and square both sides
(p
2a −3)2 = (5)
2
2a −3 = 25
2a = 28
a = 14
Check a = 14 :p
28−3−4X= 1
2.p
5x −10 =p
x +2
(p
2a −3)2 = (5)
2
5x −10 = x +2
4x = 12
x = 3
Check x = 3 :p
15−10X=p
3+2
79
Equations involving radicals
3.p
4x − x +3 = 0
(p
4x)2 = (x −3)
2
4x = x2 −6x +9
x2 −10x +9 = 0
(x −1)(x −9) = 0
x = 1, x = 9
Check x = 1 :p
4−1+3X= 0
Check x = 9 :p
36−9+3X= 0
4.p
3x +1 = 3+p
x
(p
3x +1)2 = (3+
px)
2
3x +1 = 9+6p
x + x
2x −8 = 6p
x
(x −4)2 = (3
px)
2
x2 −8x +16 = 9x
x2 −17x +16 = 0
(x −1)(x −16) = 0
x = 1, x = 16
Check x = 1 :p
3+1 6= 3+p
1 Extraneous solution
Check x = 16 :p
48+1 = 3+p
16X= 7
5.p
x +p
x +3 = 3
(p
x +3)2 = (3−
px)
2
x +3 = 9−6p
x + x
6p
x = 6p
x = 1
x = 1
Check x = 1 :p
1+p
1+3X= 3 ä
80
Chapter 2
2.6.2 Homework: Equations in Radicals
Solve the following equations in radicals and checkyour solution
2.6.1p
x +7 =p
4x +1
2.6.2p
3x +6 =p
4x +1
2.6.3p
x +9−p
2x +1 = 0
2.6.4p
3x +1+p
2x −1 = 0
2.6.5√
3y +4 = 4
2.6.6 3p
x +2 = 18
2.6.7 4p
n −6 = 0
2.6.8p
2n −5−6 =−5
2.6.9p
5x +7 =p
4x +6
2.6.10p
4x −2−p
2x −4 = 0
2.6.11√
x2 +5x +10 = 2
2.6.12√
x2 +4x +16 =p
2x +15
2.6.13 4p
n = n −5
2.6.14√
3y +4 = 4
2.6.15 x +6 =p
x +6
2.6.16p
10−x −x = 2
2.6.17p
2a −7 =p
3a +1−2
2.6.18 3p
3x −9 = 3p
x −27
2.6.19p
x +3−p
2x −1 =−1
2.6.20p
4x +9+p
x +1 =p
5x +10
2.6.21 2 =p
x −4−p
x
81
Rational Exponents and Roots
2.7 Rational Exponents and RootsIt is known that simplifying radical expressions require many rules and tricks. Thus, can we find anexponential representation of radicals since the rules of exponential expressions are familiar and theyare easier to remember?.
To answer this question, say forp
6, let us assume that the square rootp
6 can be expressed expo-nentially say 6x that is
p6 = 6x . Now, we need to find is the actual value of x that satisfies this equation.
Square both sides (p
6)2 = (6x )2, and simplify to get 6 = 62x , the powers of 6 must be equal that is, 2x = 1
and hence x = 12.
Inspired by the power rule of the nth root
(np
a)n = a
and its corresponding exponential representation
(ax)
n = ax ·n = a1
The last equality is possible, if x =1
n, then we have
(a1n )
n = ann = a
Thus we have the exponential representations of the nth roots of 3:
312 =
p3
313 = 3
p3
314 = 4
p3
315 = 5
p3
31n = n
p3
In general, we can define that any radical of order n for any positive real number a, can be expressedin exponential form
a1n = n
pa,
amn = (
np
a)m = n
pam
2.53 Example Convert each exponential number to a radical and simplify
1. 161/2
2. 641/2
3. 0.011/2
4. 641/3
5. −(9
16)1/2
6. 322/5
7. 93/2
8. (25x2)1/2
9. (27x3)1/3
10. (−8x3 y 6)1/3
Solution:
1. 161/2 =p
16 = 4
2. 641/2 =p
64 = 8
3. 0.011/2 =p
0.01 = 0.1
4. 641/3 = 3p
64 = 4
5. −(9
16)1/2 =−
√9
16=−
3
4
82
Chapter 2
6. 322/5 = (5p
32)2 = 22 = 4
7. 93/2 = (p
9)3 = 33 = 27
8. (25x2)1/2 =√
25x2 = 5x
9. (27x3)1/3 = 3√
27x3 = 3x
10. (−8x3 y 6)1/3 = 3
√
−8x3 y 6 =−2x y 2ä
2.54 Example Change the radical notation of the expressions to exponential notation
1. 3√
7x y
2.6p
4a2b
Solution:
1. 3√
7x y = (7x y)1/3
2.6p
4a2b = (4a2b)1/6ä
Rational Exponents
For a real number a1/n, the fractional power
amn = (a
1n )
m
indicates that the denominator is the index( order) of the root and numerator is its power, for example,
43/2 = (4
1/2)
3 = (p
4)3 = 2
3 = 8
Negative rational exponents
The negative fractional exponents are defined as follows:
a−m/n =1
am/n
For example
64−23 =
1
6423
=1
(3p
64)2=
1
42=
1
16
2.55 Example Evaluate the following
1. 3245
2. 932
3. 25−32
4. (27)2/3
5. (1
8x3)4/3
6. 25−1/2
7. 16−3/2
8. (−27a3)−2/3
Solution:
1. 3245 = (32
15 )4 = 24 = 16
2. 932 = (9
12 )3 = 33 = 27
3. 25−32 =
1
2532
=1
(2512 )3
=1
53=
1
125
4. (27)2/3 = (271/3)2 = 32 = 9
5. (1
8x3)4/3 = ((
1
8x3)1/3)4 = (
1
2x)4 =
1
16x4
6. 25−1/2 =1
251/2=
1
5
7. 16−3/2 =1
163/2=
1
43=
1
64
8. (−27a3)−2/3 =1
(−27a3)2/3=
1
(−3a)2=
1
9a2 ä
2.56 Example Simplify(
(16)1/2)1/2
Solution: Use the power to power rule:(
(16)1/2)1/2 =
(
(24)1/2)1/2 = 24· 1
2· 1
2 = 2 ä
83
Rational Exponents and Roots
2.57 Example Use rational exponents to simplify
1. 6√
(25x)3
2. 4√
16a4y2
Solution:
1. 6√
(25x)3 = ((25x)16 )3 = (25x)3/6 = (25x)1/2 =
p25x = 5
px
2. 4√
16a4 y2 = (16a4 y2)1/4 = 161/4(a4)1/4(y2)1/4 = 2a y1/2 = 2ap
y ä
2.58 Example Use rational exponents to express as a single radical
1. 3p
4 · 4p
4
2.√p
2
3.3√
4p
5
4.
pa
3p
a
Solution:
1. 3p
4 · 4p
4 = 413 ·4
14 = 4
4+312 = 4
712 = 12
p47
2.√p
2 = (21/2)1/2 = 21/4 = 4p
2
3.3√
4p
5 = (51/4)1/3 = 51/15 = 15p
5
4.
pa
3p
a=
a12
a13
= a12− 1
3 = a16 = 6
pa
ä
2.59 Example Simplify the following fractional expressions
1. a12 ·a
25
2. (7a12 b
32 a2b−3)0
3. (x12 + y
12 )(x
12 − y
12 )
Solution:
1. a12 ·a
25 = a
12+ 2
5 = a5+410 = a
910
2. (7a12 b
32 a2b−3)0 = 1
3. (x12 + y
12 )(x
12 − y
12 ) = x − y ä
84
Chapter 2
2.7.1 Homework: Rational Exponents and Radicals
Evaluate the following expressions
2.7.1 (−27)
1
3
2.7.2 (49)
−1
2
2.7.3 (64)
2
3
2.7.4 −(25)
−3
2
2.7.5 −9
3
2
2.7.6 (256)
3
4
2.7.7 (−1000)
1
3
2.7.8 (−8
27)
−4
3
2.7.91
32−1/5
Express in radical form
2.7.10 (3x)
1
4
2.7.11 (4a +5b)
2
3
2.7.12 (x2 + y 2)1/2
Express in exponential form
2.7.13 −43
√
x2 + y 2
2.7.14√
a2 −b2
2.7.15 6
√
6x2 y 3
Simplify the expressions without negativeexponents. Assume all variables are positive.
2.7.16 (−27x6)1/3
2.7.17 (36x4 y 6)1/2
2.7.18 [(x +64)4]1/4
2.7.19 6−2/36−1/3
2.7.2044/341/3
42/3
2.7.21 (31/2 −51/2)(31/2 +51/2)
2.7.22 (2x13 )(x
12 )
85
Scientific Notation
2.8 Scientific NotationScientific notation are used by scientists to express very large numbers and very small numbers in acompact form. For example, the distance from earth to sun is about
≈ 93,000,000 miles.
The mass of the proton is about
≈ 0.000000000000000000000000000001672 kg (26 zeros).
The national public debt as of February 28, 2007 was about
8,776,000,000,000.0
and as of January 10, 2011 was about 14,021,000,000,000.0 dollars. Scientific notationEvery decimal number q can be written in scientific notation as follows
q = a ×10n
, where 1 ≤ a < 10 and n is an integer
How to Write a number q in scientific notation?
1. Move the decimal point to a place so it will be of the form 1 ≤ a < 10, that is the point should beafter the fist significant digit from left.
2. n is the number of decimal places that the point was moved.
3. If the decimal point was moved to the left (q > 1) then n is positive and q = a ×10n, for exampleq = 309= 3.09×102
4. If the decimal point was moved to the right (q < 1) then n is negative and q = a ×10−n, for exampleq = 0.00309= 3.09×10−3
Powers of 10:
1. 10−3 Thousandth, 10−2 Hundredth, 10−1 Tenth
2. 103 Thousand, 106 Million, 109 Billion, 1012 Trillion
2.60 Example Express the distance from earth to sun in scientific notation
Solution: 93,000,000 = 9.3×107 ä
2.61 Example Express the mass of the proton in scientific notation
Solution: 0.000000000000000000000000000001672= 1.672×10−27 ä
2.62 Example Write in scientific notation
1. 1991
2. 200900.0003
3. 0.010099100
4. 47.6×103
5. 0.063×10−2
Solution:
86
Chapter 2
1. 1991= 1.991×103
2. 200900.0003= 2.009000003×105
3. 0.010099100= 1.00991×10−2
4. 47.6×103 = 4.76×102 ×103 = 4.76×105
5. 0.063×10−2 = 6.3×10−2 ×10−2 = 6.3×10−4ä
2.63 Example Convert to standard form:
1. 8.706×105
2. 1.1×10−3
Solution:
(a) 8.706×105 = 870600
(b) 1.1×10−3 = 0.0011 ä
2.64 Example Change to scientific notation and simplify(220,000)(0.000009)
0.00033.
2.65 Example Simplify(1.8×103)(8×10−2)
(3×10−5)(4×102).
2.66 Example A megabyte is 220 bites. (A gegabyte has 230 bites). How many bites in one megabyte andone gegabyte?. Express your answer in scientific notation.
2.8.1 Homework: Scientific Notation
Write in scientific notation
2.8.1 35,000,00
2.8.2 7803.021
2.8.3 0.00035
2.8.4 53,700×1000
Write in standard decimal notation
2.8.5 2.67×103
2.8.6 5.09×10−1
2.8.7 3.00601×10−4
2.8.8 3.77×107
Use scientific notation and simplify
2.8.90.000048
0.0012
2.8.10(60,000,0)(0.0012)
(0.036)(2,000,00)
2.8.11 3p
27000
2.8.12 (125000)
2
3
87
Exercises
2.9 ExercisesFind the roots
1.p
64
2.p
.49
3.√
(−9)2
4.
√1
36
5.√
9u2
6.√
16v 4
7.√
25x4z2
8.√
4x10 y 6
9.
√
100a2
9z6
10.
√9
x2
11.
√
121a4c2z8
100x4
12. 3p−216
13.3√
a12b3
14.6√
56
15. (7p
67)7
16.5√
−32x5
17.4
√
a4b8
16k12
18.3
√
−8a6
27b3
Simplify
19. x0x
1
4 x
3
4
20. (−3a3)2 + (−2a2)3
21. 3p
2(p
6−2p
8)
22. (x
2
3 )6
23. (5a−3b−2c3)−3
24. (4−3x6)
2
3
25.3√
.008x7
26. 3
√
−16
x3 y 6
27.p
5p
10
28.
p2acp
4c
29. (2p
2+p
3)(3p
2−4p
3)
30.
p2+
p5
3p
5−p
2
31.
p7−
p6
2p
6+p
7
Solve the equation
32. 2x −1 =√
2x2 +1
33. 2x −1 =−√
2x2 +1
34.p
2x −2−p
4x +5 = 3
35.p
8 = 3x +1
36. 6− y =py
37. 4x2 +xp
3 = 0
38. Multiply and simplify(
3√
2x2)(
3p−4x
)
39. Simplify(
−3x y 2 y−2x−3)2
40. Simplify 7√
8a2b −√
16a2b −5√
36a2b −10√
2a2b
Advanced ExercisesSimplify
41. (−1)32
+123
42.
√
x2
16+
x2
9, x > 0
88
Chapter 2
43. (1
4)−
1
4
44. (1
27)−
1
9
45. (4−1 −3−1)−1
46. (3−1 −2−1)−1
47.
√
x√
xp
x
48.
√
2
√
2√
2p
2
49. 55 +55 +55 +55 +55
50. 66 +66 +66 +66 +66 +66
51.1530
4515
52.
√
810 +410
84 +411
53. 44 ·94 ·49 ·99
54. Find m > 0: the points (m,2) and (1,m) lie on a line
with slope m
55. Solve
x +√
x2 −1+1
x −√
x2 −1
= 20
2.10 Review Radicals and Exponents1. Basic Definitions:
A square, cube, nth root are denoted by
pa = b,
3p
a = c ,np
a = d
These radical numbers are the same as the rational exponents
a1/2 = b, a1/3 = c , a1/n = d
Also we haveam/n = n
pam = (
np
a)m
84/3 = (
3p
8)4 = 2
4 = 16
If n is even, the radicand must be positive or zero.
2. Properties of Radicals:
np
a · np
b = np
abp
10p
40 =p
400= 20
n
√a
b=
np
anp
b
3
√
t 2
125=
3p
t 2
5
3. Rationalize the denominator: Use the property
pa ·
pa = a
to eliminate radical from the denominator
4p
3=
4p
3·p
3p
3=
4p
3
3
4. Like Radicals: Have the same index and the same radicand
4p
ab and −7p
ab but not 4p
x and −73p
x
89
Review Radicals and Exponents
5. Adding and subtracting radicals: Combine like radicals
2p
10−p
40= 2p
10−p
4 ·p
10 = 0
6. Multiplying radicals: Use distributive property
(p
x −1)(p
x +5) = x +4p
x −5
7. Conjugate and rationalizing the denominator: Conjugate pairs by changing + to − and vice versa:
px +3 and
px −3
p3−
p5 and
p3+
p5
Multiply and divide by the conjugate of denominator
1p
x +3=
1p
x +3·p
x −3p
x −3=
px −3
x −9
8. Equations in radicals: Apply the power rule to transform the equation to an equivalent equationwithout radicals
p2x = 3x
2x = 9x2
2x −9x2 = 0
x(2−9x) = 0
x = 0, x = 2/9
90
3 Quadratic Equations and Inequalities
3.1 Complex NumbersOBJECTIVES
• Imaginary numbers
• Arithmetic of complex numbers
• Powers of imaginary unit
Complex numbers are introduced in connection with solving quadratic equations. For example, theequation x2 = 1 has two solutions x =+1, x =−1, but the equation x2 =−1 has no real solution becausethe square of any real number is positive number. To resolve this dilemma, Mathematicians inventeda new notation just to handle the
p−1 in this type of equation. They called it imaginary numbers.
The introduction of these numbers has long historical development starting with the works of Car-dano (1501-1576) on cubic equations, Bombelli (1526-1572) on negative and complex numbers, andDescartes (1596-1650) who coined the name imaginary. Finally, the works of Euler (1707-1783) andGauss (1777-1855) had established the rules of complex numbers and their interpretations. Currently,complex numbers and analysis are applied to model phenomena in physics, electronics, and computerscience.
1 Definition The square root of negative one is denoted by the letter i (imaginary unit):p−1 = i and
i 2 = (p−1)2 =−1. That is,
i =p−1 and i 2 =−1
Using this definition, we will express the square root of any negative number in terms of this newnotation: p
−a =p−1 ·a =
p−1 ·
pa = i ·
pa =
pa i
For example,
1.p−16=
p−1 ·16 =
p−1 ·
p16= i ·4 = 4i
2.p−18= i
p18= i 3
p2 = 3
p2 i
3.p−7 =
p7 i
4. −p−100=−10i
3.1 Example Expression the following radicals in terms of i :
1.p−81
2. −p−8
3.
√
−98
16
Solution:
1.p−81 =
p−1 ·81 = i 9= 9i
2. −p−8 =−i
p8 =−2
p2 i
3.
√
−98
16=
7p
2
4i ä
91
Complex Numbers
With the introduction of numbers involving the imaginary unit i , we define a new type of numberscalled complex numbers.
2 Definition A complex number is a number of the form a +bi , where a and b are real numbers andi =
p−1 is the imaginary unit. The number a is called the real part, and b is called the imaginary part
of the complex number a +bi .
For example, the real part of the complex number 3+2ip
5 is 2 and the imaginary part is 2p
5. In factany real number x can be considered a complex number x +0i and a complex number in the form bi iscalled pure imaginary, such as 3i .
3.2 Example Identify the real and imaginary part of the complex numbers:
1. 3+4i
2. −6−5i
3. −7
4. ip
2
Solution:
1. Re(3+4i )= 3, Im(3+4i ) = 4
2. Re(−6−5i )=−6, Im(−6−5i ) =−5
3. Re(−7) =−7, Im(−7) = 0
4. Re(ip
2) = 0, Im(ip
2) =p
2 ä
☞ The product rule of square radicals is valid for positive numbers only. For example,
p6 ·
p10 =
p6 ·10 =
p60 =
p4 ·15 = 2
p15
but it is not true that p−3 ·
p−3 6=
√
(−3) · (−3) =p
9 = 3
The previous multiplication is false. The correct multiplication is
p−3 ·
p−3 = i
p3 · i
p3 = i 2 ·
p9 =−3
3.3 Example Multiply or divide the radicals and simplify
1.p−5 ·
p6
2.p−3 ·
p−6
3.−72p−2
4.
p−20
p−5
Solution:
1.p−5 ·
p6 = i
p5 ·
p6 = i
p30
2.p−3 ·
p−6 = i
p3 · i
p6 = i
p18 = 3i
p2
3.
p−72
p−2
=ip
72
ip
2=
√72
2=p
36 = 6
4.
p−20p
5=
ip
20p
5= i
√20
5= i
p4 = 2i
ä
92
Chapter 3
3 Definition Two complex numbers a +bi and c +d i are equal if and only if a = c and b = d . For example(3−2)+ i
p9 is equal to 3i +1
To solve linear equations involving complex numbers: Equate real and imaginary parts:
3.4 Example Find the values x and y so the complex numbers (x −2)+2yi and 8−5i are equal.
Solution: Apply the equality definition: (x −2)+2yi = 8−5i , this leads to x −2 = 8 hence x = 10
and 2y =−5 hence y =−5
2ä
3.5 Example Solve 3y +2−21i = 8+7xi
Solution: Real and imaginary parts are equal, that is: 3y +2 = 8, y = 2 and 7x =−21, x =−3 ä
4 Definition The form a +bi is called in the standard form of a complex number. For example, thefollowing complex numbers are the same number. The last form is the standard form:
1.5+3i
8
2.3i +5
8
3.3i
8+
5
8
4.5
8+
3i
8
5.5
8+
3
8i
3.1.1 Fundamental Operations with Complex Numbers
Addition and Subtraction of Complex Numbers
3.6 Example Simplify the expressions
1. (2+7x)+ (5−4x)
2. (3−5x)− (9−8x)
Solution: Combine like terms
1. (2+7x)+ (5−4x) = 2+7x +5−4x = 7+3x
2. (3−5x)− (9−8x) = 3−5x −9+8x =−6+3x ä
The sum and difference of complex numbers is performed by combining like terms:
3.7 Example Simplify the expressions
1. (2+7i )+ (5−4i )
2. (3−5i )− (9−8i )
Solution: Combine like terms
1. (2+7i )+ (5−4i )= 2+7i +5−4i = 7+3i
2. (3−5i )− (9−8i )= 3−5i −9+8i =−6+3i ä
93
Complex Numbers
5 Definition The addition and subtraction of complex numbers is defined by combining real parts andimaginary parts, as follows:
(a +bi )+ (c +d i )= (a +c)+ (b +d )i
(a +bi )− (c +d i )= (a −c)+ (b −d )i
For any real numbers: a,b,c ,d .
3.8 Example Find the sum or the difference:
1. (5+3i )+ (3+ i )
2. (−4+ i )−2(6−2i )
3. (8+4i )+ (12+8i )
4. (−6+ i )− (3+2i )
Solution:
1. (5+3i )+ (3+ i )= 5+3+3i + i = 8+4i
2. (−4+ i )−2(6−2i )=−4+ i −12+4i =−16+5i
3. (8+4i )+ (12+8i )= 8+12+ (4+8)i = 20+12i
4. (−6+ i )− (3+2i )= (−6−3)+ (1−2)i =−9− i ä
Multiplication of Complex Numbers
The product of two complex numbers involves the application of three rules:
1. Distributive property
2. Combining like terms
3. Substituting i 2 =−1
3.9 Example Evaluate
1. 5(3−2i )
2. (3i )(2+7i )
3. (−4i )(1−7i )
4. (2+3i )(4+5i )
Solution:
1. 5(3−2i )= 5 ·3−5 ·2i = 15−10i
2. (3i )(2+7i )= 6i +21i 2 = 6i +21(−1)=−21+6i
3. (−4i )(1−7i )=−4i +28i 2 =−4i +28(−1)=−28−4i
4. (2+3i )(4+5i )= 2(4+5i )+3i (4+5i )= 8+10i +12i +15i 2 = 8+22i −15 =−7+22i ä
3.10 Example Multiply the complex numbers
1. (2+ i )(5+6i )
2. (2+3i )(3−2i )
3. (3+5i )2
Solution:
94
Chapter 3
1. (2+ i )(5+6i )= 2(5+6i )+ i (5+6i )= 10+12i +5i +6i 2 = 10+17i −6 = 4+17i
2. (2+3i )(3−2i )= 2(3−2i )+3i (3−2i )= 6−4i +9i −6i 2 = 6−5i +6 = 12−5i
3. (3+5i )2 = 32 +2(3)(5i )+ (5i )2 = 9+30i −25 = 16+30i ä
6 Definition The multiplication of two complex numbers is defined as follows
(a +bi )(c +d i ) = (ac −bd )+ (ad +bc)i
For any real numbers: a,b,c ,d .
3.11 Example Multiply (−4+2i )(2+ i )
Solution: 2i (−4+2i )(2+i )= (−8i+4i 2)(2+i )= (−4−8i )(2+i )=−4(2+i )−8i (2+i )=−8−4i−16i−8i 2 =−8−20i +8 =−20i ä
Complex Conjugate
7 Definition The complex numbers a +bi and a −bi are called complex conjugates. For example,thepairs 3+4i , 3−4i and 5−7i , 5+7i are complex conjugates.
The product of complex conjugates is always positive real number:
(a +bi )(a −bi )= a2 − (bi )2
= a2 −b2(−1)
= a2 +b2
For example (2−3i )(2+3i )= (2)2 − (3i )2 = 4−9i 2 = 4−9(−1)= 4+9= 13
3.12 Example Evaluate
1. (3+ i )(3− i )
2. (1− i )(1+ i )
3. (3+5i )(3−5i )
Solution:
1. (3+ i )(3− i )= 3(3− i )+ i (3− i )= 9−3i +3i − i 2 = 9+1 = 10
2. (1− i )(1+ i )= 12 +12 = 2
3. (3+5i )(3−5i )= 32 +52 = 9+25 = 34 ä
Division of Complex Numbers
The division of complex numbers is accomplished by multiplying the numerator and denominator bythe complex conjugate of the denominator, then simplify and express the result in standard form. thatis
a +bi
c +d i=
a +bi
c +d i·
c −d i
c −d i=
(ac +bd )+ (bc −ad )i
c 2 +d 2=
(ac +bd )
c 2 +d 2+
(bc −ad )
c 2 +d 2i
3.13 Example Evaluate and write the answer in standard form
1.3+5i
8
2.7
5i
95
Complex Numbers
3.2−6i
3i
4.3i
1−3i
5.2−3i
5+4i
Solution:
1.3+5i
8=
3
8+
5
8i
2.7
5i=
7
5i·
i
i=
7i
5i 2=−
7
5i
3.2−6i
3i=
2−6i
3i·
i
i=
2i −6i 2
3i 2=
2i +6
−3=−2+
2
3i
4.3i
1−3i=
3i
1−3i·
1+3i
1+3i=
3i +9i 2
1+9=
3i −9
10=−
9
10+
3
10i
5.2−3i
5+4i=
2−3i
5+4i·
5−4i
5−4i=
10−8i −15i +12i 2
52 +42=
10−23i −12
25+16=
−2−23i
41=
−2
41−
23
41i
ä
3.14 Example Evaluate and write the answer in standard form
1.2−3i
1−2i
2.1
3+ i
3.7
2i
4.3− i
2+ i
Solution:
1.2−3i
1−2i=
2−3i
1−2i
1+2i
1+2i=
(2−3i )(1+2i )
12 − (2i )2=
2−4i −3i −6i 2
1−4i 2=
−2−7i +6
1+4=
4−7i
5=
4
5−
7
5i
2.1
3+ i=
1
3+ i·
3− i
3− i=
3− i
9+1=
3
10−
1
10i
3.7
2i=
7
2i·
i
i=
7i
−2=−
7
2i
4.3− i
2+ i=
3− i
2+ i·
2− i
2− i=
6−3i −2i + i 2
4+1=
5−5i
5= 1− i
ä
Powers of i
96
Chapter 3
The powers of i have the following pattern
i 1 = i
i 2 =−1
i 3 = i 2 · i =−i
i 4 = i 2 · i 2 =−1 · (−1) = 1
i 5 = i 4 · i = i
i 6 = i 4 · i 2 = 1 · (−1) =−1
i 7 = i 4 · i 3 = 1 · (−i ) =−i
i 8 = i 4 · i 4 = 1 ·1 = 1
i 4k = 1
i 4k+1 = i
i 4k+2 =−1
i 4k+3 =−i
3.15 Example Simplify
1. i 44
2. i 55
3. i 102
4. i 2010
Solution: Express the powers as a multiple of 4 and a remainder:
1. i 44 = i 4(11) = 1
2. i 55 = i 4(13)+3 =−i
3. i 102 = i 4(25)+2 =−1
4. i 2010 = i 4(502)+2 =−1 ä
3.16 Example Simplify
1. (1+ ip
2)2
2.5+
p−2
1−p−2
3.
p3−6i
p3−2i
4. (1+ i + i 2)3
5. (1− i + i 2 + i 3)2
Solution:
1. (1+ ip
2)2 =
(1+ i )2
2=
1+2i + i 2
2= i
2.5+
p−2
1−p−2
=5+ i
p2
1− ip
2·
1+ ip
2
1+ ip
2=
5+5p
2i +p
2i −2
1+2=
3+6p
2i
3= 1+2
p2i
3.
p3−6i
p3−2i
=p
3−6ip
3−2i·p
3+2ip
3+2i=
3+2p
3i −6p
3i −12i 2
3+4=
15−4p
3i
7=
15
7−
4p
3
7i
4. (1+ i + i 2)3 = (1+ i −1)3 = i 3 =−i
5. (1− i + i 2 + i 3)2 = (1− i −1− i )2 = (−2i )2 = (−2)2i 2 = 4(−1) =−4 ä
97
Complex Numbers
3.1.2 Homework: Complex Numbers
Simplify the radical expressions
3.1.1p−48
3.1.2p−20
3.1.3 −p−1000
3.1.4p−64
3.1.5p−22
3.1.6p−6
p−16
Perform the operations and write in standard forma +bi
3.1.7 (2+3i )+ (7+5i )
3.1.8 (−5+4i )− (3−2i )
3.1.9 (12−22i )− (−5−6i )
3.1.10 (3+5i )+ (−5−7i )− (3+2i )
3.1.11 (4+3i )(5−7i )
3.1.12 (−6i )(−3+2i )
3.1.13 (2− i )(2+ i )
3.1.14 (3+4i )(3−4i )
3.1.15 (2−3i )2
3.1.165
i
3.1.17−3
2i
3.1.182+ i
3i
3.1.1910
1−2i
3.1.20−20i
3+4i
3.1.217
p3+2i
3.1.22−3i
−3−5i
3.1.23 i 20
3.1.24 i 33
98
Chapter 3
3.2 Quadratic EquationsOBJECTIVES
• Factoring method to solve quadratic equations
• The square root method
In this chapter, we will introduce various techniques to solve quadratic equations in one variable.
8 Definition A quadratic equation is an equation of the form ax2 +bx + c = 0 where a 6= 0,b,c are realnumbers. Such as 3x2 −12 = 0 and x2 −5x +6 = 0.
3.2.1 Solving equations by factoring
The basic principle for solving factored equations is the zero-factor property:
If a ·b = 0, then a = 0 or b = 0
That is, if the product of factors is zero, then at least one of them is zero. For example if 3(x −5) = 0,then one factor is zero: x−5 = 0, x = 5. The factored equation: (x+3)(x−4) = 0 is satisfied for x+3 = 0, x =−3
and x −4 = 0, x = 4. To solve equations by factoring, we apply the following procedure:
1. Get zero on one side and all terms on one the other side of the equation.
2. Factor the expression into linear factors.
3. Set each factor to zero and solve the simpler equations.
3.17 Example Solve the quadratic equation x2 −2x = 3
Solution:
x2 −2x −3 = 3−3 Subtract 3 from both sides
x2 −2x −3 = 0 Factor
(x +1)(x −3) = 0, Set each factor to zero
x +1 = 0, x −3 = 0,
x =−1, x = 3
The solutions are x =−1 and x = 3. ä
3.18 Example Solve the equation x2 −4 = 0
Solution:
x2 −4 = 0 Factor
(x +2)(x −2) = 0
x +2 = 0, x −2 = 0
x =−2, x = 2
ä
3.19 Example Solve the equation x2 +3x −18= 0
Solution:
x2 +3x −18 = 0 Factor
(x +6)(x −3) = 0
x +6 = 0, x −3 = 0,
x =−6, x = 3 ä
99
Quadratic Equations
3.20 Example Solve the quadratic equations by factoring
1. 4(x −5)(2x −7) = 0
2. x2 +12 = 7x
3. 6x2 −7x = 3
Solution:
1. 4(x −5)(2x −7) = 0
4(x −5)(2x −7) = 0, Set each factor to zero
x −5 = 0, 2x −7 = 0,
x = 5, x = 7/2
2. x2 +12 = 7x
x2 +12−7x = 7x −7x
x2 −7x +12 = 0
(x −4)(x −3) = 0
x −4 = 0, x −3 = 0,
x = 4, x = 3
3. 6x2 −7x = 3
6x2 −7x −3 = 3−3
6x2 −7x −3 = 0
(2x −3)(3x +1) = 0
2x −3 = 0, 3x +1 = 0,
x = 3/2, x =−1/3 ä
3.21 Example Solve the quadratic equation (2x +3)(x +1)= 1
Solution: Distribute left side and combined with 1:
(2x +3)(x +1) = 1
2x2 +2x +3x +3 = 1
2x2 +5x +3−1 = 1−1
2x2 +5x +2 = 0
(2x +1)(x +2) = 0
2x +1 = 0, x +2 = 0,
x =−1/2, x =−2 ä
3.22 Example Solve the cubic equation 5x3 −20x = 0
Solution: Factor the expression and use the zero factor property:
5x3 −20x = 0
5x(x2 −4) = 0
5x(x −2)(x +2) = 0
5x = 0, x −2 = 0, x +2 = 0,
x = 0, x = 2, x =−2 ä
100
Chapter 3
3.2.2 The square root property
Many quadratic equations cannot be factored over integer coefficients such as the equation 2x2+4x+1 = 0
and x2 −10 = 0. To resolve such difficulty, we introduce a new method of solving quadratic equationsbased on the square root property:
if x2 = c , then x =+p
c , or x =−p
c
if (x −a)2 = c , then x = a +
pc , or x = a −
pc
For any real numbers a,c.
☞ You may ask, The square root of any number is positive (we called it the principal root) so why wehave two of them? Let me show why there is no inconsistency here.
x2 = c, Take the square root of both sides√
x2 =p
c , Apply the definition of the square root
|x| =+p
c , Both sides are positive. Solve the absolute value equation
x =p
c , x =−p
c , Two solutions
There is a standard notation to combine these two solutions:
x =+p
c , or x =−p
c : x =∓p
c =±p
c
3.23 Example Solve the quadratic equations by the square root method
1. x2 = 25
2. y2 = 81
3. x2 +16 = 0
4. (x +3)2 = 16
5. (2x +3)2 = 49
Solution:
1. x =∓5
2. y =∓9
3. x2 =−16, x =∓p−16 =∓4i
4.
(x +3)2 = 16, Take the the square root
x +3 =∓4, Solve for x
x =−3∓4, Separate the solutions
x =−3+4, x =−3−4
x = 1, x =−7
5. (2x +3)2 = 49
(2x +3)2 = 49, Take the the square root
2x +3 =∓7, Solve for x
2x =−3∓7, Separate the solutions
2x =−3+7, 2x =−3−7
x = 2, x =−5 ä
3.24 Example Solve the quadratic equations by the square root method
101
Quadratic Equations
1. x2 = 15
2. 2y2 = 36
3. x2 +8 = 0
4. 3(x +2)2 = 18
5. (2x +6)2 +4 = 0
6. 9(x −4)2 −11 = 0
Solution:
1. x =∓p
15
2.
2y2 = 36
y2 = 18
y =∓p
18
y =∓3p
2
3.
x2 +8 = 0
x2 =−8
x =∓p−8
x =∓2p
2i
4.
3(x +2)2 = 18
(x +2)2 = 6
x +2 =∓p
6
x =−2∓p
6
5.
(2x +6)2 +4 = 0
(2x +6)2 =−4
2x +6 =∓p−4
2x =−6∓2i
x =−3∓ i
6.
9(x −4)2 −11 = 0
(x −4)2 =
11
9
x −4 =∓p
11
3
x = 4∓p
11
3
ä
102
Chapter 3
3.2.3 Homework: Factoring and Square Root Method
Solve the following equations by factoring
3.2.1 x2 +x −20 = 0
3.2.2 12x2 +16x −3 = 0
3.2.3 x2 +4kx = 0
3.2.4 x2 −25x = 0
3.2.5 2x2 −3x = 9
3.2.6 10x2 −7x +1 = 0
3.2.7 20−x = 12x2
3.2.8 2x3 −50x = 0
3.2.9 x3 −4x2 +3x = 0
Solve the following equations by the Square RootMethod
3.2.10 x2 = 36
3.2.11 x2 = 44
3.2.12 12x2 = 20
3.2.13 (2−3x)2 =−16
3.2.14 5x2 +20 = 0
3.2.15 5(2x +1)2 −25 = 0
3.2.16 4x2 −8 = 0
3.2.17 x2 +12 = 0
3.2.18 5(x −1)2 = 45
3.2.19 4(x +3)2 +4 = 0
3.2.20 18x2 −3 = 0
3.2.21 In isosceles right triangle (a = b ), find the hy-potenuse c if b = 8
3.2.22 In a right triangle , find the side a if the hy-potenuse c = 10 and b = 4
103
Completing the square method
3.3 Completing the square methodPerfect quadratic trinomials can be written as squared quantity. These trinomials are recognizablefrom the two special products:
x2 +2ax +a2 = (x +a)2
x2 −2ax +a2 = (x −a)2
For example,
x2 +6x +9 = (x +3)2
x2 −6x +9 = (x −3)2
x2 +10x +25 = (x +5)2
x2 + x +1
4= (x +
1
2)
2
x2 −14x +49 = (x −7)2
☞ The last term in the trinomial is a positive number and it is equal to the square of half the coefficientof x. Thus, if the perfect trinomial is x2 +p x +q = 0 then
q = (p
2)
2 =1
4P 2
.
For the previous examples,
9 = (6
2)
2 =1
46
2 =36
4= 9
9 = (−6
2)
2
25 = (10
2)
2
1
4= (
1
2)
2
49 = (−14
2)
2
3.25 Example Identify the coefficient of x and find half of the coefficient of x and calculate its square:
1. x2 −12x +8 = 0
2. x2 +7x +12 = 0
3. x2 − 1
2x = 10
4. x2 + 35
x = 2
Solution: Let b be the coefficient of x, one half of it be1
2b, its square be (
b
2)2 =
b2
4
1. x2 −12x +8 = 0: b =−12, its half,1
2b = −12
2=−6, its square (−6)2 = 36
2. x2 +7x +12= 0: b = 7, its half, 7
2, its square (
7
2)2 = 49
4
3. x2 − 1
2x = 10: b =− 1
2, its half, 1
2· −1
2=− 1
4, its square (− 1
4)2 = 1
16
4. x2 + 3
5x = 2: b = 3
5, its half, 1
2· 3
5= 3
10, its square (
3
10)2 = 9
100 ä
3.26 Example Add a third term to the binomial, so the resulting trinomial is a perfect square:
104
Chapter 3
1. x2 −12x
2. x2 +30x
3. x2 − 2
3x
4. x2 + x
Solution: The third term must be the square of one-half the coefficient of b:
1. x2 −12x +(
12· (−12)
)2 = x2 −12x +36 = (x −6)2
2. x2 +30x +(
12· (30)
)2 = x2 +30x +225= (x +15)2
3. x2 − 23
x +(
12· (− 2
3))2 = x2 − 2
3x + 1
9= (x − 1
3)2
4. x2 + x + (12
)2 = x2 + x + 14= (x + 1
2)2 ä
Quadratic equations that are not perfect squares can be made perfect by adding a suitable positivenumber to both sides of the equation. The process of solving quadratic equations by completing thesquare consists of the following steps:
The method of completing the square
Any quadratic equations of the formax2 +bx +c = 0
can be solved by the method of completing the square.
1. Terms involving the variable (x) should be on one side (left side) and the constant on the otherside (right).
2. The coefficient of x2 should be 1, divide both sides by a.
3. Completing the square: Add to both sides the square of one-half the coefficient of x.
4. Write the perfect trinomial (left side) as a squared quantity, and write the right side as one numberif possible.
5. Apply the square root method and solve for x.
3.27 Example Solve x2 +8x +7 = 0 by completing the square method
Solution:
x2 +8x =−7, Prepare the equation
x2 +8x + (8/2)2 =−7+ (8/2)
2, Add the missing constant
x2 +8x +16 =−7+16
(x +4)2 = 9, Apply the square root method
x +4 = 3, x +4 =−3
x =−1, x =−7 ä
3.28 Example Solve the quadratic equation x2 −6x +4 = 0 by completing the square
Solution:
x2 −6x +4 = 0
x2 −6x =−4, -6, -3, (−3)2 = 9, Add 9
x2 −6x +9 = 9−4
(x −3)2 = 5
x −3 =±p
5
x = 3±p
5 ä
105
Completing the square method
3.29 Example Solve 4x2 −8x −3 = 0 by completing the square
Solution:
4x2 −8x =+3, Divide by 4
x2 −2x =3
4, Prepare the equation
x2 −2x + (−1)2 =
3
4+ (−1)
2, Add the missing constant
x2 −2x +1 =3
4+1
(x −1)2 =
7
4, Apply the square root method
x −1 =p
7
2, x −1 =−
p7
2
x = 1+p
7
2, x = 1−
p7
2 ä
3.30 Example Solve the quadratic equation 2x2 +4x +1 = 0 by completing the square
Solution:
2x2 +4x =−1
x2 +2x = 11
2
x2 +2x + (1)2 =−
1
2+ (1)
2
x2 +2x +1 =−1
2+1
(x +1)2 =
1
2
x +1 =±√
1
2
x =−1±1p
2=−1±
p2
2ä
3.31 Example Solve the quadratic equation x2 +18x = 19 by completing the square
Solution:
x2 +18x = 19
x2 +18x + (9)2 = 19+ (9)
2
x2 +18x +81 = 19+81
(x +9)2 = 100
x +9 = 10, x +9 =−10
x = 1, x =−19ä
3.32 Example Solve 6x2 +5x −6 = 0
106
Chapter 3
Solution:
6x2 +5x = 6
x2 +5
6x = 1
x2 +5
6x + (
1
2·
5
6)
2 = 1+ (1
2·
5
6)
2
(x +5
12)
2 =144+25
144=
169
144
x +5
12=∓
13
12
x =2
3, x =−
3
2ä
3.33 Example Solve the quadratic equations by completing the square
1. x2 − x = 1
2. 2u2 +9u +9 = 0
3. 2y2 −7y +8 = 0
Solution:
1. x2 − x = 1
x2 − x = 1
x2 − x +1
4= 1+
1
4
(x −1
2)
2 =5
4
x −1
2=∓
p5
2
x =1
2∓p
5
2
2. 2u2 +9u +9 = 0
2u2 +9u =−9
u2 +9
2u =−
9
2
u2 +9
2u +
81
16=−
9
2+
81
16
(u +9
4)
2 =9
16
u +9
4=∓
9
4
u =−9
2, u =−3
107
Completing the square method
3. 2y2 −7y +8 = 0
2y2 −7y =−8
y2 −7
2y =−4
y2 −7
2y +
49
16=−4+
49
16
(y −7
4)
2 =−15
16
y −7
4=∓
p15
4i
y =7
4∓p
15
4i ä
108
Chapter 3
3.3.1 Homework: Completing the Square Method
Solve the equations by completing the square
3.3.1 x2 −3x −18 = 0
3.3.2 3n2 −14n −5 = 0
3.3.3 x2 −4x −1 = 0
3.3.4 x2 −8x = 8
3.3.5 x2 −4 =−x
3.3.6 3x2 −9x =−6
3.3.7 y 2−8y =−17
3.3.8 n2 +2n +12 = 0
3.3.9 x2 −7x −1 = 0
3.3.10 x2 +6x −7 = 0
3.3.11 (3x −2)(2x +9) = 0
3.3.12 (x +5)2 = 18
3.3.13 2n2 +4n −3 = 0
3.3.14 x2 −7 = 2x
3.3.15 4x2 +4x = 1
3.3.16 x2 +8x +10 = 0
3.3.17 3x2 +6x +2 = 0
3.3.18 x2 −5x −5 = 0
109
The Quadratic Formula
3.4 The Quadratic FormulaThe quadratic formula is the short cut for completing the square method of a general quadratic equa-tion. The derivation of quadratic formula for solving quadratic equation in standard form
ax2 +bx +c = 0
is parallel to solving by completing the square:
1. Divide both sides by a
2. Separate the variables from constants
3. Complete the square
Identify rule that has bee applied to the equation in each step:
x2 +bx
a+
c
a=
0
a, Divide by a
x2 +bx
a=−
c
a, Constant on right side
x2 +bx
a+ (
b
2a)
2 = (b
2a)
2 −c
a, Add the missing constant to both sides
x2 +bx
a+ (
b2
4a2) =
b2
4a2−
c
a, Common denominator is 4a2
(x +b
2a)
2 =b2 −4ac
4a2, Apply the square root
x +b
2a=∓
pb2 −4acp
4a2
x =−b
2a∓p
b2 −4ac
2a, Combine as one fraction
x =−b ∓
pb2 −4ac
2a, The quadratic formula
The most famous quadratic formula is
x =−b ∓
pb2 −4ac
2a
it says, give the coefficients of a quadratic equation in standard form, and I give you the solution byjust an arithmetic operations.
3.34 Example Solve the quadratic equation x2 + x −1 = 0 by use of the quadratic formula
Solution: The equation is in standard form, so the coefficients are a = 1,b = 1,c = −1,substitute in the quadratic formula
x =−1∓
√
(1)2 −4(1)(−1)
2(1)
x =−1∓
p1+4
2
x =−1∓
p5
2
ä
3.35 Example Solve the quadratic equation x2 −2x −3 = 0 by using the quadratic formula
110
Chapter 3
Solution: The equation is in standard form, so the coefficients are a = 1,b = −2,c = −3,substitute in the quadratic formula
x =−(−2)∓
√
(−2)2 −4(1)(−3)
2(1)
x =2∓
p4+12
2
x =2∓
p16
2
x =2∓4
2
x =2+4
2, =
2−4
2
x = 3, x =−1ä
3.36 Example Solve the quadratic equation x2 −12x +36 = 0 by using the quadratic formula
Solution: The equation is in standard form, so the coefficients are a = 1,b = −12,c = 36,substitute in the quadratic formula
x =−(−12)∓
√
(−12)2 −4(1)(36)
2(1)
x =12∓
p144−144
2
x =12∓
p0
2
x =12∓0
2
x = 6ä
3.37 Example Solve the quadratic equation 3x2 + x −10 = 0 by use of the quadratic formula
Solution: The equation is in standard form, so the coefficients are a = 3,b = 1,c = −10,substitute in the quadratic formula
x =−1∓
√
(1)2 −4(3)(−10)
2(3)
x =−1∓
p1+120
6
x =−1∓
p121
6
x =−1∓11
6
x =−1+11
6, =
−1−11
6
x =5
3, x =−2
ä
3.38 Example Solve the quadratic equation x2 +30 = 11x by use of the quadratic formula
Solution: Write the equation in standard form that is x2 −11x +30 = 0, the coefficients are
111
The Quadratic Formula
a = 1,b =−11,c = 30, substitute in the quadratic formula
x =11∓
√
(11)2 −4(1)(30)
2(1)
x =11∓
p121−120
2
x =11∓
p1
2
x =11∓1
2
x =11+1
2, =
11−1
2
x = 6, x = 5 ä
3.39 Example Solve x2 +2x +9 = 0 by use of the quadratic formula
Solution: We have a = 1,b = 2,c = 9, substitute in the quadratic formula
x =−2∓
√
(2)2 −4(1)(9)
2(1)
x =−2∓
p4−36
2
x =−2∓
p−32
2
x =−2∓4
p2i
2
x =−2
2∓
4p
2i
2
x =−1∓2p
2i ä
3.40 Example Solve 6x2 − x −5 = 0 by using quadratic formula
Solution: We have a = 6,b =−1,c =−5, substitute in the quadratic formula
x =−(−1)∓
√
(−1)2 −4(6)(−5)
2(12)
x =+1∓
p1+120
24
x =1∓
p121
24
x =1∓11
24
x =1+11
24, =
1−11
24
x =1
2, x =−
5
12 ä
3.41 Example Solve x2 −5
3=−
11
6x
Solution: It is easy to rewrite the equation without fractions by multiplying both sides by 6,then substitute in the quadratic formula
112
Chapter 3
x2 −5
3=−
11
6x, Rewrite in standard form
x2 +11
6x −
5
3= 0, Multiply both sides by 6
6x2 +11x −10 = 0, The coefficients a=6, b=11, c=-10
x =−(11)∓
√
(11)2 −4(6)(−10)
2(6)
x =−11∓
p121+240
12
x =−11∓
p361
12
x =−11∓19
12
x =−11+19
12, =
−11−19
12
x =2
3, x =−
5
2 ä
3.42 Example Solve 2x2 =−4x −1 by using quadratic formula
Solution: Rewrite in standard form 2x2 +4x +1 = 0, so that a = 2,b = 4,c = 1, substitute in thequadratic formula
x =−(4)∓
√
(4)2 −4(2)(1)
2(2)
x =−4∓
p16−8
4
x =−4∓
p8
4
x =−4∓2
p2
4
x =−2∓
p2
2 ä
3.43 Example Use quadratic equations to solve
1. −2x2 +4x −3 = 0
2. 3y2 +7y −6 = 0
3. 5x2 −9x +3 = 0
4. 11s2 −7s +1 = 0
Solution:
1. −2x2+4x −3 = 0, Multiply by −1 to get 2x2 −4x +3 = 0, the solution is x =4∓
p16−24
4= 1∓
p2i
2
2. 3y2 +7y −6 = 0, the solution is x =−7∓
p49+72
6=
−7+∓p
121
6, x =−3, x = 2/3
3. 5x2 −9x +3 = 0, the solution is x =9∓
p81−60
10=
9+∓p
21
10
4. 11s2 −7s +1 = 0, the solution is x =7∓
p49−44
22=
7+∓p
5
22 ä
113
The Quadratic Formula
3.44 Example Find two consecutive positive integers whose product is 240.
Solution: Assume the first number is x, so the next number must be x + 1. These twonumbers satisfy the equation x(x + 1) = 240. Put in standard form x2 + x − 240 = 0, solve byfactoring (x −16)(x +15)= 0. The pair of solutions are: 15,16. ä
3.45 Example The sum of a number and its reciprocal is25
12. Find the number.
Solution: Assume the number is x, it satisfies the equation x+1
x=
25
12, this fractional equation
is equivalent to 12x2 −25x +12 = 0, by quadratic formula x = 3/4. ä
114
Chapter 3
3.4.1 Homework: Quadratic Formula
Solve the equations by the quadratic formula.Identify the coefficients a,b,c.
3.4.1 4x2 −4x +1 = 0
3.4.2 3x2 =−4x +2
3.4.3 a2 −6a = 3
3.4.4 x2 −16x +60 = 0
3.4.5 2x2 −x −4 = 0
3.4.6 −5n2 +3n −5 = 0
3.4.7 4x2 −x = 4
3.4.8 5x2 =−3
3.4.9 −x2 +5x −3 = 0
3.4.10 3x2 −4x −3 = 0
3.4.11 7x2 −4x −20 = 0
3.4.12 x2 +2 = 8x
3.4.13 x2 = 4−12x
3.4.14 2x2 = 4x −1
3.4.15 13x2 −4x = 4
3.4.16 x2 +x +1 = 0
3.4.17 x2 +4 = x +2
3.4.18 x2 −5x +4 = 5x +4
3.4.19 3x2 −3x +5 = x2 −2x +1
3.4.20 12
x2 −3x −5 = 0
3.4.21 x2 +4x +4 = 2x2 +1
115
Equations in Quadratic Form and the Discriminant
3.5 Equations in Quadratic Form and the DiscriminantEquations of the form
1. x4 −5x2 +6 = 0
2. 2x6 + x3 −8 = 0
3. x2/3 −4x1/3 +4 = 0
4. (x2 −9)2 −3(x2 −9)−10 = 0
are not quadratic equations, but it is easy to rewrite them in quadratic form with the help of a substi-tution. This happens because the power of the middle term is half of the first term. Thus
1. x4 −5x2 +6 = 0 becomes u2 −5u +6 = 0 with substitution u = x2
2. 2x6 + x3 −8 = 0 becomes 2u2 +u −8 = 0 with substitution u = x3
3. x2/3 −4x1/3 +4 = 0 becomes u2 −4u +4 = 0 with substitution u = x1/3
4. (x2 −9)2 −3(x2 −9)−10 = 0 becomes u2 −3u +6 = 0 with substitution u = x2 −9
3.46 Example Solve x4 −10x2 +9 = 0
Solution:
x4 −10x2 +9 = 0
u2 −10u +9 = 0, Substitute u = x2
(u −1)(u −9) = 0, Solve for u by factoring
u = 1 or u = 9
x2 = 1 or x2 = 9, Solve for x
x =∓1 x =∓3 ä
3.47 Example Solve x4 + x2 −20 = 0
Solution:
x4 + x2 −20 = 0
u2 +u −10 = 0, Substitute u = x2
(u −4)(u +5) = 0, Solve for u by factoring
u = 4 or u =−5
x2 = 4 or x2 =−5, Solve for x
x =∓2 x =∓ip
5 ä
3.48 Example Solve x2/3 −5x1/3 +6 = 0
Solution:
x2/3 −5x1/3 +6 = 0
u2 −5u +6 = 0, Substitute u = x1/3
(u −2)(u +3) = 0, Solve for u by factoring
u = 2 or u =−3
x1/3 = 2 or x1/3 =−3, Solve for x
x = 8 x =−27 ä
116
Chapter 3
3.49 Example Solve (2x +3)2 −6(2x +3)−7 = 0
Solution:
(2x +3)2 −6(2x +3)−7 = 0
u2 −6u −7 = 0, Substitute u = 2x +3
(u +1)(u −7) = 0, Solve for u by factoring
u =−1 or u = 7
2x +3 =−1 or 2x +3 = 7, Solve for x
x =−2 x = 2 ä
3.50 Example Solve x5 −81x = 0
Solution:
x5 −81x = 0
x(x4 −81)= 0, Factor
x(x2 −9)(x2 +9) = 0
x(x −3)(x +3)(x2 +9) = 0, Solve for x
x = 0, x = 3, x =−3, x =∓ip
3 ä
3.51 Example Solve x4 −5x2 +4 = 0
Solution: We factor the equation in terms of x2:
x4 −5x2 +4 = 0
(x2 −1)(x2 −4) = 0
x2 −1 = 0, x2 −4 = 0
x =∓1, x =∓2 ä
3.52 Example Solve the fourth order equations x4 −8x2 +7 = 0
Solution: We factor the equation in terms of x2:
x4 −8x2 +7 = 0
(x2 −1)(x2 −7) = 0
x2 −1 = 0, x2 −7 = 0
x =∓1, x =∓p
7 ä
3.5.1 The Discriminant
9 Definition The expression b2 − 4ac from the quadratic formula is called the discriminant and it isdenoted by d = b2 −4ac.
3.53 Example Compute the discriminant of the following quadratic equations
1. x2 −7x +12 = 0
2. x2 +10x +25 = 0
3. x2 + x +1 = 0
Solution:
117
Equations in Quadratic Form and the Discriminant
1. x2 −7x +12= 0, d = (−7)2 −4(1)(12)= 49−48 = 1
2. x2 +10x +25 = 0, d = 102 −4(1)(25)= 100−100= 0
3. x2 + x +1 = 0, d = 12 −4(1)(1)= 1−4 =−3 ä
The discriminant gives information about the number of solutions and the nature of the solutions
Discriminant Nature of solutions
d = b2 −4ac > 0 Two real solutions
d = b2 −4ac = 0 One real solution
d = b2 −4ac < 0 Two complex solutions
3.54 Example Compute the discriminant and identify the nature of solutions of the following quadraticequations
1. x2 +7x +1 = 0
2. 12x2 +14x +5 = 0
3. x2 −8x +16 = 0
Solution:
1. x2 +7x +1 = 0, d = (7)2 −4(1)(1)= 49−4 = 45 > 0. The equation has two real solutions.
2. 12x2 +14x +5 = 0, d = 142 −4(12)(5)= 196−480< 0. The equation has two complex solutions.
3. x2 −8x +16= 0, d = (−8)2 −4(1)(16)= 64−64 = 0. The equation has one real solution. ä
3.5.2 Homework: Equations in Quadratic Form
Solve the following equations
3.5.1 x4 −x2 −12 = 0
3.5.2 x4 −10x2 +25 = 0
3.5.3 x4 −6x3 = 0
3.5.4 x2/3 −x1/3 −6 = 0
3.5.5 (4x +5)2 − (4x +5)−2 = 0
3.5.6 2x4 −32 = 0
3.5.7 x4 +x2 −90 = 0
Find the number of distinct real roots of theequation without solving it.
3.5.8 x2 −50x −250 = 0
3.5.9 x2 −20x +200 = 0
3.5.10 10x2 −30x −1 = 0
118
Chapter 3
3.6 Quadratic and Rational InequalitiesIn this section we will study quadratic inequalities and linear rational inequalities of the form
x2 −4x +3 > 0,2x −6
x +4≤ 0
The process of solving these inequalities involve several ideas and it is accomplished by following a stepby step procedure.
☞1. There are many variations of performing the procedure of solving quadratic and rational inequalities,
such as sign chart and graphical method.
2. The procedures of solving inequalities can be applied and generalized to any polynomial and anyrational inequality.
We will review some of the inequalities basics by true or false examples.
3.55 Example True or false? Explain
1. The solution of the equation 4x +5 = 17 is x = 3
2. The solution of the inequality 4x +5 < 17 is x < 12
3. The solution of the inequality 4x +5 ≥ 17 is x ≥ 3
4. The solution of the inequality −x < 2 is the interval (2,∞)
5. The solution of the inequality −10 ≤ 2x ≤ 4 is the interval [2,−5]
6. The inequality x2 −3x ≤ 4 is equivalent to x2 −3x −4 ≤ 0
7. The inequality 10 > 5x is equivalent to 5x −10 > 0
8. The inequality 0 > 2x2 −5x +3 is equivalent to 2x2 −5x +3 < 0
9. The inequality −x2 +7 ≥ 0 is equivalent to x2 −7 ≥ 0
Solution:
1. True
2. False, 4x < 12, x < 3
3. True
4. True
5. False, it is equivalent to the interval [−5,2]. There is an order to the interval notation.
6. True
7. False, the correct answer is 5x −10< 0
8. True
9. False, x2 −7 ≤ 0 ä
3.56 Example True or false? Explain
1. The solution of the inequality 4x2−5x+15 > 0 is the set of all real numbers that make the expression4x2 −5x +15 positive upon substitution.
2. The solution of the inequality −x2 +3x <−8 is the set of all real numbers that make the expression−x2 +3x +8 negative upon substitution.
3. The solution of the inequality x2 +1 > 0 is the interval (−∞,∞)
119
Quadratic and Rational Inequalities
4. The solution of the inequality x2 +4 < 0 is the interval (−∞,∞)
Solution:
1. True
2. True
3. True
4. False. The square of any real number plus 4 is a positive number. ä
3.6.1 Quadratic Inequality
We will illustrate the steps of solving quadratic inequality by example with detailed explanations.
3.57 Example Solve the quadratic inequality x2 − x < 6
Solution:
1. Get zero on one side of the inequality (It does not matter left or right side). This is accom-plished by moving all terms to one side of the inequality.
x2 − x −6 < 0
Note that the expression x2 − x −6 is a quadratic polynomial. The zeros of this polynomialare the solutions of the equation: x2 − x −6 = 0
2. Factor the quadratic polynomial (if possible) and find its zeros that is solve the equation:
x2 − x −6 = 0
(x +2)(x −3) = 0
x +2 = 0, x −3 = 0,
x =−2, x = 3
The solutions −2, 3 divide number line into three intervals. There is a theorem in algebra,states that the sign of a polynomial is positive or negative in each interval that is boundedby its zeros. To find out the signs of the polynomial (quadratic expression), it is sufficientto pick a single point from each interval and substitute in the expression.
3. Graph the zeros (solution points) on a number line (schematically), as in the Figure 5.1.Pick a test point in each interval ( it does not matter which point so select easy numbers,as long as these points are not the zeros (dividers), and substitute in the expression, thenrecord the signs only.
(a) Test point x =−3, substitute in the expression (x +2)(x −3) : (−3+2)(−3−3)= (−1)(−6)> 0.The sign is positive +
(b) Test point x = 0: (0+2)(0−3) = (2)(−3)< 0. The sign is negative −(c) Test point x = 4: (4+2)(4−3) = (6)(1)> 0. The sign is positive +
4. Select the intervals that satisfy the inequality. The inequality is satisfied for negative signsonly, that is for −2 < x < 3 and in interval notation (−2,3). The boundaries are not included,because the inequality is strictly less <. The solution is the interval (−2,3). ä
0 1 2 3 40−1−2−3
3.58 Example Solve the quadratic inequality x2 − x −6 > 0
Solution: Using the previous procedure, we conclude that the solution is the two intervals(−∞,−2) and (3,∞). Note that these two intervals are separated and they cannot be written asdouble inequality. ä
120
Chapter 3
−2 3
Test Points x=−3 x=0 x=4
Evaluate (−1)(−6) >0 (0+2)(0−3) < 0 (4+2)(4−3) >0 Signs + + + − − − + + +Ranges x < − 2 −2 < x < 3 x > 3 Intervals (−∞ , −2) ( −2 , 3 ) ( 3 , ∞ )
Figure A: Solution Diagram of x 2 −x−6 < 0
1 3
Test Points x=0 x=2 x=4
Evaluate (−1)(−3) >0 (1)(−2) < 0 (3)(1) >0 Signs + + + − − − + + +Ranges x < 1 1 < x < 3 x > 3 Intervals (−∞ , 1) ( 1 , 3 ) ( 3 , ∞ )
Figure B: Solution Diagram of x 2 −4x+3 > 0
Figure 3.1: Quadratic Inequalities
3.59 Example Solve the quadratic inequality 2x2 −12 ≤ 2x
Solution: Note this inequality is equivalent to 2x2 −2x −12 ≤ 0. Divide both sides by 2 to getx2−x−6 ≤ 0. This is the same inequality as the previous detailed example, with allowed equalityat the end point. That is the zeros are acceptable solutions so the solution of this inequality is[−2,3] ä
Method of solving quadratic inequalities
1. Zero: Move all terms to one side of the inequality and zero to the other side.
2. Factor and solve: Factor the quadratic polynomial, set it to zero and solve the equation.
3. Number line: Mark the solution points on a number line.
4. Test Points: Select a number in each interval, substitute these numbers in the inequality, anddetermine its signs (+ or −).
5. Select the solution interval based on the sign diagram.
6. Graph the solution on a number line if requested.
3.60 Example Solve the quadratic inequality x2 −4x >−3
Solution:
1. Zero: x2 −4x +3 >−3+3,
x2 −4x +3 > 0
121
Quadratic and Rational Inequalities
2. Factor and solve:
x2 −4x +3 = 0
(x −1)(x −3) = 0
x −1 = 0, x −3 = 0,
x = 1, x = 3
3. Number line: Mark x = 1 and x = 3 on a number line.
4. Test Points:
x = 0, x = 2, x = 4
(0−1)(0−3) > 0, (2−1)(2−3) < 0, (4−1)(4−3) > 0
+++, −−−, +++
5. The solution is: (−∞,1) and (3,∞) ä
0 1 2 3 40−1
3.61 Example Solve the quadratic inequality x2 −4x +3 ≤ 0
Solution: From the previous diagram, the solution interval is [1,3], note the boundary pointsx = 1 and x = 3 are included in the solution because they satisfy the inequality. ä
3.62 Example Solve the quadratic inequality x2 − x −12 < 0
Solution:
1. Zero: x2 − x −12 < 0
2. Factor and solve:
x2 − x −12 = 0
(x +3)(x −4) = 0
x +3 = 0, x −4 = 0,
x =−3, x = 4
3. Number line: Mark x =−3 and x = 4 on a number line.
4. Test Points:
x =−5, x = 0, x = 5
(−5+3)(−5−4)> 0, (0+3)(0−4)< 0, (5+3)(5−4)> 0
+++, −−−, +++
5. The solution is the interval (−3,4) ä
0 1 2 3 4 50−1−2−3−4
3.63 Example Solve the quadratic inequality x3 −2x2 −8x > 0
Solution:
122
Chapter 3
1. Zero: x3 −2x2 −8x > 0
2. Factor and solve:
x3 −2x2 −8x = 0
x(x2 −2x −8) = 0
x(x +2)(x −4) = 0
x = 0, x +2 = 0, x −4 = 0,
x = 0, x =−2, x = 4
3. Number line: Mark x =−2, 0 and x = 4 on a number line.
4. Test Points:
x =−3, x =−1, x = 1, x = 5
−3(−3+2)(−3−4)=−21 < 0, −1(−1+2)(−1−4)= 5 > 0, (1+2)(1−4)=−6 < 0, 5(5+2)(5−4)= 35> 0
−−−−, ++++, −−−, +++
5. The solution is the interval (−2,0) or (4,∞) ä
0 1 2 3 4 50−1−2−3−4
3.64 Example Solve the inequality x2 ≤ 9
Solution:
1. Rewrite as x2 −9 ≤ 0
2. Factor and solve (x +3)(x −3) = 0, x =−3 and x = 3
3. Mark the points, −3,3 on a number line.
4. Test points: x =−4, (−4+3)(−4−3)> 0, x = 0, (0+3)(0−3)< 0, x = 4, (4+3)(4−3) > 0.
5. The solution is −3 ≤ x ≤ 3 or [−3,3] ä
3.65 Example Solve x2 +16 ≥ 0
Solution: The polynomial x2 +16 does not have any real zero and it is positive for any realnumber, so the inequality is satisfies for all real number, that is the solution is (−∞,∞) ä
3.66 Example Solve x2 > 16
Solution:
1. Rewrite as x2 −16> 0
2. Factor and solve (x +4)(x −4) = 0, x =−4 and x = 4
3. Mark the points, −4,4 on a number line.
4. Test points: x =−5, (−5+4)(−5−4)> 0, x = 0, (0+4)(0−4)< 0, x = 5, (5+4)(5−4) > 0.
5. The solution set is the intervals (−∞,−4) and (4,∞) ä
3.67 Example Solve 2x2 < 9x −4
Solution:
1. Zero: 2x2 −9x +4 < 0
123
Quadratic and Rational Inequalities
2. Factor and solve:
2x2 −9x +4 = 0
(2x −1)(x −4) = 0
2x −1 = 0, x −4 = 0,
x = 1/2, x = 4
3. Number line: Mark x = 1/2 and x = 4 on a number line.
4. Test Points:
x = 0, x = 1, x = 5
(0−1)(0−4)> 0, (2−1)(1−4)< 0, (10−1)(5−4)> 0
+++, −−−, +++
5. The solution is the interval (0.5,4) ä
3.6.2 Rational Inequality
The process of solving inequalities that involve rational expressions such as
x
3+ x>
1
3+ x,
3x
x −2−2 < 0
is similar to the process of solving quadratic equations.Method of solving rational inequalities
1. Zero: Move all terms to one side of the inequality and zero to the other side.
2. Combine all terms into a single fraction.
3. Determine the numbers that make the numerator and denominator zero.
4. Number line: Mark the numbers from previous step on a number line.
5. Test Points: Select a number in each interval, substitute these numbers in the inequality, anddetermine its signs (+ or −).
6. Select the solution interval based on the sign diagram.
7. Graph the solution on a number line if requested.
3.68 Example Solve the rational inequality3x
x −2> 2
Solution:
1. Zero:3x
x −2−2 > 0
2. Combine:3x
x −2−2
x −2
x −2=
x +4
x −2> 0
3. Factor and solve:
x +4 = 0, x −2 = 0,
x =−4, x = 2
4. Number line: Mark x =−4 and x = 2 on a number line to form three regions x <−4, −4 < x <2, x > 2.
124
Chapter 3
5. Test Points:
(−∞,−4) (−4,2) (2,∞)
x =−5, x = 0, x = 5
−5+4
−5−2> 0,
0+4
0−2< 0,
5+4
5−2> 0
+++, −−−, +++
6. The solution set is the regions x <−4, x > 2 or in interval notation (−∞,−4) and (2,∞) ä
0 1 2 3 40−1−2−3−4−5
3.69 Example Solve the rational inequality3
x −4≤ 1
Solution:
1. Zero:3
x −4−1 ≤ 0
2. Combine:3− x +4
x −4=
7− x
x −4≤ 0
3. Factor and solve:
x −4 = 0, 7− x = 0,
x = 4, x = 7
4. Number line: Mark x = 4 and x = 7 on a number line to form three regions x ≤ 4, 4 ≤ x ≤ 7, 7 ≤ x.
5. Test Points:
(−∞,4] [4,7] [7,∞)
x = 0, x = 5, x = 8
7−0
0−4< 0,
7−5
5−4> 0,
7−8
8−4< 0
−−−, +++, −−−
6. The solution set is (−∞,4] and [7,∞) ä
0 1 2 3 4 5 6 7 8
3.70 Example Solve the rational inequalityx
3+ x<
1
3+ x
Solution:
1. Zero:x
3+ x−
1
3+ x< 0
2. Combine:x −1
3+ x< 0
3. Factor and solve:
3+ x = 0, x −1 = 0,
x =−3, x = 1
125
Quadratic and Rational Inequalities
4. Number line: Mark x =−3 and x = 1 on a number line to form three regions x <−3, −3 < x <1, 1 < x.
5. Test Points:
(−∞,−3) (−3,1) (1,∞)
x =−4, x = 0, x = 2
−4−1
3−4> 0,
0−1
3+0< 0,
2−1
3+2> 0
+++, −−−, +++
6. The solution set is (−3,1) ä
3.71 Example Solve the rational inequality8
x> 3
Solution:
1. Zero:8
x−3 > 0
2. Combine:8−3x
x> 0
3. Factor and solve:
x = 0, 8−3x = 0,
x = 0, x = 8/3
4. Number line: Mark x = 0 and x = 8/3 on a number line to form three regions x < 0, 0 < x <8/3, 8/3< x.
5. Test Points:
(−∞,0) (0,8/3) (8/3,∞)
x =−2, x = 1, x = 5
8+4
−2< 0,
8−3
1> 0,
8−15
5< 0
−−−, +++, −−−
6. The solution set is (0,8/3) ä
0 1 2 30−1
3.72 Example Solvex +15
3x −5≤−3
Solution:
1. Zero:x +15
3x −5+3 ≤ 0
2. Combine:x +15+9x −15
3x −5=
10x
3x −5≤ 0
126
Chapter 3
3. Factor and solve:
10x = 0, 3x −5 = 0,
x = 0, x = 5/3
4. Number line: Mark x = 0 and x = 5/3 on a number line to form three regions x < 0, 0 < x <5/3, 5/3< x.
5. Test Points:
(−∞,0] [0,5/3] [5/3,∞)
x =−1, x = 1, x = 5
−10
−3−5> 0,
10
3−5< 0,
50
15−5> 0
+++, −−−, +++
6. The solution set is [0,5/3] ä
0 1 2 30−1
3.73 Example Solve the rational inequality1
(x +1)2< 0
Solution: There is no real number that satisfies this inequality, because left side is alwayspositive. The solution does not exist. ä
127
Quadratic and Rational Inequalities
3.6.3 Homework: Quadratic and Rational Inequalities
Solve the following inequalities. Express the solutionin interval notation and graph the solution on anumber line
3.6.1 (x +3)(2x −7) ≥ 0
3.6.2 (x −2)(x +4) < 0
3.6.3 x(x +5) ≤ 0
3.6.4 x2 +2x −3 < 0
3.6.5 x(x −1) < 30
3.6.6 7(x2 −9) < 0
3.6.7 −16(x2 −16) ≥ 0
3.6.8 4x2 +20x +25 > 0
3.6.9 2x2 > x +10
3.6.10 x2 +6x ≤ 0
3.6.11 x2 −6x +5 < 0
3.6.12(x +2)
(x −2)< 0
3.6.13x
x −2< 0
3.6.143
x> 6
3.6.154
x −2+
x
x −2> 0
3.6.16(3x +10)
(x)≥ 0
3.6.17(1−x)
(x +6)≤ 0
3.6.18(x +3)
(x +6)≤ 2
3.6.193y −5
y +1< 3
3.6.202
x −1+
3
x +1> 0
3.6.215x
(2x +7)2≥ 0
3.6.22 1−6
x> 0
3.6.23x
x2 +1< 0
3.6.24 (x +3)(x +1)(x −3) > 0
128
Chapter 3
3.7 ExercisesSolve the following equations by the method of
square root or factoring1. (x +2)2 −1 = 0
2. 3(x −4)2 =−36
3. x2 +8 = 6x
4. 3x2 =27
4
5. 10x2 −60x =−80
6. x(x −1) = 6x −10
Solve the following equations by the method of complet-ing the square7. x2 +8x = 33
8. x2 −6x = 7
9. x2 +12 = 8x
10. 3x2 −6x +5 = 0
11. 4x2 −8x = 3
12. x2 +2x −10 = 0
13. 4x2 −12x +9 = 0
14. x2 −4x +6 = 0
15. 2x2 −4 = 3x
Solve the following equations using the quadratic for-mula16. x2 −4x −12 = 0
17. −x2 +10x −25 = 0
18. 2x2 −x = 1
19. (x −3)2 =−10x +5
20. x2 +5x +1 = 0
21. 4x2 +9 = 12x
22. 16x2 −25 = 0
23. 2x2 +3−8x = 0
24. 25x2 +4 = 20x
25. x2 +p
5x −5 = 0
26. x2 −2p
3+12 = 0
Solve by factoring, by completing a square, and by thequadratic formula27. 2x2 +5x −3 = 0
28. (x +6)(5x −1) = 5x −25
29. (3x −4)2 −45 = 0
30. x2 +x = 42
31. 3x2 +5 = 16x
32. The sum of two numbers is 16 and the sum of their
squares is 200. What are the numbers?33. Find two numbers whose difference is 5 and the
difference of their squares is 45.34. Divide 45 into two parts whose product is 434.
35. Find two consecutive integers whose product is
306.36. Find two consecutive odd positive integers whose
product is 323.37. Find the length of a side of a square where a diag-
onal is 4 feet longer than a side.38. Simplify (i + i−1)−1
39. Simplify (i − i−1)−1
Solve the following inequalities40. x(x −4) ≥ 0
41. x2 −x −2 < 0
42. (2x −5)2 ≤ 100
43. x2 > 2x +3
44.2x −8
x +8< 0
45.x −1
x +2> 1
129
Review Quadratic Equations
46.2
x −6≤ 7
3.8 Review Quadratic Equations1. Complex Numbers
Imaginary unit is i =p−1 and i 2 =−1 and
p−a = i
pa if a > 0
Complex numbers have theorem a +bi such as 3−4i
2. Arithmetic Operations on complex numbers:To add or subtract complex numbers: combine real parts and imaginary parts.To multiply complex numbers, use distribution and the fact that i 2 =−1.To divide complex numbers, multiply and divide by the conjugate of the denominator. For example,
(3+2i )− (5−6i )=−2+8i
(−2+3i )(2+5i )=−19−6i
4
2+3i=
4(2−3i )
(2+3i )(2−3i )=
8
13−
12
13i
3. Quadratic Equations Methods of solving quadratic equations are
(a) Factoring: Solve x2 +2x −8 = 0
Solution: (x +4)(x −2) = 0, so x =−4,2ä
(b) Square root: Solve 4x2 −24 = 0
Solution: x2 = 6, so x =∓p
6ä
(c) Completing the square: Solve 2x2 −20x −48 = 0
2x2 −20x = 48
x2 −10x = 24
x2 −10x + (1
2·10)
2 = 24+ (1
2·10)
2
x2 −10x +25 = 24+25
(x −5)2 = 49
x −5 =∓7
x = 12, x =−2
(d) Quadratic Formula:
x =−b ∓
pb2 −4ac
2a
Solve 2x2 −7x +3 = 0
Solution: a = 2,b =−7,c = 3 substitute in the formula to get
x =7∓
√
(−7)2 −4(2)(3)
2(2)
Simplify to obtain x =1
2,3
ä
130
4 The Distance Formula and Circles
4.1 The Distance Formula and CircleOBJECTIVES
• The distance Formula
• The Midpoint Formula
• The equation of a circle
• Graph circles with center and radius
4.1.1 Coordinate System
A rectangular Cartesian system of coordinates on the plane consists of two perpendicular axis iden-tified as x-axis and y-axis. Both are given fixed positive direction and a scale. Each point P of theplane is defined by a pair of numbers (x, y) referred as the coordinates of the point. These coordinatesrepresent the distances with proper sign of the point from both axis. The axes divide the plane intofour quadrants.
4.1 Example Draw a system of coordinates and plot the points A(−4,3), B (0,1.5) and C (2,−4)
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
b A
b B
b C
Figure 4.1: Plotting Points
4.2 Example Find the coordinates of the vertices of the given rectangle in the figure.
1
2
3
4
−1
−2
−3
1 2 3 4−1−2−3
Figure 4.2: Vertices of Rectangle
131
The Distance Formula and Circle
Solution: The vertices are ((4,2),(−2,2),(−2,−1),(4,−1). ä
4.1.2 The Distance Formula
The distance between two points is a positive number. For any two points with coordinates x1 and x2
on a number line, the distance between the two points is |x2 − x1|, the absolute value is to insure theanswer is always positive. For example, the distance between the points x1 = 1 and x2 = 5 is clearly|5−1| = 4
0 1 2 3 4 50−1−2−3−4
4.3 Example Plot the two points on a number line and find the distance between them.
1. x1 = 3; x2 = 7
2. x1 =−2; x2 = 4
3. x1 = 1; x2 =−5
4. x1 =−3; x2 =−8
Solution:
1. x1 = 3; x2 = 7: d = |7−3| = 4
2. x1 =−2; x2 = 4: d = |4− (−2)| = 6
3. x1 = 1; x2 =−5: d = |−5−1| = 6
4. x1 =−3; x2 =−8: d = |−8− (−3)| = |−5| = 5 ä
The distance formula between any two points P1(x1, y1) and P2(x2, y2) is based on the PythagoreanTheorem for the right triangle with hypotenuse c and sides a and b and it has the familiar form
a
bc
Figure 4.3: a2 +b2 = c 2
a2 +b2 = c 2
Applying this formula for the right triangle P1P2Q:
P1Q2 +P2Q
2 = P1P2
2
or in terms of the coordinated of the points and letting d = P1P2, we have (x2 − x1)2 + (y2 − y1)2 = d 2.
132
Chapter 4
This leads to the distance formula
d =√
(x2 − x1)2 + (y2 − y1)2
b
b
P1
P2
Q
b
|y2−
y1|
|x2 − x1|
Figure 4.4: Distance between two points
4.4 Example Find the distance between the points (3,−2) and (1,8)
Solution: Substitute carefully in the formula, making sure the order of coordinates and theirsigns are correct. Let (x1, y1) = (3,−2) and (x2, y2) = (1,8). It does not matter which point is calledfirst.
d =√
(1−3)2 + (8− (−2))2 =√
(−2)2 + (10)2 =p
4+100=p
104= 2p
26 ä
4.5 Example Find the distance between the points (−2,3) and (4,5)
Solution: Let (x1, y1) = (−2,3) and (x2, y2) = (4,5). Then
d =√
(4− (−2))2 + (5−3)2
=√
(4+2)2 + (2)2
=p
36+4
=p
40
= 2p
5 ä
4.6 Example Find the distance between the points (0,5) and (−3,−3)
Solution: Let (x1, y1) = (0,5) and (x2, y2) = (−3,−3). Then
d =√
(−3−0)2 + (−3−5)2
=√
(−3)2 + (−8)2
=p
9+64
=p
73 ä
4.7 Example Graph and find the distance between (−6,3),(−2,0)
Solution: Let (x1, y1) = (−6,3) and (x2, y2) = (−2,0). Then
d =√
(−3+6)2 + (0−3)2
=√
(3)2 + (−3)2
=p
9+9
= 3p
2 ä
133
The Distance Formula and Circle
Another useful formula is the midpoint formula between two points. For example the midpointbetween the points A(1) and B (5) on a number line below is the point M(3), with distance 2 to A and B .
0 1 2 3 4 50−1−2−3−4
In fact, for any two points on a number line x1, x2, the midpoint between them is given by x =x1 + x2
2.
4.8 Example Find the midpoint of the given two points.
1. x1 = 3; x2 = 7
2. x1 =−2; x2 = 4
3. x1 = 1; x2 =−5
4. x1 =−3; x2 =−8
Solution:
1. x1 = 3; x2 = 7: x =3+7
2= 5
2. x1 =−2; x2 = 4: x =−2+4
2= 1
3. x1 = 1; x2 =−5: x =1−5
2=−2
4. x1 =−3; x2 =−8: x =−3−8
2=−11/2 ä
1 Definition The midpoint M of the segment line joining two points P1(x1, y1) and P2(x2, y2) is a pointon the segment line P1P2 which is equidistant from its endpoints, the coordinates of the midpoint aregiven by the formula
M(x1 + x2
2,
y1 + y2
2)
4.9 Example Find the midpoint of the pair (6,−4),(−8,7)
Solution: Use the midpoint formula to get (6−8
2,−4+7
2) = (−1,
3
2) ä
4.10 Example Find the midpoint of the pair (0,5),(4,−3)
Solution: Use the midpoint formula to get (0+4
2,
5−3
2) = (2,1) ä
134
Chapter 4
4.1.3 The Circle
2 Definition A circle is the set of all points in a plane that are a fixed distance from a point called thecenter. The fixed distance is called the radius of the circle.
(h,k)
r
Figure 4.5: A circle with centre (h,k) and radius r
To develop the general equation of a circle with center (h,k) and radius r . Note that the distancebetween the center and any point on the circle (x, y) is = r :
r =√
(x −h)2 + (y −k)2
(x −h)2 + (y −k)
2 = r 2
This is the standard form of the general equation of a circle.If the center is at (0,0), then the equation is x2 + y2 = r 2.
4.11 Example Find the equation of the circle with
1. Center (4,2) and radius 5
2. Center (−1,0) and radiusp
6
3. Center (0,0) and radius 10
Solution:
1. Center (4,2) and radius 5: (x −4)2 + (y −2)2 = 25
2. Center (−1,0) and radiusp
6: (x +1)2 + y2 = 6
3. Center (0,0) and radius 10: x2 + y2 = 100 ä
4.12 Example Identify the radius and center of the circles
1. x2 + y2 = 1
2. (x −4)2 + y2 = 8
3. (x +3)2 + (y −9)2 = 7
Solution:
1. The circle x2 + y2 = 1 has center (0,0) and radius r = 1
2. The circle (x −4)2 + y2 = 8 has center (4,0) and radius r = 2p
2
3. The circle (x +3)2 + (y −9)2 = 7 has center (−3,9) and radius r =p
7 ä
4.13 Example Graph the circle x2+y2 = 25 by identifying the center and five opposite points on the circle.
135
The Distance Formula and Circle
4.14 Example Find the equation of the circle with radius 5 and center at (3,2), by identifying the centerand five opposite points on the circle.
2
4
6
8
−2
−4
−6
−8
−10
2 4 6 8−2−4−6−8−10
b b
b
b
b
Figure 4.6: x2 + y2 = 25
2
4
6
8
−2
−4
−6
−8
−10
2 4 6 8−2−4−6−8−10
bb b
b
b
Figure 4.7: center (3,2), r=5
4.15 Example Find the center and the radius of the circle x2 +2x + y2 = 15
Solution: Write the equation in standard form by using completing the square method asfollows:
x2 +2x +1+ y2 = 15+1
(x +1)2 + y2 = 16 ä
Thus the center is (−1,0) and the radius is r = 4
4.16 Example Find the center and the radius of the circle x2 + y2 −8y = 0
Solution: Write the equation in standard form by using completing the square method asfollows:
x2 + y2 −8y +16 = 0+16
x2 + (y −4)2 = 16 ä
Thus the center is (0,4) and the radius is r = 4
4.17 Example Find the center, the radius and graph the circle x2 + y2 +10x +6y = 0
Solution: Write the equation in standard form as follows:
x2 +10x + y2 +6y = 0 Separate the variables and complete the squares
x2 +10x +52 + y2 +6y +3
2 = 0+52 +3
2
(x +5)2 + (y +3)
2 = 25+9
(x +5)2 + (y +3)
2 = 34
ä
Thus the center is (−5,−3) and the radius is r =p
34
4.18 Example Find the center, the radius and graph the circle x2 + y2 −4x +2y = 20
Solution: Write the equation in standard form by using completing the square method asfollows:
x2 −4x + y2 +2y = 20 Separate the variables and complete the squares
x2 −4x +4+ y2 +2y +1 = 20+4+1
(x −2)2 + (y +1)
2 = 25 ä
136
Chapter 4
Thus the center is (2,−1) and the radius is r = 5
2
4
6
8
−2
−4
−6
−8
−10
2 4 6 8−2−4−6−8−10bb b
b
b
Figure 4.8: (x −2)2 + (y +1)2 = 25
4.19 Example Find the center, the radius and graph the circle x2 + y2 = 20x −91
Solution: Write the equation in standard form as follows:
x2 −20x + y2 =−91 Separate the variables and complete the squares
x2 −20x +102 + y2 =−91+10
2
(x −10)2 + y2 = 9 ä
Thus the center is (10,0) and the radius is r = 3
4.20 Example Write the equation of the circle with center (1,−3) and passing through the point (4,2)
Solution: To write the equation of the circle with the given center (1,−3), we need to find itsradius. The radius of this circle is the distance between the center and the point (4,2). That is
r =√
(1−4)2 + (−3−2)2 =p
9+25=p
34. Using the standard form, the equation of the circle is
(x −1)2 + (y +3)
2 = 34 ä
4.21 Example Sketch the graph of the circle (x −1)2 + (y −2)2 = 4
2
4
6
−2
−4
−6
2 4 6−2−4−6
bb b
b
b
Figure 4.9: (x −1)2 + (y −2)2 = 4
137
The Distance Formula and Circle
4.1.4 Homework: The Distance Formula and Circles
Find the distance between the points
4.1.1 (3,1) and (1,0)
4.1.2 (−2,5) and (−8,−2)
4.1.3 (0,11) and (1,12)
Find the midpoint between the pair of points
4.1.4 (−3,2) and (1,0)
4.1.5 (−7,6) and (3,−6)
4.1.6 (2,−3 and (6,9)
Find equations of the circles
4.1.7 Center (−2,−3); r = 3
4.1.8 Center (5,−4); r = 3p
2
4.1.9 Center (0,0); r = 8
4.1.10 Center (0,0); r =p
12
4.1.11 Write the equation of the circle that satisfies:tangent to both axes, center is in the second quadrantand r = 6
Write the equations in standard form(x −h)2 + (y −k)2 = r 2 and identify the center and theradius
4.1.12 x2 + y 2 = 49
4.1.13 x2 = 16− y 2
4.1.14 x2 +2x + y 2 = 8
4.1.15 x2 + y 2 +6x = 0
4.1.16 x2 + y 2 −8x −14y =−62
4.1.17 x2 + y 2 −6x +8y −55 = 0
4.1.18 x2 + y 2 −12y = 0
4.1.19 4x2 +4y 2 = 100
4.1.20 2x2 +2y 2 −8x = 0
138
5 Functions
5.1 Introduction to FunctionsOBJECTIVES
• Representing Functions
• Function notation
• Finding the domain and range of a function
• The vertical line test
• Graph functions with a calculator
Functions are the most fundamental concept in higher mathematics. Discovering any relationshipbetween quantities in nature is advanced with understanding and related to the idea of functionalrelationships. Roughly speaking, if a function exists between two quantities, then it implies that anyinformation about one quantity leads to concrete information about the other quantity. Such as, therelationship between the area of a square and its side. For some people, the word function is associatedwith the performance of concrete tasks, such as the functions of various organs (heart, kidneys) in thebody. In fact, there is a humorous joke about Mathematicians that asks: Why old mathematiciansnever die? and the answer is: because they just lose some of their functions.
Relevant vocabularies
1. A Set is a collection of well defined objects. The objects are called elements or members of the set.Capital letters are used to name the sets. The list of elements are given within braces { }. Forexample,
A = {1,2,3,4, book, box, s}
Familiar sets are the sets of natural numbers N , integer numbers Z , rational numbers Q, and realnumbers R, or part of them.
2. A variable is a symbol such as x, y, t which represents a number that can vary over a set ofnumbers. For example, the room temperature and the stock market index are variables.
3. A constant is a symbol that represents a number which is fixed for all time, such as 5,8,π.In expression such as 3x2 −5x +2 = 0, the symbol x is assumed to be a variable until it is beendetermined by the equation.
4. A formula is an equation that relates variables and constants such as the circumference of a
circle C = 2πr and the area of a triangle A =hb
2.
5. An ordered pair (x, y) is a pair of real numbers x, y enclosed in parentheses, where x is calledthe first coordinate or the first component (element) and y is called the second coordinate or thesecond component (element) of the pair (x, y). For example, the three pairs: (2,5),(5,2),(0,7) aredifferent pairs.
6. A rectangular Cartesian system of coordinates on the plane consists of two perpendicular axisidentified as x-axis and y-axis. Both are given fixed positive direction and a scale. Each point P
of the plane is defined by a pair of numbers (x, y) referred as the coordinates of the point. Thesecoordinates represent the distances with proper sign of the point from both axis.
139
Introduction to Functions
5.1.1 Basic definitions
There are different approaches to defining functions. Some of these definitions are Mathematiciansdefine functions according to their own abstract considerations. We list some of these definitions.
Interdependency approach
The concept of function started with observations that the magnitude of some natural quantities areuniquely defined by the magnitude of other quantities. These interdependencies were discovered andcalled formulas or laws of nature. Some of these quantities are: length, area, volume, time, tempera-ture, distance, speed, pressure and so on. For example,
1. the area of a circle is uniquely defined by its radius r :
A =πr 2
2. the volume of a cube is completely defined by the length of its side s:
V = s3
3. the distance of a free falling object from a high point is determined by the time t:
d =g
2t 2
where g is the earth gravitational constant,
4. the roots (or solutions) of the quadratic equation x2+2x+n = 0 are given by the formula x = 1∓p
1−n.
The interdependency leads to calling the area, volume, distance and roots dependent magnitudesand the radius, side, time and coefficient n independent magnitudes.
1 Definition We say there is a function between a dependent variable y and an independent variablex, if there is a rule between them such that, for every value of x from a well defined set of numbers,
there corresponds a unique value of y.
The set of values x is called the domain of the function, the set of values y is called the range of thefunction. The rule is denoted by a letter such as f , the function (interdependence) is denoted by
y = f (x)
which is read as y is a function of x.
Usually, the names of functions are denoted by f , g ,h,K ,F,G, H ,T or any other letter. For example,f (x) = 3x +4.
Correspondence approach
This approach is purely formal and it has been by many authors.
2 Definition A function has three components:
1. A set X called the domain,
2. a set Y called the range, and
3. a rule denoted by f : X → Y , such that for every element x in X , there is a unique element y inY .
We write y = f (x). This equation is read y equals f of x.
140
Chapter 5
For example, let X ,Y be the sets of all humans. The function f assigns to each x in X a father (definedin concrete way) y in F , f : X → F , where F is the set of all fathers which is a subset in Y . that isy = f (x) is the father of x. Another example, m : X → M, where y = m(x) mother of x and M is the setof all mothers. In both examples, the father and the mother are well defined functions and satisfy theconditions of functions. An example of a rule that is not a function is the relationship child: son ordaughter. Let us call the relationship child g , that is y = g (x) is a child of x from X . y = g (x) is not afunction, Because x may not have a child, or may have many children. However by restriction the seton which g to the set of all humans having children, and modifying the rule to associate y = g (x) to bethe oldest child, or the youngest child. In such a case, this modified g is a well defined function.
Another variant of the above definition is expressed in terms of ordered pairs.
3 Definition A function is a set of ordered pairs such that no two different ordered pairs have the samefirst component.
Ordered pairs can be graphed in the so called mapping diagram, where it is easy to observe whethera relation is a function or not, we use , as shown in figures (5.1-5.4). For example, we can express thefigure 5.2 as the set of ordered pairs {(0,0),(1,2),(2,4),(2,8)}. Clearly the first component in the distinctpairs (1,2),(1,4) is the same, which violates the definition of the function.
domain
rangerule
total setb
b
b
b
b
b
b
b
Figure 5.1: Mapping diagram of a function
.
2 41 20 0
8
Figure 5.2: Not a function
10
2
2
4
Figure 5.3: Not a function
10
2
2
4
Figure 5.4: A function
☞ A relation is a set of ordered pairs, where the set of the first components is the domain and the
set of of its second components is the range. Note, there is no restriction on the domain as we did in thefunction definition. Thus, every function is a relation, but not every relation is a function.
Black Box analogy
In this approach, it is assumed that the function is a machine, the domain of the function is calledthe input and the range is called the output, with the requirement that the machine gives exactly oneoutput for every valid input.
Input x → Function f → Output y
We write y = f (x).
141
Introduction to Functions
5.1.2 Representations of a Function
The function is a rule between two sets. The rule can be expressed verbally (words), numerically (tables),visually (Graphs), and abstractly (formulas). We will give examples illustrating these possibilities.
1. Verbal format: the function f is "Triple the input". Thus if the input is 5, then the output is3×5 = 15.
2. Table format: The data are given in a table format in which the domain is the set {1,3,5,7,9} andthe range is the set {3,6,15,21,27} as in the following table
x 1 3 5 7 9
y 3 9 15 21 27
3. Relation format: The set of all ordered pairs are given explicitly. For example, The function isidentified by the set {(3,5),(−3,5),(0,8)}.
4. Algebraic format: The function is identified by a formula. For example y = 3x and y = x2.
5. Graphical format: In this representation the ordered pairs are plotted on a system of coordinates.For example, The graph of the function F = {(−4,−2),(−3,−1),(2,4),(3,−3)} is shown below.
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
bb
b
b
Figure 5.5: The graph of function F
5.1 Example The function F is described by the following set of ordered pairs: F = {(0,1),(1,3),(2,5),(3,7)}.Express the function F in table, algebraic, verbal and graphical representations.
Solution:
1. Table representation:
x F(x)
0 1
1 3
2 5
3 7
142
Chapter 5
2. Algebraic (Symbolic) representation: We observe that as we change the input by one, theoutput is increased by two. This indicates a linear relationship with slope two and y-intercept five. Thus, y = 2x +1 or F (x) = 2x +1.
3. Verbal representation: "Double the input and add one".
4. Graphical representation:
−1
0
1
2
3
4
5
6
7
8
−1 0 1 2 3 4 5
b
b
b
b
Figure 5.6: Graph of f = 2x +1
ä
5.2 Example Give the domain (the set of all independent first component) and the range (the set of alldependent second component) and determine whether each relation defines a function.
1. F = {(2,0),(4,1),(6,0)}
2. G = {(2,1),(3,−1),(2,−1)}
3. H = {(5,0),(7,0),(8,0)}
Solution:
1. Domain: {2,4,6}, Range: {0,1}, F represents a function. Each first component has a uniquesecond component.
2. Domain: {2,3}, Range: {1,−1}, G is not a function, The input 2 has two different outputs.
3. Domain: {5,7,8}, Range: {0}, H is a function. In the definition of the function, there is norestriction on the output. ä
5.3 Example Give the domain and the range and determine whether each relation defines a function.
−4
−3
−2
−1
0
1
2
3
4
−4 −3 −2 −1 0 1 2 3 4
b
b b
Figure 5.7: f
−4
−3
−2
−1
0
1
2
3
4
−4 −3 −2 −1 0 1 2 3 4
b
b
b
Figure 5.8: g
−4
−3
−2
−1
0
1
2
3
4
−4 −3 −2 −1 0 1 2 3 4
b b b
Figure 5.9: h
Solution:
143
Introduction to Functions
1. Domain: {−2,0,1}, Range: {−1,3}, f represents a function.
2. Domain: {−2,1}, Range: {0,3,−1}, g is not a function, The x-component −2 has two differenty-components 0,3.
3. Domain: {−2,1,3}, Range: {2}, h is a function. ä
5.4 Example Express the numerical data of the function as an equation between x and y.
1.x 0 1 2 3 4 5
y -1 1 3 5 7 9
2.x -2 -1 0 1 2 3
y -8 -5 -2 1 4 7
Solution:
1. The difference between consecutive y is 2 as x changes by 1. This indicates a linearformula with slope 2, that is y = 2x +b, when x = 0, y =−1, thus y = 2x −1.
2. Comparing the changes in x and y, leads to the linear formula y = 3x −2. ä
5.5 Example The function f is given by the formula f (x) =1
2x+1. Its domain consists of all positive even
integers less than 13. List the elements of the domain and the range of the function f .
Solution:
1. Domain: {2,4,6,8,10,12}, Range: {2,3,4,5,6,7} ä
5.6 Example Find the functional relationship between the diagonal d of a square with side s.
Solution: Using the Pythagoras Theorem: d =p
2s ä
5.7 Example The price of gasoline in a city is $3.45 per gallon. Express the cost C of a quantity ofgasoline as a function of the number x of gallons bought.
Solution: C = 3.45x. So the cost of ten gallons is C = 3.45(10)= 34.5. ä
5.1.3 Evaluation and equality of Functions
To evaluate a function f (x) of a value x = a is to replace or to substitute x by a and use the formulaor the table or the graph to find the value f (a).
5.8 Example For f (x) = x2 −3x, evaluate the following:
1. f (−3)
2. f (1
2)
3. f (a)
4. f (a −2)
5. f (t)
144
Chapter 5
6. f (1
t)
Solution:
1. f (−3) = (−3)2 −3(−3)= 9+9 = 18, " f of −3 equals 18,
2. f (1
2) =
1
4−
3
2=−
5
4
3. f (a) = a2 −3a
4. f (a −2) = (a −2)2 −3a = a2 −4a +4−3a = a2 −7a +4
5. f (t) = t 2 −3t
6. f (1
t) =
1
t 2−
3
t=
1−3t
t 2 ä
5.9 Example For the function g (x) defined by g = {(1,2),(1.4,3),(2,2),(3,−1),(4,0),(5,6)}.
1. Evaluate g (3) and g (1.4).
2. Find two different values of x such that g (x) = 2.
Solution:
1. g (3)=−1 and g (1.4) = 3.
2. By looking at the set, we see that g (1)= 2 and g (2)= 2. ä
5.10 Example For the function f (x) = 3x2 −2x evaluate f (2), f (−2) and f (3)− f (1).
Solution:
1. f (2) = 3(22)−2(2) = 12−4 = 8
2. f (−2) = 3(−2)2 −2(−2) = 12+4 = 16
3. f (3)− f (1) = 3(32)−2(3)− (3(1)2 −2(1))= 27−6−3+2 = 20 ä
5.11 Example Let g (x) = x2 − x. Find
1. g (s)
2. g (2s)
3. g (3)− g (2)
Solution:
1. g (s)= s2 − s
2. g (2s)= (2s)2 −2s = 4s2 −2s
3. g (3)− g (2)= 32 −3− (22 −2) = 9−3−4+2 = 4 ä
Equality of Functions
4 Definition Two functions f (x) and g (t) are equal if they have the same domain and for every input,they produce same output. For example, the functions f (x) = 3x +1 is equal to the function g (t) = 3t +1
defined on the set of real numbers. Take the input 5 and check that
f (5) = 3(5)+1 = 16, g (5)= 3(5)+1 = 16
5.12 Example Are the following pair of functions equal, why?
145
Introduction to Functions
1. f (x) = x2 −1 and g (x) = (x +1)(x −1). Both are defined on the set of all real numbers.
2. f (t) = 3t +3 defined on the set t > 0 and g (t) = 3t +3 defined on the set t < 0
Solution:
1. f (x) is equal to the function g (x). They have the same domain and the same expression.
2. f (t) is not equal to g (t). They are defined on different domains, although, they have thesame expression. ä
5.1.4 Domain of Functions
The domain of a function is the set of all elements where the function is well defined. If the functionis given as a set of ordered pairs, then the domain is the set of all first components as seen before.However, if the function is given as an expression (formula), then the domain is either given explicitlyor implicitly. For example,
1. Explicit domain: The function f (x) = 3x − 5 is defined on [0,9]. Clearly the domain is the giveninterval.
2. Implicit domain: Given a function g (x) = 3x−5. Here, the domain of g is not specified, so we alwaysassume that the domain is all real numbers that is (−∞,∞) except where the function fails tobe well defined "real number". The domain of any function in algebraic form is all real numbersexcept:
(a) The values that make denominator zero. For example, the domain of f (x) =2x
x −3is x 6= 3.
(b) The values that make the radicand of square root ("even roots") a negative number. Forexample, the domain of f (x) =
px is x ≥ 0.
5.13 Example Find the domain of the following functions
1. f (x) = 3x2 +4x −5
2. g (x) =x +3
x +6
3. h(x) =p
x +4
4. f (x) = 7x4 −2x3 + 2
5x
5. f (x) =2x +7
(x −2)(3x +8)
6. g (x) =x2 −4
x2 −7x +12
7. h(x) =p
3x −5
Solution:
1. f (x) = 3x2 + 4x − 5. The expression is defined for all real number, that is the domain is(−∞,∞).
2. g (x) =x +3
x +6. The expression makes sense for all real numbers except when x =−6, thus the
domain can be expressed as x 6= −6, or in interval notation: (−∞,−6) and (−6,∞).
3. h(x) =p
x +4. The formula defines real number if x +4 ≥ 0, that is x ≥−4.
4. f (x) = 7x4 −2x3 + 25
x. The domain is all numbers (−∞,∞).
5. f (x) =2x +7
(x −2)(3x +8). The domain is all numbers except {2,−8/3}.
146
Chapter 5
6. g (x) =x2 −4
x2 −7x +12. The domain is all numbers except where x2 − 7x + 12 = 0, by factoring
(x −3)(x −4) = 0 we solve x = 3,4, that is the domain is all numbers except {3,4}.
7. h(x) =p
3x −5: The domain is all numbers for which 3x −5 ≥ 0 or x < 5, that is x ≥5
3, or in
interval notation [5
3,∞).
ä
5.14 Example Find the domain of the following functions
1. F = {(−2,4),(0,6),(2,8)}
2. f (x) = 7x2 +5−|3−2x|
3. f (x) =1
x −2
4. g (x) =3x −12
2x +5
5. h(x) =p
4+2x
6. h(x) =p
x2 +1
Solution:
1. F = {(−2,4),(0,6),(2,8)}. D = {−2,0,2}
2. f (x) = 7x2 +5−|3−2x|. D = (−∞,∞)
3. f (x) =1
x −2. D = {x|x 6= 2}
4. g (x) =3x −12
2x +5. D = {x|x 6= −
5
2}
5. h(x) =p
4+2x. D = {x|x ≥−2}
6. h(x) =p
x2 +1. D = (−∞,∞) ä
5.1.5 Graphical Representation of Functions
The graphical representation of a function is to visualize the correspondence between the input valuesand the output values based on the given rule of the function. This visualization is accomplished withthe help of the familiar Cartesian coordinate system. The approach is to plot enough points whosecoordinates satisfy the functional relation. We already saw in Figure 5.5 the graph of the functionF = {(−4,−2),(−3,−1),(2,4),(3,−3)} is represented by four points.
5 Definition The graph of a function y = f (x) defined on a domain D is the set of all points of the form(x, y) where x varies over the domain D. The definition will be illustrated by the following example.
5.15 Example Sketch the function f (x) =−2x +2 on the interval [0,3]
Solution: General steps to draw graphs by hand:
1. Rewrite the function in the familiar format y =−2x +3
2. Select several values of x and compute the corresponding y values. The values of x usu-
ally are whole numbers such as −5,−2,0,1,4, simple fractions such as1
2,
1
3,
1
4,
3
4or decimals
0.1,0.2,0.4,0.6. The choice of these inputs is up to the user, and the number of these pointsshould be sufficient to draw the main features of the graph and this is gained by experi-menting with graphing functions. The simpler the better.
147
Introduction to Functions
3. Create a table
x y=f(x)=-2x+2 (x,y)
0 2 (0,2)
1 0 (1,0)
2 -2 (2,-2)
3 -4 (3,-4)
4. Draw a system of coordinates
5. Plot the points
6. Connect the points
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
b
b
b
b
Figure 5.10: f =−2x +2
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
b
b
b
b
Figure 5.11: f =−2x +2
ä
5.16 Example Sketch the function f (x) = x2 −2x
Solution: General steps to draw graphs by hand:
1. Rewrite the function in the familiar format y = x2 −2x
2. Create a table
x -2 -1 0 1 2 3 4
y 8 3 0 -1 0 3 8
(x,y) (-2,8) (-1,3) (0,0) (1,-1) (2,0) (3,3) (4,8)
148
Chapter 5
3. Draw a system of coordinates
4. Plot the points
5. Connect the points
−3
−2
−1
0
1
2
3
4
5
6
7
8
−3 −2 −1 0 1 2 3 4 5 6 7 8
b
b
bb
b
b
b
Figure 5.12: The graph of points of (x,f(x))
1
2
3
4
5
6
7
−1
−2
−3
1 2 3 4 5 6 7−1−2−3
Figure 5.13: The graph of function f(x)ä
5.17 Example Sketch the function f (x) = |x| on the interval [−5,5]
Solution: Rewrite as y = |x|. Select several values of x and compute the corresponding y
values:
x -5 -2 -1 0 1 3 5
y 5 2 1 0 1 3 5
(x,y) (-2,5) (-2,2) (-1,1) (0,0) (1,1) (3,3) (5,5)
Plot the points and connect them.
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
b
bbbb
b
b
Figure 5.14: f = |x|
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
b
bb
bb
b
b
Figure 5.15: f = |x|ä
149
Introduction to Functions
Interpretations of Graphs
If the graph of a function is the only given representation. That is no formula or numerical values,then the graph can be used to determine the values, the domain and the range of the function. Wealso, can determine if a given graph represent a function or just a relation between variables.
1. Finding the values of a function from its graph We illustrate the techniques of estimating thevalues of the function from its graph by example.
5.18 Example The graph of function f is given in Figure (5.15) and the graph of function g is givenin Figure (5.16).
(a) Find f (1)
(b) Find the value of x for which f (x) =−2
(c) Is −1.5 in the domain of f ?
(d) Find g (1)
(e) Find the value of x for which g (x) = 1
(f) Is −3 in the domain of g ?
1
2
3
4
−1
−2
−3
−4
1 2 3 4−1−2−3−4
Figure 5.16
1
2
3
4
5
6
7
−1
−2
−3
−4
−5
1 2 3 4 5 6 7−1−2−3−4−5
Figure 5.17
Solution:
(a) To find the value f (1):
i. Locate 1 on the x-axis.
ii. Move vertically until you reach the graph.
iii. Move horizontally to the y-axis.
iv. Estimate the value f (1) = 2.
Note the point of intersection on the graph is (1,2).
(b) To find the value of x for which f (x) =−2:
150
Chapter 5
i. Locate −2 on the y-axis.
ii. Move horizontally until you reach the graph.
iii. Move vertically to the x-axis.
iv. Estimate the value x =−3. The graph point is (−3,−2)
(c) To check if the input x =−1.5 in the domain of f , If the vertical line passing through theinput point, intersects the graph, then the input is in the domain. That is x =−1.5 is inthe domain of f .
(d) We follow the previous procedure to estimate that g (1)=−3
(e) We estimate that x ≈ 2.25 for which g (x) = 1
(f) The input −3 is not in the domain of g , because there is no intersection with the verticalline passing through x =−3. ä
2. Finding the domain and range of a function from its graph We note that the domain is assumedto be on x-axis and the range on y-axis. It is possible to estimate the domain by combining allx values such that the vertical line passing through it intersects the graph this approach can besaid by considering the projection of the graph on x-axis. Similarly, the range can be estimatedby identifying the projection of the graph on y-axis.
5.19 Example The graph of function f is given in Figure (5.15) and the graph of function g is givenin Figure (5.16).
(a) Find the domain and range of the function f (x) from Figure (5.15)
(b) Find the domain and range of the function g (x) from Figure (5.16)
Solution:
(a) The graph of the function f (x) is extended in both direction. The projection on x-axiscovers the entire line. Thus the domain is the set of all real numbers. The range is alsothe set of all real numbers.
(b) The projection of the graph of g (x) can be seen to be the interval [−2,3] which representsthe domain. The range is the interval [−4,5]
ä
5.20 Example The graph of function f is given in Figure (5.17) and the graph of function g is givenin Figure (5.18).
(a) Find the domain and range of the function f (x) from Figure (5.17)
(b) Find the domain and range of the function g (x) from Figure (5.18)
1
2
3
4
5
−1
−2
−3
−4
−5
1 2 3 4 5−1−2−3−4−5
Figure 5.18
1
2
3
4
5
−1
−2
−3
−4
−5
1 2 3 4 5−1−2−3−4−5
Figure 5.19
151
Introduction to Functions
Solution:
(a) By considering the projection of the graph, we conclude that the domain of f(x) is [−4,∞)
and the range is [−2,∞)
(b) The domain of g (x) is the interval [−3,3] and the range is [0,3] ä
3. Does every graph represent a function?
Some graphs represent functions and some graphs do not. To test for functional relationship, wehave the so called the vertical line test: If a vertical line intersects the graph at more thanone point, then the graph does not define a function. For example, the graph of Figures 5.19and 5.20 do not represent a function because they fail the vertical line test.
Figure 5.20 Figure 5.21
152
Chapter 5
5.1.6 Homework: Introduction to Functions
Determine which of the relations are functions.
5.1.1 {(2,3),(4,5),(6,9), (2,11)}
5.1.2 {(1,0),(2,1),(3,0), (4,1)}
5.1.3 y = 3x +1
5.1.4 x2 + y 2 = 9
5.1.5 Given h(x) =1
3x −
2
3, find
1. h(2)
2. h(3)
3. h(1
3)
4. h(a)
5.1.6 Given f (x) =−3x +1, find
1. f (a)
2. f (a +3)
3. f (a +h)
5.1.7 Given f (x) = x2 −5x, find
1. f (a)
2. f (a −1)
3. f (a +h)
5.1.8 Given f (x) = x2 −6x +9, find
1. f (−a)
2. f (3)
3. f (a −3)
4. f (3x)
5. f (x2)
5.1.9 Given f (x) = x2 +5x −3, find
1. f (−2)+ f (5)
2. f (−2+5)
3. f (a)+3
4. f (x)− f (3)
5. 4f (x)
Determine the domain of the following functions
5.1.10 f (x) =p
x
5.1.11 f (x) =2
3x
5.1.12 f (x) = (x −3)2
5.1.13 f (x) =p
4x −9
5.1.14 f (x) = x3 −5x2 +7
5.1.15 f (x) = 2−p
4−2x
5.1.16 f (x) =6x
5x +2
5.1.17 f (x) =3x
(x +2)(3x −1)
5.1.18 F = {(1,2),(2,5),(3,10), (0, 1)}
5.1.19 F = {(1,−1),(2,1),(3,−1),(4,1), · · · }
5.1.20 f (x) =3x
x2 +3
153
Quadratic Functions: Parabolas
5.2 Quadratic Functions: ParabolasOBJECTIVES
• Quadratic functions and parabolas
• The vertex and axis of symmetry
• Finding the zeros
6 Definition A quadratic function f in one variable x is a second degree polynomial function of the form
f (x) = ax2 +bx +c ,
where a,b, and c are real numbers and a 6= 0.
For example, f (x) = x2 −2x −3 and f (x) = 4x2 −1.
The simplest quadratic function is f (x) = x2. The graph of this function in the x-y plane, can be studiedby plotting enough points y = x2 as demonstrated in the two tables and the corresponding figures:
x -3 -2 -1 0 1 2 3
y 9 4 1 0 1 4 9
(x,y) (-3,9 (-2,4) (-1,1) (0,0) (1,1) (2,4) (3,9)
and the table with additional points
x -3 -5
2-2 -
3
2-1 -
1
20
1
21
3
22
5
23
y 925
44
9
41
1
40
1
41
9
44
25
49
−2
−1
0
1
2
3
4
5
6
7
8
9
−4 −3 −2 −1 0 1 2 3 4
b
b
b
b
b
b
b
Figure 5.22: 7 points of (x, y = x2)
−2
−1
0
1
2
3
4
5
6
7
8
9
−4 −3 −2 −1 0 1 2 3 4
b
b
b
b
b
b
b
b
b
b
b
b
b
Figure 5.23: 13 points of (x, y = x2)
154
Chapter 5
1
2
3
4
5
6
7
8
−1
−2
1 2 3−1−2−3−4
Figure 5.24: The graph of quadratic function f (x) = x2
The graph of y = x2 is called parabola. It has a U shaped graph. The parabola has a line of symmetrythat passes through its vertex. The vertex of a vertical parabola is its lowest point or highest point andits axis of symmetry is a vertical line passing through the vertex.
☞ Parabolas have very interesting properties and can be observed in many applications
1. Any point on the parabolas is equidistant from a fixed point (focus) and a straight line (directrix).
2. The surface of a liquid in a rotating container is a paraboloid of revolution. This can be observed bystirring a glass of water with a spoon.
3. A tossed stone travels along a parabola.
4. Parabolic mirrors are used in telescope construction.
0 2 4 6 8 10 12 14 160
1
2
3
4
5
6
7
8
Distance
Hei
ght
Balls with different initial velocities
Figure 5.25: Parabolas
Every quadratic function in general form f (x) = ax2 +bx +c can be rewritten in the form of
f (x) = a(x −h)2 +k ,
155
Quadratic Functions: Parabolas
where h and k will be specified and derived in the following section. For example, the function f (x) =3x2 −6x +5 can be rewritten as f (x) = 3(x −1)2 +2, so that h = 1,k = 2. Thus, the study of the graph ofany quadratic function can be reduced to the investigation of the effects of the parameters a,h,k on thegraph of the function.
5.2.1 Graphs of f (x) = ax2
How is the graph of the parabola y = ax2 changes as a varies?
5.21 Example 1. Compare the table values of y = x2, y = 2x2, y = 3x2
2. Graph y = x2, y = 2x2, y = 3x2 on the same system of coordinates
3. Graph y = x2, y =1
4x2, y =
1
16x2 on the same system of coordinates
4. Graph y = x2, y =−x2, y =−4x2, y =1
4x2, y =
1
16x2 on the same system of coordinates
5. Discuss the effect of changing a on their graphs.
Solution:
x 0 ∓ 1 ∓ 2 ∓ 3 ∓ 4
y = x2 0 1 4 9 16
y = 2x2 0 2 8 18 32
y = 3x2 0 3 12 27 48
1. The output in the table values are multiplied by a factor a as compared to y = f (x).
2. The graphs of y = 2x2, y = 3x2 are vertically stretched by a factor of a
3. The graph y =1
2x2, y =
1
16x2 are vertically compressed by a factor of a
4. The graphs of y = −x2, y = −4x2, y = −1
4x2 are reflected with respect to x-axis with vertical
stretching and compression.
1
2
3
4
5
6
7
8
9
10
11
−1 1 2 3−1−2−3−4
Figure 5.26
1
2
3
4
5
6
7
8
9
10
11
−1 1 2 3−1−2−3−4
Figure 5.27ä
156
Chapter 5
7 Definition The the vertex of the parabolas y = ax2 is the origin (0,0), and the axis of symmetry isthe vertical line passing through the vertex: x = 0.
Main conclusions
The effects of a on the graph of y = ax2 are
1. The parabola opens upward when a > 0
2. The parabola opens downward when a < 0
3. The parabola stretches vertically when |a| > 1 and compresses vertically when |a| < 1
4. The parabola has a vertex (0,0) and axis of symmetry x = 0
1
2
3
4
5
−1
−2
−3
−4
1 2 3−1−2−3−4
Figure 5.28: y = x2, y =−x2, y =−4x2, y =−1/4x2
5.2.2 Graphs of f (x) = ax2 +k
What are the effects of adding or subtracting k units to the output of the function f = ax2?
5.22 Example Graph the parabolas and compare their graphs
1. y = x2,
2. y = x2 +3,
3. y = x2 −4
Figure 5.29: y = x2, x2 +3, x2 −4
157
Quadratic Functions: Parabolas
The graph of y = x2 +3 is a vertical translation (or vertical shift) of the graph y = x2. Its vertex is (0,3).The graph of y = x2 −4 is shifted downward compared to the graph of y = x2. Its vertex is (0,−4).
Main conclusions
Suppose the graph of y = ax2 is given. The graph of y = ax2 +k is a vertical shift, such that
1. If k > 0, then the shift is k units upward.
2. If k < 0, then the shift is k units downward.
5.2.3 Graphs of f (x) = a(x −h)2
What are the effects of adding or subtracting h units to the input of the function f = ax2?
5.23 Example Graph the parabolas and compare their graphs
1. y = x2
2. y = (x −3)2
3. y = (x +4)2
y = x2
Figure 5.30: y = (x +4)2, y = x2, y = (x −3)2
The graph of y = (x +4)2 is a horizontal translation (or horizontal shift) of the graph y = x2, the shiftis to the left by 4 units. Its vertex is (−4,0). The graph of y = (x −3)2 is shifted to the right by 3 units,compared to the graph of y = x2. Its vertex is (3,0).
Main conclusions
Suppose the graph of y = ax2 is given. The graph of y = a(x −h)2 is a horizontal shift, such that
1. If h > 0, then the shift is h units to the left.
2. If h < 0, then the shift is h units to the right.
3. The graph of y = a(x −h)2 has vertex (h,0) and axis of symmetry x = h.
5.2.4 Graphs of f (x) = a(x −h)2 +k
The graph of the quadratic function f (x) = a(x −h)2 +k is a parabola with vertex (h,k) and axis of sym-metry x = h. The parabola opens up when the parameter a > 0 and opens down when a < 0.
5.24 Example Graph y = 2(x −3)2 −4 by identifying the vertex and axis of symmetry.
Solution: To find the vertex we let −h =−3,h = 3 and k =−4 so the vertex is (3,−4) and theaxis of symmetry is x = 3. The parabola opens up because a = 2 > 0. ä
158
Chapter 5
5.25 Example Determine the vertex, the axis of symmetry and the direction of y =−(x +1)2 −4
Solution: The vertex is (−1,−4), The axis of symmetry is x = −1, and the parabola opensdown (a =−1< 0). ä
5.26 Example Determine the vertex and the axis of symmetry of y = 2(x −5)2 +3
Solution: The vertex is (5,3), the axis of symmetry is x = 5, and the parabola opens up(a = 2 > 0). ä
159
Quadratic Functions: Parabolas
5.2.5 Homework: Quadratic Functions
Graph the functions by setting up a table of valuesfor x and y
5.2.1 y = 2x2 −4
5.2.2 y =−x2 +4x +4
Graph the reference function f (x) = x2, then graphthe following functions by identifying theirtransformations
5.2.3 f (x) = x2 +2
5.2.4 f (x) =−x2 +2
5.2.5 f (x) = (x −3)2 −4
5.2.6 f (x) =−(x +5)2 +6
5.2.7 f (x) =−(x −3)2 −1
Graph the reference function f (x) = x2, then graphthe following functions by identifying theirtransformations
5.2.8 f (x) =1
2x2
5.2.9 f (x) =−4x2
5.2.10 f (x) =−1
4x2
5.2.11 f (x) = 5x2 −8
Identify the vertex and line of symmetry
5.2.12 f (x) =−3x2 −5
5.2.13 f (x) = 5(x +4)2 −2
5.2.14 f (x) = 2(x −3)2 +9
5.2.15 f (x) = (x −5)2 +5
6
5.2.16 f (x) =−(x +1)2 +4
5.2.17 f (x) =−(x −6)2 −7
5.2.18 f (x) = 3(x +2)2
160
Chapter 5
5.3 Quadratic Functions: General formIt is important to recognize that the graph of ay quadratic function of the form f (x) = ax2 +bx + c is avertical parabola.The characteristics of a parabola are
1. The vertex point
2. The direction of its opening: Up or down
3. The axis of symmetry
4. The x-intercept points which are also called zeros or roots of the quadratic function
5. The y-intercept point
These values will be easily identified once we express the general form in standard form that is to writef (x) = a(x −h)2 +k. This will be accomplished by completing the square method. We will illustratethese properties by the following examples:
5.27 Example Find the vertex and axis of symmetry of the quadratic function f (x) = x2 +2x
Solution: We use completing the square method
f (x) = x2 +2x
= x2 +2x +1−1 Add a number to complete the square and subtract it
= (x +1)2 −1
Compare with f (x) = a(x −h)2 +k to conclude that a = 1, h =−1, k =−1. The vertex is (−1,−1) andthe axis of symmetry is x =−1. ä
5.28 Example Find the vertex and axis of symmetry of the quadratic function f (x) = x2 −3x
Solution: We use completing the square method
f (x) = x2 −3x
= x2 −3x +9
4−
9
4Add a number to complete the square and subtract it
= (x −3
2)
2 −9
4
The vertex is (3/2,−9/4) and the axis of symmetry is x =−3/2. ä
5.29 Example Find the vertex and axis of symmetry of the quadratic function f (x) = x2 −6x +5
Solution: We use completing the square method
f (x) = x2 −6x +5
= x2 −6x +9−9+5 Add a number to complete the square and subtract it
= (x −3)2 −4
Compare with f (x) = a(x−h)2+k to conclude that a = 1, h = 3, k =−4. The vertex is (3,−4) and theaxis of symmetry is x = 3. ä
161
Quadratic Functions: General form
5.30 Example Find the vertex and axis of symmetry of the quadratic function f (x) = 2x2 +4x −6
Solution: We use completing the square method
f (x) = 2x2 +4x −6
= 2(x2 +2x)−6 Factor the coefficient of x2
= 2(x2 +2x +1−1)−6 Add a number to complete the square and subtract it
= 2(x2 +2x +1)−2−6
f (x) = 2(x +1)2 −8
Compare with f (x) = a(x −h)2 +k to conclude that a = 2, h =−1, k =−8. The vertex is (−1,−8) andthe axis of symmetry is x =−1. ä
The general formula for finding the vertex of the quadratic function f (x) = ax2 +bx +c is given by
(−b
2a, f (−
b
2a))
For example, the parameters of the parabola f (x) = 2x2+4x−6 are a = 2,b = 4,c =−6, thus the x-coordinate
of the vertex is x =−b
2a=−
4
4=−1, the y-coordinate is computed as follows: y = f (−1) = 2(−1)2 +4(−1)−6 =
2−4−6 =−8, that is the vertex is (−1,−8) as derived in the previous example. To see how this formula isobtained we use the complete of square to the general form:
f (x) = ax2 +bx +c
= a(x2 +b
ax)+c
= a(x2 +b
ax +
b2
4a2−
b2
4a2)+c
= a(x +b
2a)
2 −b2
4a+
4c a
4a
f (x) = a
(
x +b
2a
)2
+4ac −b2
4a
The last equation is the equation of quadratic function in standard form. By comparison, we have
h = −b
2aand k =
4ac −b2
4a. The value of k is hard to memorize, instead we use an equivalent but an
easier substitution k = f (h) = f (−b
2a) =
4ac −b2
4a.
5.31 Example Find the vertex of the following parabolas:
1. y =−3x2 −12x −1
2. f (x) = 4x2 −6x +7
3. y = 2x2 +5x
Solution: Use the formulas: h =−b
2aand k = f (h) = f (−
b
2a)
1. y =−3x2 −12x −1: a =−3,b =−12,c =−1, h =−−12
−6=−2, k = y(−2) =−3(−2)2 −12(−2)−1= 11
2. f (x) = 4x2 −6x +7: a = 4,b =−6,c = 7, h =−−6
8=
3
4, k = f (
3
4) = 4(
3
4)2 −6(
3
4)+7 =
9
4−
18
4+7
4
4=
19
4
3. y = 5x2 +15x: a = 2,b = 15,c = 0, h =−15
10=−
3
2, k = y() = 5(−
3
2)2 +15(−
3
2) =
45
4−
45
2=−
45
4 ä
162
Chapter 5
☞ The y-coordinate of the vertex represent the minimum of upward parabola, or the maximum ofdownward parabola.
5.32 Example Find the length and width of a rectangle, whose sum is 18, so the area is a maximum.
Solution: Let L = x be the length then the width will be W = 18 − x. The area is theirproduct and can be considered as a function of x: A = x(18− x) = −x2 + 18x. The parameter
a =−1 indicates the parabola opens down, and its vertex is the maximal value. x =−18
−2= 9 and
A =−(9)2 +18(9)= 81 ä
5.3.1 Finding the x- and y- intercepts
To find the intercepts of f (x) = ax2 +bx +c, we proceed as follows
1. The y-intercept is easily computed y = f (0)
2. The x-intercepts or (zeros) are the solution of the equation f (x) = 0
5.33 Example Find the intercepts of y = x2 +5x
Solution:
1. The y-intercept: y = f (0) = 02 +5(0) = 0, so the y-intercept is the point (0,0)
2. The x-intercepts: Set x2+5x = 0, solve by factoring x(x+5) = 0, hence x = 0,−5, so the interceptpoints are (0,0) and (−5,0) ä
5.34 Example Find the intercepts of y = x2 −6x +8
Solution:
1. The y-intercept: y = f (0) = 02 −6(0)+8 = 0, so the y-intercept is the point (0,8)
2. The x-intercepts: Set x2 −6x +8 = 0, solve by factoring (x −4)(x −2) = 0, hence x = 2,4, so theintercept points are (2,0) and (4,0) ä
5.35 Example Find the zeros of the following quadratic functions
1. y = x2 + x −2
2. y =−x2 +4x −5
Solution: To find the zeros, set y = 0 and solve the equations for x
1. x2 + x −2 = 0. Use quadratic formula with a = 1,b = 1,c =−1: x =−1∓
p1+4
2=
−1∓p
5
2
2. −x2 + 4x − 5 = 0, or x2 − 4x + 5 = 0. Use quadratic formula with a = 1,b = −4,c = 5: x =4∓
p16+25
2=
4∓p
41
2 ä
5.36 Example Find the vertex and the zeros of the parabola y = 4x2 −3x +5
Solution: Identify a = 4,b =−3,c = 5 and use the vertex formula and the quadratic formula
1. The vertex: x =−−3
8=
3
8, y = f (x) = 4(
3
8)2 −3
3
8+5 =
53
16, (
3
8,
53
16)
2. 4x2−3x+5, x =3∓
p9−80
8=
3∓p−71
8, the zeros are complex numbers, so the parabola does
not intersect x-axis. ä
163
Quadratic Functions: General form
5.3.2 Homework: Quadratic Functions: General Form
Write each quadratic function in the formy = a(x −h)2 +k and sketch its graph.
5.3.1 y = x2 −4x
5.3.2 y = x2 +2x
5.3.3 y = x2 −6x +9
5.3.4 y = x2 +14x +50
5.3.5 f (x) =−x2 −5x −2
5.3.6 f (x) = 2x2 +8x −5
5.3.7 f (x) =−3x2 −30x −70
Find the vertex, axis of symmetry, y-intercept,x-intercepts, and direction of the parabola, thensketch the graph.
5.3.8 y = x2 −4
5.3.9 y =−x2 −1
5.3.10 y = 9−x2
5.3.11 f (x) = x2 −2x +3
5.3.12 f (x) = x2 +10x +25
5.3.13 f (x) = x2 −8x +12
5.3.14 f (x) = 3x2 −6x +3
5.3.15 f (x) =−x2 +4x +1
5.3.16 f (x) =−x2 −6
5.3.17 f (x) = x2 +5x
5.3.18 y = (x −4)2 −1
164
Chapter 5
5.4 Function Transformations and their Graphs8 Definition The graph of a function y = f (x) is the set of all points of the form (x, f (x)) plotted on theplane and x is any value in the domain of the function.
5.4.1 Graphs of basic functions
In this section, we will list the basic functions that are encountered in algebra and its applications.In this section, using graphing calculators is encouraged as an appropriate tool to investigate theproperties of functions.
1. The power functions are the reference (base) functions for polynomials: f (x) = x0, x, x2, x3, x4, · · ·
(a) The constant function has the form f (x) = a, where a is a real number. The graph of thisfunction is a horizontal line. For example, y = 3 and y =−4 are two horizontal lines.
(b) The linear function is represented by the function f (x) = x. Its graph is a straight line. Thegeneral form of linear functions is f (x) = mx +b, for example, y = 2x +5 and y = −3x +1. Theproperties of linear functions are studied in the chapter on straight lines.
(c) The quadratic function is represented by the function f (x) = x2. We already studied theparabola and its graph.
(d) The cubic function is represented by the function f (x) = x3. Its graph has two branchesextending in opposite direction to ∞
Note that the similarity of graphs of even and odd powers. In the following sections, we will restrictour examples and illustrations to polynomials of degree three or less.
Figure 5.31: y = x2 Figure 5.32: y = x3 Figure 5.33: y = x4 Figure 5.34: y = x5
2. The root functions are the reference functions for radicals: f (x) =p
x , 3p
x , 4p
x , · · ·
(a) The square root function is represented by the function f (x) =p
x. Its graph is a halfhorizontal parabola and its domain is restricted to positive numbers.
(b) The cubic root function is represented by the function f (x) = 3p
x. Its graph is a rotatedgraph of the cube function and its domain is the entire number line.
165
Function Transformations and their Graphs
3. The absolute value function is represented by the function f (x) = |x|. It has a V shaped graph.
4. The rational function is represented by the function f (x) =1
x. This function is not defined at
x = 0 and it has two branches and it is called a hyperbola.
Figure 5.35: y =p
x Figure 5.36: y = 3p
x Figure 5.37: y = |x| Figure 5.38: y =1
x
5.4.2 Transformations of Functions and their Graphs
In this section we will study the following questions:
1. What are the effects of shifting the output by a number on the graph of the function y = f (x)?That is how the graph of y = f (x)∓k is compared to the graph of y = f (x), where k is a real number?
2. What are the effects of shifting the input on the graph of the function y = f (x)?That is how the graph of y = f (x∓h) is compared to the graph of y = f (x) , where h is a real number?
3. What are the effects of scaling the output on the graph of the function y = f (x)?That is how the graph of y = a f (x) is compared to the graph of y = f (x) , where a is a real number?
☞ The words: shift, transformation, translation have the same definition in this text.
In the following discussion, we will assume that the function y = f (x) is one of the basic functionsand the numbers k and h are real positive numbers.
Vertical Shifts (Vertical Translations)
The graph of a function y = f (x) is a curve in the xy-plane. The coordinates of each point are the pairsof numbers (x, f (x). The graph of the function y = f (x)+k is also a curve. The coordinates of each pointare the pairs (x, f (x)+k). Comparing the points (x, f (x) and (x, f (x)+k) shows a vertical shift upward bya k units. This leads to conclude that the graph of y = f (x)+k is a mere vertical shift of the graph ofy = f (x) by k units upward. Similarly the graph of y = f (x)−k is a vertical shift of the graph of y = f (x)
by k units downward.
5.37 Example For the function y = |x|, sketch the graph of the following functions and compare them tothe reference function.
1. y = |x|
2. y = |x|+2
3. y = |x|−4
5.38 Example Graph the function y = f (x) = x3. Sketch and write the formula of the graph that it isobtained from y = x3 by the following transformations:
166
Chapter 5
1. Shift the graph up by 2.5 units
2. Shift the graph down by 2 units
3. Shift the graph up by 1 units
y = |x|−4
y = |x|+2
Figure 5.39: y = |x|, y = |x|+2, y = |x|−4
y = x3 +2.5
y = x3 −2
Figure 5.40: y = x3, y = x3 +2.5, y = x3 +1, y = x3 −2
Horizontal Shifts (Horizontal Translations)
Comparing the graphs of y = f (x) and y = f (x +h) shows that the graph of y = f (x +h) is the same as they = f (x) but horizontally shifted to the left by h units. Similarly, the graph of y = f (x −h) is the same asgraph y = f (x) but horizontally shifted to the right by h units.
☞ Note that the shift in the input moves in the opposite sign:
1. The shift of f (x+h) is h units to the left. For example, the functions y = (x−2)2 , y = (x−3)3 , y = |x−2|, y =px −2 are shifted two units to the right.
2. The shift of f (x−h) is h units to the right. For example, the functions y = (x−2)2 , y = (x−3)3 , y = |x−2|, y =px −2 are shifted two units to the right.
5.39 Example Plot the function y =p
x, then use the shifting techniques to sketch the graphs of
1. y =p
x −4, y =p
x +4
2. y =p
x +5, y =p
x −5
y =p
x +5
y =p
x +5
y =p
x −4
y =p
x −4
Figure 5.41: y =p
x , y =p
x +5, y =p
x +5, y =p
x −4, y =p
x −4
167
Function Transformations and their Graphs
Stretching and Compression
9 Definition For a well defined function y = f (x), we have
1. The graph of y = a f (x) is stretched away from x-axis if a > 1
2. The graph of y = a f (x) is compressed towards x-axis if a < 1
5.40 Example Graph the function y = |x| and sketch the graph of the scaled functions
1. y = 2|x|
2. y =1
4|x|
5.41 Example Graph the function y = f (x) =p
x and sketch the graphs of the scaled functions
1. y = 4p
x
2. y =1
2
px
y =1
4|x|
y = 2|x|
Figure 5.42: y = |x|, y = 2|x|, y =1
4|x|
y = 4p
x
y =1
2
px
Figure 5.43: y =p
x, y = 4p
x, y =1
2
px
Reflections
The graph of a function y = f (x) can be reflected vertically and horizontally. We start with the reflectionof a point P(x, y):
1. The point V (x,−y) is called the vertical reflection or simply the reflection of P across x-axis. Forexample The points P(2,3) and V (2,−3) are vertical reflections.
2. The point H(−x, y) is called the horizontal reflection or simply the reflection of P across x-axis.For example The points P(2,3) and H(−2,3) are horizontal reflections.
b P
b V
b H
Figure 5.44: The point P and its reflections
The reflections of a function follow the same constructions.
168
Chapter 5
1. The graph of y =− f (x) is the vertical reflection (with respect to x-axis) of the graph of y = f (x).
2. The graph of y = f (−x) is the horizontal reflection (with respect to y-axis) of the graph of y = f (x).
5.42 Example Graph the function y = f (x) =p
x and sketch the graphs of the following functions
1. y =−p
x
2. y =p−x
y =p
x
y =−p
x
y =p−x
Figure 5.45: y =p
x , y =−p
x , y =p−x
Combining shifts and reflections
For a given graph of a function y = f (x), the graph of y = f (x −h)+k is
1. a vertical shift of k units and
2. a horizontal shift of h units of the graph of y = f (x).
The graph of y =− f (x −h)+k is a vertical shift of k units, a horizontal shift of h units, and a reflection ofthe graph of y = f (x).
5.43 Example Assume the graph of y = f (x) is given. Identify the translations and reflections that canbe applied to y = f (x) to graph the following functions:
1. y = f (x)+4
2. y = f (x −4)
3. y = f (x)−6
4. y = f (x +6)
5. y =− f (x)+5
6. y =− f (x −5)−7
7. y =− f (x +1)−3
8. y = 3 f (x −2)+8
Solution:
1. y = f (x)+4: 4 units up
2. y = f (x −4): 4 units to the right
3. y = f (x)−6: 6 units down
4. y = f (x +6): 6 units to the left
5. y =− f (x)+5: 5 units up and a vertical reflection
169
Function Transformations and their Graphs
6. y =− f (x −5)−7: 7 units down, 5 units to the right, and a vertical reflection
7. y =− f (x +1)−3: 3 units down, 1 units to the left, and a vertical reflection
8. y = 3 f (x −2)+8: 8 units up, 2 units to the right, and a vertical stretching ä
5.44 Example The graphs of y = f (x) and y = g (x) are given in the figures. Find an expression for thefunction g (x).
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
f = x2
g
Figure 5.46: a
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
f = x2
g
Figure 5.47: b
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
f = x2
g
Figure 5.48: c
Solution: The formulas of the function g are:
1. Figure a: g (x) = (x −2)2 +1
2. Figure b: g (x) =−(x +3)2 +1
3. Figure c: g (x) = (x −3)2 −2 ä
5.45 Example The graphs of y = f (x) and y = g (x) are given in the figures. Find an expression for thefunction g (x).
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
f = x3
g
Figure 5.49: a
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
f =p
x
g
Figure 5.50: b
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
f = |x|
g
Figure 5.51: c
Solution: The formulas of the function g are:
1. Figure a: g (x) = x3 +2
2. Figure b: g (x) =−p
x +3−2
3. Figure c: g (x) =−|x −2|+5 ä
170
Chapter 5
5.46 Example The graph of the function y = f (x) is given. Match the following functions to the graphsin the figure.
1. y = f (x −2)−3
2. y =− f (x)+4
3. y = f (x +4)+2
4. y =− f (x +2.5)−2.5
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
f
(a)
(b)
(c) (d )
Figure 5.52: Functions Matching
Solution:
1. y = f (x −2)−3: (d)
2. y =− f (x)+4: (a)
3. y = f (x +4)+2: (c)
4. y =− f (x +2.5)−2.5: (b) ä
171
Function Transformations and their Graphs
5.4.3 Homework: Function Transformations
Identify the function y = f (x) and list thetransformations in each function.
5.4.1 f (x) = (x −3)2 +5
5.4.2 f (x) =−(x +4)3
5.4.3 f (x) = |x −5|−3
5.4.4 f (x) =−p
x −2−4
Write the equation of the graph after the requestedtransformations.
5.4.5 The graph of y = |x| is shifted two units down-ward.
5.4.6 The graph of y = |x| is shifted five units to the leftand one unit upward.
5.4.7 The graph of y = x3 is shifted three units to theright and three unit upward.
5.4.8 The graph of y = x3 is shifted four units down-ward and is reflected with respect x-axis.
5.4.9 The graph of y =p
x is shifted two units upward,seven units to the right and is reflected with respect x-axis.
Use transformations of the function y = x3 to graphthe following functions.
5.4.10 f (x) = (x +6)3 −3
5.4.11 f (x) =−x3 +2
5.4.12 f (x) =−(x +1)3 −3
5.4.13 f (x) =−0.5x3
5.4.14 f (x) =−2x3 +8
Use transformations of the function y = |x| to graphthe following functions.
5.4.15 f (x) = |x −2|
5.4.16 f (x) = |x +3|+4
5.4.17 f (x) =−|x|+5
5.4.18 f (x) = |x −6|+2
5.4.19 f (x) = 3|x|
5.4.20 f (x) =−1
4|x|−1
Use transformations of the function y =p
x to graphthe following functions.
5.4.21 f (x) =−p
x
5.4.22 f (x) =p
4+x
5.4.23 f (x) =p
x −6+3
5.4.24 f (x) =−p
x +2+2
5.4.25 f (x) =p
x −3−4
5.4.26 f (x) = 4p
x
172
Chapter 5
5.5 The Algebra of Functions and CompositionNumbers can be added, subtracted, multiplied and divided to produce new numbers. Functions canalso be combined using arithmetic operations to produce new functions. In addition, there is a newway of combining functions called composition will be introduced in this section. Questions related tothe domains of the resulting functions will be discussed.
5.5.1 The Algebra of Functions
In this section, we introduce some of the common techniques of construction functions and combiningthem. The standard algebraic operations are well defined for functions as long as there are commonelements between their domains. These operations are the sums, differences, products, quotients androots of functions.
For any two functions f (x) defined on a domain D f and g (x) defined on a domain Dg , such that theintersection of these domains is not empty; that is D f ∩Dg 6=∅, we define the four arithmetic operationson the functions f , g for any common point in their domains, that is for any x ∈ D f ∩Dg :
• Sum of functions: ( f + g )(x) = f (x)+ g (x)
• Difference of functions: ( f − g )(x) = f (x)− g (x)
• Product of functions: ( f ·g )(x) = f (x) ·g (x)
• Quotient of functions:
(f
g
)
(x) =f (x)
g (x)given that g (x) 6= 0
5.47 Example Let f (x) = 2x +4 and g (x) = x2 −2x. Find the following functions and their domains.
1. f + g
2. f − g
3. f g
4.f
g
Solution:
1. ( f + g )(x) = f (x)+ g (x) = 2x +4+ x2 −2x = x2 +4. Domain is the set of all real numbers R
2. ( f − g )(x) = f (x)− g (x) = 2x +4− (x2 −2x) =−x2 +4x +4. Domain is R
3. ( f g )(x) = f (x)g (x) = (2x +4)(x2 −2x) = 2x3 −8x. Domain is R
4.
(f
g
)
(x) =f (x)
g (x)=
2x +4
x(x −2). Domain is x 6= 0,2
ä
5.48 Example Let f (x) = x −5 and g (x) = x2 + x. Find the following values.
1. ( f + g )(3)
2. (2 f − g )(−1)
3. ( f ·g )(2)
4. (f
g)(a)
Solution:
1. ( f + g )(3)= f (3)+ g (3)= 3−5+32 +3 = 10
2. (2 f − g )(−1)= 2 f (−1)− g (−1)= 2(−1−5)− ((−1)2−1) =−12
173
The Algebra of Functions and Composition
3. ( f ·g )(2) = f (2) ·g (2)= (2−5)(22 +2) =−18
4. (f
g)(a) =
f (a)
g (a)=
a −5
a2 +a ä
5.49 Example Let f (x) = x2 and g (x)=2
x. Find the following functions and their domains.
1. f + g
2. f − g
3. f g
4.f
g
Solution:
1. ( f + g )(x) = f (x)+ g (x) = x2 +2
x. Domain {x 6= 0}
2. ( f − g )(x) = f (x)− g (x) = x2 −2
x. Domain {x 6= 0}
3. ( f g )(x) = f (x)g (x) = (x2)(2
x) = 2x. Domain {x 6= 0}. This is because ( f g )(0) is undefined.
4.
(f
g
)
(x) =f (x)
g (x)=
x2
2
x
=x3
2. Domain {x 6= 0}
ä
5.5.2 Difference Quotients
10 Definition Ratios of the formf (a +h)− f (a)
hare called difference quotients. These quotients are very
important in calculus, since they represent the average change of the function f between the pointsa +h and a and they lead to the definition of derivative.
5.50 Example Find the difference quotientf (a +h)− f (a)
hfor f (x) = 3x +5
Solution: We divide the work into the following steps
1. f (a) = 3a +5
2. f (a +h) = 3(a +h)+5 = 3a +3h +5
3. f (a +h)− f (a) = 3(a +h)+5 = 3a +3h +5−3a −5 = 3h
4.f (a +h)− f (a)
h=
3h
h= 3 ä
5.51 Example Find the difference quotientf (a +h)− f (a)
hfor f (x) =−7x −2
Solution: We follow the steps
1. f (a) =−7a −2
2. f (a +h) =−7(a +h)−2 =−7a −7h −2
3. f (a +h)− f (a) =−7a −7h −2− (−7a −2)=−7a −7h −2+7a +2 =−7h
4.f (a +h)− f (a)
h=
−7h
h=−7 ä
5.52 Example Find the difference quotientf (a +h)− f (a)
hfor f (x) = x2
174
Chapter 5
Solution:
1. f (a) = a2
2. f (a +h) = (a +h)2 = a2 +2ah +h2
3. f (a +h)− f (a) = a2 +2ah +h2 −a2 = 2ah +h2 = (2a +h)h
4.f (a +h)− f (a)
h=
(2a +h)h
h= 2a +h
ä
5.53 Example Findf (a +h)− f (a)
hfor f (x) =
1
x
Solution:
1. f (a) =1
a
2. f (a +h) =1
a +h
3. f (a +h)− f (a) =1
a +h−
1
a=
a − (a +h)
a(a +h)=
−h
a(a +h)
4.f (a +h)− f (a)
h=
−h
a(a +h)
h=
−h
a(a +h)×
1
h=−
1
a(a +h) ä
5.54 Example Find the difference quotientf (x +h)− f (x)
hfor f (x) = 3+6x
Solution: We follow the steps
1. f (x +h) = 3+6(x +h) = 3+6x +6h
2. f (x +h)− f (x) = 3+6x +6h − (3+6x) = 3+6x +6h −3−6x = 6h
3.f (x +h)− f (x)
h=
6h
h= 6
ä
5.5.3 Composition of Functions
The composition of functions is a central idea in higher mathematics. This concept allows us to simplifythe calculations on complicated expression with patterns and allows us to perform the operations insequential process. First, we will introduce several illustrations and examples of the advantage of thisconcept, then we present the formal definition and finally we present additional examples and problemsof composition of functions.
5.55 Example The function
f (x) =x2 +1
1+ (x2 +1)3
has the repeated pattern x2 +1, thus by introducing a new function g (x) = x2 +1 and substituting it inthe function f , we obtain a new function which is simpler to work with:
f (g )=g
1+ g 3
5.56 Example The BP Gulf oil spill is the worst environmental disaster in US history. Assume the oilspill moves in circular motion on the water surface with a constant speed of 0.8 meters/min. How largeis the oil spill in one hour, in one day?
175
The Algebra of Functions and Composition
Solution: The area of the circular spill is given by the formula A =πr 2 and the spread ofthe oil spill is given by the distance formula r = 0.8t for t in minutes. The distance is the radiusof the circle. In this problem, we have two functions A = A(r ) depending on radius, and r = r (t)
depending on time. Thus, after one hour we have r = 1.2×60 = 48 meters, and A =π(48)2 ≈ 7235
square meters. After one hour, r = 0.8×60×60÷1000= 2.88 km with spill area A =π(10.8)2 ≈ 26
square kilometres. The radius of spill after a day, r = 69 km with spill area A =π(692) ≈ 14.900
square kilometres. ä
5.57 Example A stone is dropped into a calm pond, causing ripples in the form of concentric circles. Afigure of circular illusion of moving circles is given below. The radius (in feet) of the outer ripple (circle)is given by r (t) = 0.6t, where t is time in seconds after the stone strikes the water. The area of the circleis given by the function A(r ) =πr 2.
1. Find the area of the circle after 5 seconds.
2. Find and interpret the area as a function of t: A(r (t)).
Solution:
1. At 5 seconds the radius is r =).6 = 3 meters, and the area of the circle with radius 3 isA =π32 ≈ 28.3 square meters.
2. To compute the area as a function of time is to make the substitution:
A(t) = A(r (t)) =πr 2(t) =π(0.6t)
2 = 0.36πt 2
ä
Figure 5.53: Circular Illusion
5.58 Example Assumef (x) = 3x −2 and g (x) = 5x +1
Compute the following values by substitution:
1. f (2)
2. f (a)
3. f (4a)
4. g (1)
176
Chapter 5
5. f (g )
6. f (g (1))
7. g ( f )
Solution:
1. f (2) = 3(2)−2 = 6−2 = 4
2. f (a) = 3a −2
3. f (4a +2) = 3(4a +2)−2= 12a +6−2= 12a +4
4. g (1)= 5(1)+1= 6
5. f (g ) = 3g −2
6. f (g (1))= 3g (1)−2= 3×6−2 = 18−2= 16
7. g ( f ) = 5 f +1 ä
The substitution (or feeding) of a function into another function is called composition of functions.
11 Definition Assume f and g are two functions such that the domain of f contains the range of g . Thecomposition of f and g is a new function denoted by f ◦g , and it is defined by
f ◦g = ( f ◦g )(x) = f (g (x))
The function f ◦g is read
1. f circle g or
2. f of g or
3. f composed with g .
5.59 Example Assume f (x) = x2 and g (x)= 3x +2
1. evaluate g (−3)
2. evaluate ( f ◦g )(−3)
3. evaluate f (−2)
4. evaluate (g ◦ f )(−2)
5. Find a formula for the composition f ◦g
6. Find a formula for the composition g ◦ f
Solution:
1. g (−3)= 3(−3)+2=−9+2 =−7
2. ( f ◦g )(−3) = f (g (−3))= f (−7) = (−7)2 = 49
3. f (−2) = (−2)2 = 4
4. (g ◦ f )(−2) = g ( f (−2))= g (4) = 3(4)+2= 12+2 = 14
5. f ◦g = ( f ◦g )(x) = f (g (x)) = f (3x +2) = (3x +2)2 = 9x2 +12x +4
6. g ◦ f = (g ◦ f )(x) = g ( f (x)) = g (x2) = 3x2 +2 ä
5.60 Example Assume f (x) =p
x and g (x) = x2 −9. Find formulas for f ◦ g and g ◦ f . Use these formulasto evaluate ( f ◦g )(5) and (g ◦ f )(100).
Solution:
177
The Algebra of Functions and Composition
1. ( f ◦g )(x) = f (g (x)) = f (x2 −9) =p
x2 −9
2. (g ◦ f )(x) = g ( f (x)) = g (p
x) = (p
x)2 −9 = x −9
3. ( f ◦g )(5) = f (g (5))=p
52 −9 =p
25−9= 4
4. (g ◦ f )(100)= g ( f (100))= 100−9= 91 ä
5.61 Example Assume f (x) = x2 −4, g (x)=2
x, h(x) =−5x, evaluate the following expressions
1. ( f ◦g )(5)
2. ( f ◦h)(4)
3. (h ◦ f )(4)
4. ( f ◦ f )(0)
5. ( f ◦g ◦h)(2)
6. ( f ◦g ◦ f )(0)
Solution:
1. ( f ◦g )(5) = f (g (5))= f (2
5) = (
2
5)2 −4 =
4
25−4 =−
96
25
2. ( f ◦h)(4) = f (h(4)) = f (−5×4) = f (−20) = (−20)2 −4 = 400−4= 396
3. (h ◦ f )(4) = h( f (4)) = h(42 −4) = h(12) =−5×12=−60
4. ( f ◦ f )(0) = f ( f (0))= f (02 −4) = f (−4) = (−4)2 −4 = 16−4 = 12
5. ( f ◦g ◦h)(2) = f (g (h(2)))= f (g (−10))= f (−2
10) = f (−
1
5) =
1
25−4 =−
99
25
6. ( f ◦g ◦ f )(0) = f (g ( f (0)))= f (g (−4))= f (−2
4) = f (−
1
2) =
1
4−4 =−
15
4 ä
5.62 Example Assume f (x) =x
x +2, g (x) = x −1, h(x) =
1
x, find formulas for the following compositions
1. ( f ◦g )(x)
2. (g ◦h)(x)
3. (h ◦ f )(x)
4. ( f ◦g ◦h)(x)
5. (h ◦h)(x)
6. (h ◦h ◦h)(x)
Solution:
1. ( f ◦g )(x) = f (g (x)) = f (x −1)=x −1
x −1+2=
x −1
x +1
2. (g ◦h)(x) = g (h(x)) = g (1
x) =
1
x−1 =
1− x
x
3. (h ◦ f )(x) = h((x)) = h(x
x +2) =
x +2
x
4. ( f ◦g ◦h)(x) = f (g (h(x)) = f (1− x
x) =
1− x
x1− x
x+2
=
1− x
x1+ x
x
=1− x
x×
x
1+ x=
1− x
1+ x
5. (h ◦h)(x) = h(h(x)) = h(1
x) = x
178
Chapter 5
6. (h ◦h ◦h)(x) = h(h(h(x)) = h(x) =1
x ä
5.63 Example Express the following functions as a composition of two functions in the form f ◦g .
1. u = (x +7)4
2. u =p
2x +3
3. u =x4
1+ x4
Solution:
1. u = (x +7)4, g (x) = x +7, f (x) = x4, u = f ◦g .
2. u =p
2x +3, g (x) = 2x +3, f (x) =p
x, u = f ◦g .
3. u =x4
1+ x4, g (x) = x4, f (x) =
x
1+ x, u = f ◦g . ä
179
The Algebra of Functions and Composition
5.5.4 Homework: Algebra of Functions and Compositions
Find f +g , f −g , f ·g , f /g and specify the domain.
5.5.1 f (x) = 2x +3, g (x) = 2x −3
5.5.2 f (x) = x2 −x, g (x) = x2 −6
5.5.3 f (x) =p
x +1, g (x) =p
x
5.5.4 f (x) =−x2 +5x −6, g (x) = 5x −x2 −6
Find (f ◦g )(x) and (g ◦ f )(x)
5.5.5 f (x) = 3x +3, g (x) =−2x
5.5.6 f (x) = x2 +1, g (x) =−x2 +1
5.5.7 f (x) = 2x +4, g (x) =p
x −4
5.5.8 f (x) = x +4, g (x) = x2 −1
5.5.9 f (x) =1
x, g (x) = 2x +5
5.5.10 f (x) =1
x, g (x) =−
1
x2
5.5.11 Given f (x) = 3x+1, g (x) = x2−2, find (f ◦g )(−1),(f ◦g )(3)
5.5.12 Given f (x) =1
x, g (x) = 3x−1, find (f ◦g )(2),(f ◦g )(1)
5.5.13 Given f (x) = |x −2|, g (x) = x −2, find (f ◦g )(−4),(g ◦f )(−4)
5.5.14 Given f (x) = 2x2 −3, find
1. f (0)
2. f (x +h)
3.f (x +h)− f (x)
h
5.5.15 Given f (x) = 3x −6, find
1. f (h)
2. f (a)
3.f (a)− f (h)
a −h
5.5.16 Given f (x) =2
x, find
f (x +h)− f (x)
h
180
Chapter 5
5.6 Inverse FunctionsThe concept of inverse process indicates the reversal process. For example, the inverse of addition issubtraction, the inverse of multiplication is division, the inverse of clockwise rotation is anticlockwiserotation. In mathematics, it is important to define the inverse of every operation. This includes theconcept of functions. The function is a rule relating an input to an output, the inverse function is alsoa rule relating the output to the input. The following examples illustrate the concept of the inversefunctions.
1. The process of buying and item. The input is paying money, say $10 for an item A (output). Theinverse process is returning the item A and getting the initial refund for $10.
2. The price of gasoline is $4 per gallon. The cost of x gallons is given by the function C = 4x. So 10gallons cost $40. The inverse problem is how many gallons can one buy for $30?. So the inversefunction is x =C /4 or x = 30/4= 7.5 gallons.
3. Let the function f be defined by the set of ordered pairs: f = {(1,3),(2,5),(3,7),(4,9)}, then its inverseis another function g that reverses the rule and it is defined by the pairs: g = {(3,1),(5,2),(7,3),(9,4)}.However, the function h defined by h = {(1,3),(2,3),(5,1)} does not have an inverse function becausethe reversal set H = {(3,1),(3,2),(1,5)} has an input 3 with two different outputs 1 and 2 so therelation H is not a function.
From the previous examples, we observe that not every function has an inverse and that finding theinverse requires a procedure.
5.6.1 One-to-one Functions
12 Definition A function f is called one-to-one if for each number in the range (output) there is onlynumber in the domain (input). That is for every y in the range there is only one x such that f (x) = y.
For example,
1. The function f = {(1,−1),(2,−2),(3,−3),(4,−4)} is a one-to-one function, each output appears onlyonce.
2. The function g = {(1,−1),(2,1),(3,1),(4,−4)} is not a one-to-one function, because the output 1 appearsmore than once.
3. The function f (x) = x2 is not a one-to-one function, because the output 16 correspond to twoinputs: x2 = 16 has two solutions x = 4,−4.
Graphically, we have the so called horizontal line test:
13 Definition A function f is one-to-one if no horizontal line intersects the graph of the function atmore than one point.
181
Inverse Functions
Applying the horizontal line test to the graphs below, we conclude that the graph in figure a is aon-to-one function, but the graphs of figures b and c are not one=to-one functions since they fail thehorizontal test.
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
Figure 5.54: a
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
Figure 5.55: b
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
Figure 5.56: c
5.64 Example Use the horizontal line test to determine whether the following graphs are graphs of aone-to-one functions.
Figure 5.57: a Figure 5.58: b Figure 5.59: c
Solution: The formulas of the function g are:
1. Figure a: not a one-to-one function
2. Figure b: one-to-one function
3. Figure c: not a one-to-one function
ä
5.6.2 Inverse Functions
The inverse of a function f (x) is the function that reverses the rule of assignment and it is denoted by
f −1(x). For example, the inverse of f (x) = 2x is the function f −1(x) =1
2x. Given a table of few function
values and by interchanging the x- and y- coordinates to obtain a table for the inverse function:
182
Chapter 5
x f (x) = 2x
-2 -4
0 0
1 2
3 6
x f −1(x) =1
2x
-4 -2
0 0
2 1
6 3
It is simple to check that for any input x, we have the compositions f ◦ f −1(x) = x and f −1 ◦ f (x) = x.Based on this observation we define the inverse function as follows:
14 Definition Suppose f (x) is a one-to-one function. The function f −1(x) is the inverse function of f , iff ◦ f −1(x) = x for every x in the domain of f −1 and f −1 ◦ f (x) = x for every x in the domain of f .
☞ Note that exponent −1 of any function denotes the inverse function, unlike the exponents of numbers
and expressions. For example 4−1 =1
4but the inverse f −1 6=
1
f.
Finding the inverse function
1. If the function is given by a set of ordered pairs of numbers, then interchange the the x- andy- coordinates to obtain a set for the inverse function. For example, the inverse function off = {(2,3),(3,4),(4,0)} is the function f −1 = {(3,2),(4,3),(0,4)}.
2. If the function is given by a formula such as f (x) = 2x +4 , then we follow the steps
(a) Substitute f (x) by y, that is y = f (x)
(b) Interchange x and y, that is x = f (y)
(c) Solve for y in terms of x
(d) Substitute f −1 for y
5.65 Example Find the inverse of the function f (x) = 2x +4.
Solution:
1. Substitute f (x) by y: y = 2x +4
2. Interchange x and y: x = 2y +4
183
Inverse Functions
3. Solve for y in terms of x:
x = 2y +4
x −4 = 2y
x −4
2= y
y =x −4
2
4. Substitute f −1 for y: f −1 =x −4
2 ä
5.66 Example Find the inverse of the function f (x) = x3.
Solution: We follow the standard steps:
y = x3
x = y3
3p
x = y
y = 3p
x
f −1 = 3p
x ä
5.67 Example Find the inverse of the function f (x) = 1−6x.
Solution: We follow the standard steps:
y = 1−6x
x = 1−6y
x −1 =−6y
x −1
−6= y
y =−x −1
6
f −1 =−x −1
6 ä
5.68 Example Find the inverse of the function f (x) =p
x +4.
Solution: We follow the standard steps:
y =p
x +4
x =√
y +4
x2 = y +4
x2 −4 = y
y = x2 −4
f −1 = x2 −4 ä
5.69 Example Find the inverse of the function f (x) =1
8x3 −10.
184
Chapter 5
Solution: We follow the steps:
y =1
8x3 −10
x =1
8y3 −10
x +10 =1
8y3
8(x +10) = y3
23p
x +10 = y
y = 23p
x +10
f −1 = 23p
x +10 ä
Graphical representation of inverse functions
In the figure below, we plot the function f (x) =1
2x+1 in black and its inverse function f −1(x) = 2(x−1)
in red.
Figure 5.60: Function and its inverse
We note that the graphs of f and its inverse f −1 are symmetric with respect to the line y = x. Ingeneral every point (a,b) on a one-to-one function, corresponds to a point (b, a) on its inverse. Thesepoints are symmetric with respect to the line y = x. Thus, the graph of the inverse function f −1 can beobtained from the graph of the function f . As illustrated in the figures
Figure 5.61: Function and itsinverse
Figure 5.62: Function and itsinverse
b
bb
b
b b
b
b
Figure 5.63: Function and itsinverse
185
Inverse Functions
5.6.3 Homework: Inverse Functions
Determine the inverse of the following functions
5.6.1 F = {(1,0),(2,3),(3, 8), (4, 15)}
5.6.2 G = {(−1, a),(0,b),(1, c))}
5.6.3 f (x) = 4x
5.6.4 f (x) = 3x −2
5.6.5 f (x) = x
5.6.6 f (x) =1
27x3
5.6.7 f (x) =p
x −4
5.6.8 f (x) =3
5x +10
5.6.9 f (x) =2
x
5.6.10 f (x) =1
3x −2
5.6.11 f (x) = x2+16 where the domain of f equals (0,∞)
Graph the following functions, then sketch thegraph of its inverse using the symmetry property.
5.6.12 f (x) = 2x −4
5.6.13 f (x) =−3x +6
5.6.14 f (x) =p
x +3
5.6.15 f (x) =−p
x −4
186
Chapter 5
5.7 ExercisesDetermine the domain of the functions1. f (x) = x3 − x +1
2. g (x) =2
x2 −4x +3
3. f (x) =x +2
2x +9
4. f (x) =p
x +4
5. h(x) =−p
x2 −25
Evaluate the following
Given f (x) = x2 −2x −3 and g (x) =3
x −36. f (−3) and g (−3)
7. f (u) and g (3u)
8. f (t + s) and g (s2 − t 2) and g (3x +3)
9. ( f g )(x) and (g / f )(x)
10. Given f (x) = x2 − x4 and g (x) =p
x +2. Find a. ( f ◦g )(x), b. (g ◦ f )(x)
Graph the functions11. f (x) =−3+
px2 +5
12. g (x) = 3−2 | x −4 | and h(x) =−3+2 | x −4 | on the same system
13. f (x) = x2 +4x −7 and find its vertex and axis of symmetry
14. y = x2 −5x +3, on the interval (−1,6)
15. y = x3 + x −1, on the interval (−3,3)
16. y = x4 −7x2 +3x −12, on the interval (−3,4)
17. y = xp
100− x2, on the interval (2,8)
18. f (x) = xp
25− x2, on the interval (0,6) and show that f (3) = f (4)
19. A golf ball is hit into the air. The height h in feet is given by h(t) =−16t 2 +96t, where t is time in
seconds.
1. Find the time at which the ball reached its maximum height.
2. Find the maximum height.
3. After how many seconds did the ball hit the field. Assuming the field to be flat.
187
Exercises
20. Let the function f defined by f (x) = ax2 −1. Find the value a, given that f ( f (p−1) = 1
Answer:a(a −1)2 −1 = 1, a3 −2a2 +a −1−1 = 0, a2(a −2)+ (a −2)= 0, a = 2.
21. If f (2x) =2(x +3)
3. Find 3 f (x).
Answer: f (2x) =2x
3+2 so f (x) =
x
3+2 and 3 f (x) = x +6
188
6 System of Linear Equations
OBJECTIVES
• The Graphing Method
• Substitution Method
• Addition Method
A puzzle problemA woman has two daughters. The daughters are twins and they have the same height. The height ofthe woman and one daughter is 9 feet. The total height of the three women is 12 feet. What are theheights of the woman and her daughters?
Solution: Assume the height of the woman is w and the height of a daughter is d then wehave the relations: w +d = 9 and w +2d = 12 we observe that an extra d in the second equationincreases the right side by 3 so d must be 3 and then w = 6 Now instead suppose we have thecombination {
w +d = 9
w +2d = 12
in this case d = 3 and w = 6 feet. This system of equations can be rewritten in terms of thevariables x = w and y = d :
{
x + y = 9
x +2y = 12 ä
6.1 Example Check that x = 5 and y = 3 solve the equations
{
x + y = 8
3x +5y = 30
We are interested in solving a pair of equations simultaneously. such as the system of two linearequations.
{
x +2y = 4
2x − y = 3
There are many ways of solving system of linear equations. In this section, we present three methodsof solving system of equations:
• Graphical method: using graphing calculators.
• Algebraic method I: Substitution method.
• Algebraic method II: Elimination (Addition) method.
6.1 The Graphing method1. Graph each line.
2. Find the coordinates of the intersection point. These coordinates are the solution.
3. If the graphs are parallel, then the system has no solution.
189
The Graphing method
4. If the graphs are identical, then there are infinite many solutions. Every point on the line issolution to both equations.
5. Check your solution
6.2 Example Solve the system of equations using graphing calculator
{
x +2y = 4
2x − y = 3
Solution:
1. Solve each equation for y:
{
y = (−x +4)/2
y = 2x −3
2. Graph the lines: Y1 = (−X +4)/2 and Y2 = 2X −3
3. Press 2nd, CALC, select from the menu 5: intersect
4. press Enter three times
5. copy the coordinates of the intersection point: X = 2,Y = 1 ä
1
2
3
4
−1
−2
−3
−4
1 2 3 4−1−2−3−4
x +2y = 4
2x − y = 3
Figure 6.1: Graphic solution of the system of equations
6.3 Example Solve the system by graphing method
{
x + y = 3
x −2y =−3
Solution:
1. Solve each equation for y:
{
y =−x +3
y = (x +3)/2
2. Graph the lines: Y1 =−X +3 and Y2 = (X +3)/2
3. Press 2nd, CALC, select from the menu 5: intersect
4. press Enter three times
5. copy the coordinates of the intersection point: X = 1,Y = 2
190
Chapter 6
6. check that the coordinates of the intersection point (1,2) satisfy both equation.
1
2
3
4
−1
−2
−3
−4
1 2 3 4−1−2−3−4
x + y = 3x −2y =−3
Figure 6.2: Graphic solution
ä
6.4 Example Solve the system by graphing method
{
6x −2y = 4
y = 3x +1
Solution: To graph the lines, we solve both for the variable y:
{
y = (6x −4)/2
y = 3x +1
The two lines are parallel so they do not have any common point. The system of equation doesnot have a solution. System of equations that does not have a solution is called inconsistentsystem of equations.
1
2
3
4
−1
−2
−3
−4
1 2 3 4−1−2−3−4
6x −2y = 4
y = 3x +1
Figure 6.3: Parallel Lines
ä
6.5 Example Solve the system by graphing
{
4x +2y = 4
y =−2x +2
191
The Graphing method
Solution: The two lines are the same line. Thus, there are infinitely many solutions. Everypoint on one line satisfies the other line.
1
2
3
4
−1
−2
−3
−4
1 2 3 4−1−2−3−4
Figure 6.4: Same Line
ä
6.6 Example Solve the system by graphing
{
2x +3y = 6
4x +6y = 24If possible.
Solution: The lines are parallel. There is no solution to this system of equations. ä
6.7 Example Solve the system by graphing
5
2x + y =
1
2
2x −1
2y = 3
If possible.
Solution:
Graph the two lines:
y =−5
2x +
1
2
y = 4x −6
The solution is the intersection point (1,−2).
1
2
3
4
−1
−2
−3
−4
1 2 3 4−1−2−3−4
Figure 6.5: Intersecting Linesä
192
Chapter 6
6.2 The Substitution Method1. Solve on equation for one variable (preferably with coefficient 1).
2. Substitute the resulting expression into the other equation and solve it.
3. Find the other value of the variable from any equation.
4. State and check the solution.
6.8 Example Solve the system of linear equations by substitution method
{
−6x + y = 1
5x + y = 12
Solution:
1. Solve the first equation for y: y = 1+6x
2. Substitute in the second equation: 5x + (1+6x) = 12. Solve for x: x = 1.
3. Substitute back in any equation to find y: y = 1+6(1)= 7.
4. The solution is x = 1, y = 7, or written as a point (1,7). ä
6.9 Example Solve the system of linear equations by substitution method
{
3x +2y = 1
4x + y =−2
Solution:
1. Solve the second equation for y: y =−4x −2.
2. Substitute in the first equation: 3x +2(−4x −2) = 1 or 3x −8x −4 = 1. Solve for x: x = 1.
3. Substitute back in any equation to find y: y =−4(1)−2=−6.
4. The solution is (1,−6) ä
6.10 Example Solve the system by substitution
{
x −4y = 6
2x + y =−6
Solution:
1. Solve the first equation for x: x = 4y +6.
2. Substitute in the second equation: 2(4y +6)+ y =−6 or 8y +12+ y =−6. Solve for y: y =−2.
3. Substitute back in any equation to find x: x = 4(−2)+6 =−2.
4. The solution is (−2,−2) ä
6.11 Example Solve the system substitution
{
4x + y = 13
−2x +3y =−17
Solution:
1. Solve the first equation for y: y =−4x +13.
2. Substitute in the second equation: −2x +3(−4x +13) = −17 or −14x +39 = −17. Solve for x:x = 4.
3. Substitute back in any equation to find y: y =−4(4)+13=−3.
4. The solution is (4,−3) ä
193
The Addition Method
6.3 The Addition MethodThe addition method is also referred to as the elimination method. The procedure of this method isbased on the following steps.
1. Write both equations in general form Ax +B y =C .
2. Multiply the terms of one or both so the coefficients differ only in sign.
3. Add the equations and solve for one variable.
4. Substitute the value in either equation and find the other variable.
5. State the solution and check your solution.
6.12 Example Solve the system by Addition Method
{
x − y = 6
x + y = 10
Solution:
1. Add both equations to get 2x = 16, solve for x: x = 8.
2. Substitute in the second equation: 8+ y = 10. Solve for y: y = 2.
3. The solution is x = 8, y = 2 ä
6.13 Example Solve the system by Addition Method
{
4x + y = 13
−2x +3y =−17
Solution:
1. Multiply the second equation by 2 to get:
{
4x + y = 13
−4x +6y =−34
2. Add both equations to get 7y =−21, solve for y: y =−3.
3. Substitute in the first equation: 4x −3 = 13. Solve for x: x = 4.
4. The solution is x = 4, y =−3 ä
6.14 Example Solve the system
{
4x = 3(2+ y)
3(x −10) =−2y
Solution:
1. Rewrite both equations in standard form:
{
4x −3y = 6
3x +2y = 30
2. Multiply the first equation by 2 and the second equation by 3 to get:
{
8x −6y = 12
9x +6y = 90
3. Add both equations to get 17x = 102, solve for x: x = 6.
4. Substitute in the second equation: 18+2y = 30. Solve for y: y = 6.
5. The solution is x = 6, y = 6 ä
6.15 Example Solve the system
{
y = 2x +4
8x −4y = 7
Solution: Substitute the first equation in the second to get: 8x − 4(2x + 4) = 7, simplify8x −8x −16 = 7 or −16 = 7 which is contradiction. That is the system is inconsistent (No solution).ä
194
Chapter 6
6.16 Example Solve the system
{
4x +6y = 12
−2x −3y =−6
Solution: Multiply the second equation by 2 and add to the first to get 0 = 0 which is anidentity. That is there are infinitely many solutions. Geometrically, the two lines are the sameline. ä
6.17 Example Solve by any method
{
5x − y = 9
x +2y =−7
Solution: We suggest addition method:
1. Multiply the first equation by 2 to get:
{
10x −2y = 9
x +2y =−7
2. Add both equations to get 11x = 2, solve for x: y = 2/11.
3. Substitute in the second equation: 2/11+2y =−7. Solve for y: y =−79/22.
4. The solution is x = 2/11, y =−79/22 ä
6.18 Example Solve by any method
{
2x − y = 7
4x + y = 17
Solution: We suggest addition method:
1. Add both equations to get 6x = 24, solve for x: x = 12.
2. Substitute in the second equation: 4(12)+ y = 17. Solve for y: y =−31.
3. The solution is x = 12, y =−31 ä
6.19 Example Solve by any method
{
3x −8 = 0
x −5y −10 = 0
6.20 Example Solve by any method
x
2+
y
6= 1
x
3−
y
2= 2
Solution:
1. We multiply the equations by common multiples so all coefficient become integers:
{
3x + y = 6
2x −3y = 12
2. Multiply the first equation by 3 to get:
{
9x +3y = 18
2x −3y = 12
3. Add both equations to get 11x = 30, solve for x: x = 30/11.
4. Substitute in the first equation: y = 6−90/11. Solve for y: y =−24/11.
5. The solution is x = 30/11, y =−24/11 ä
195
The Addition Method
6.21 Example The sum of two numbers is 53 and their difference is 19. Find the numbers.
Solution:
1. We assume the numbers to be x and y. They satisfy the following linear system of equa-tions:{
x + y = 53
x − y = 19
2. Add both equations to get 2x = 72, solve for x: x = 36.
3. Substitute in the first equation: 36+ y = 53. Solve for y: y = 17.
4. The solution is x = 36, y =−57 ä
6.22 Example The income from a student production was $ 10,000. The price of a student ticket was $3, and non student tickets were sold at $ 5 each. Three thousand tickets were sold. How many ticketsof each kind were sold?
Solution: Solve
{
3x +5y = 10000
x + y = 3000x=2500, y=500
ä
196
Chapter 6
6.3.1 Homework: System of Equations
Solve the system of equations graphically
6.3.1{
x + y = 1
3x − y = 11
6.3.2{
2x −3y = 6
−4x +6y =−12
6.3.3{
5x − y = 5
y = 5
6.3.4
1
2x −
1
4y =−4
2x − y = 7
Solve the system by substitution method
6.3.5{
2x +3y = 15
y = x +2
6.3.6{
x +4y = 6
3x +5y = 20
6.3.7
y =1
2x +3
6x −7y = 9
6.3.8
2x −3y = 5
y =1
3x −
2
3
6.3.9
{
5x −2y = 8
y = 5x +7
Solve the system by addition method
6.3.10
{
2x +3y =−5
7x −3y =−4
6.3.11
{
3x +5y = 2
7x −3y =−1
6.3.12
2
3x +
1
3y =−2
4x −2y = 3
Solve the system by either method
6.3.13
{
y =−5x +2
7x + y =−1
6.3.14
{
x −2y = 10
−3x +6y = 10
6.3.15 The sum of two numbers is 54, and their differ-
ence is 20. Find the numbers.
197
A Answers and Hints
1.1.15
6
1.1.2 −3
5
1.1.3 24a2
1.1.4 a +b
1.1.5 −4x
3y 2
1.1.6 8t 4
1.1.7x −3
x
1.1.8x
x − y
1.1.93x −2
2x −1
1.1.104x2 −2
x +1
1.1.115n +2
2
1.1.12n −8
n −6
1.1.13 −1
1.1.14 1
1.1.15x −1
2
1.1.16y 2 +x y +x2
y +x
1.1.17x +3
x2 +3x +9
1.1.18 x +1
1.1.191
3x +1
1.1.20 −1
2
1.2.1 −1
2
1.2.25
3
1.2.3 a2
1.2.41
9
1.2.525
49
1.2.610
9a
1.2.72x4
3
1.2.81
2y
1.2.93b5
2
1.2.10x +2
x −4
1.2.11 a +2b
1.2.12x −2
x +4
1.2.13b
b −6
1.2.14x +2
3(x +3)
1.2.154a
3
1.2.16 1
1.2.172x y
3(x +3)
1.2.182t 2 +9t
6t +9
1.2.19x +3
x −1
1.2.20u2 −3u +2
u +3
1.3.1 −1
4
1.3.21
2
1.3.3s
t
1.3.4 1
1.3.51
x −2
1.3.6 1
198
Appendix A
1.3.7u −1
u +1
1.3.8 −11
10
1.3.9 −2
15
1.3.103
4
1.3.117x +7
12
1.3.124x −11
10
1.3.13−17
40x
1.3.148x +9
30x2
1.3.1586n +9
18n2
1.3.16 −5x −9
x2 −3x
1.3.171
7x +1
1.3.18 −b2 −b
b2 −4
1.3.19x y −x2 +x
y 2 −2x y +x2
1.3.20 −y +x
x y
1.3.213
3−2x
1.3.22 −2x +2
2x2 +5x
1.3.237n −12
5(n +4)(n −4)
1.3.245x
(x2 +1)(x +1)
1.3.252x
(x −1)(x +1)2
1.3.262
(x −2)(x −3)
1.3.27−13x −3
5(x −1)
1.3.284x +2
x2 −1
1.3.29 −4x3 −8x
x4 −5x2 +4
1.3.30a2 +b2 +c2
abc
1.4.1 2x
1.4.28b
ac
1.4.3 m
1.4.42x2 +10x
5x −25
1.4.5 2
1.4.663
50
1.4.72a −1
2
1.4.8y 2
3y −2
1.4.9x +2
x
1.4.102a −3b
ab −2a2
1.4.113n −7
5n −13
1.4.12y −5x −10
5x +7
1.4.13y −x
y +x
1.4.141
x2 +x −2
1.4.15 1
1.4.16−1
ht
1.4.171
x + y
1.4.18x +3
x +2
1.4.19x2 −2x −3
x2 +2x −3
1.4.20 1
1.4.21a +1
a −1
1.4.2216
a
1.5.117
3
1.5.3 −3
1.5.4 −58
21
1.5.57
11
199
Answers and Hints
1.5.6−11
5
1.5.7 None
1.5.83
2
1.5.9 2,6
1.5.102
3
1.5.11 3,5
1.5.12−1
2,−2
1.5.13 −1,1
1.5.145
9
1.5.15 4
1.5.16 6,−5
1.5.173∓3
p7i
4
1.5.18−7∓
p33
2
1.5.19 −3
1.5.20 2
1.5.21 11/3
1.6 x =∓8
2.1.1 −72
2.1.2 59
2.1.3 49
2.1.481
16
2.1.5 32x5 y 15
2.1.6 ab3
2.1.78
11bc
2.1.8 27x6 y 3
2.1.9 −1
121
2.1.101
4000
2.1.11 8/9
2.1.12 64
2.1.13 −3/2
2.1.14 9
2.1.15 a4
2.1.16x4
y 2
2.1.1715
x2 y 6
2.1.181
a8
2.1.19 a4b10
2.1.20 0
2.1.21 2x y
2.1.22 2y
x
2.1.23a12
b16
2.1.24x +2
x3
2.2.1 −9
2.2.2 −3
2.2.3 81
2.2.4 3p
2
2.2.5p
22/4
2.2.6p
3/2
2.2.7p
3
2.2.8 14p
3
2.2.11 3x2 y 3
2.2.12 2a2b
2.2.135x4
4y 2
2.2.14 5p
3
2.2.15 6x2p
3
2.2.16 2a4b6p
7
2.2.17 −2x y 2p
2
2.2.18 x y 2px y z
2.2.19 x y z2 3
√
y 2z2
2.2.20 2y z4√
3x2z2
2.2.21x2
3y 2
p5x
2.2.22 10x
200
Appendix A
2.2.23 12a
2.2.24 3ab2p
5
2.2.25 2x
2.2.26 10x3p
2x
2.2.27 6x y 2 3p
y
2.2.28 2yp
7x
2.3.1 7p
3
2.3.2 −2p
6
2.3.3 −113p
2
2.3.4 −183p
3
2.3.5 3a2b4p
6a
2.3.6 4y 3√
5x y
2.3.7
p5
4
2.3.8 23√
4x2
2.3.9 14bp
a −13ap
b
2.3.10 −4xp
6x
2.3.11 15x2 y√
3x y
2.3.12 7x y3√
2x2 −6x2 y 3√
2y
2.3.13 10x −4x2 3p
2
2.3.14 8xp
10x
2.3.15 2xx√
2x +3x2
2.3.16 7x2p
10x +2x2p
5x
2.3.17 −10x3p
3
2.3.18 −6p
5a
2.3.19 14 3p
a
2.3.20 −5p
6a
2.4.1 −30p
2
2.4.2 60
2.4.3p
6+p
21
2.4.4 12p
2−6
2.4.5 −40p
6
2.4.6 −1−8p
30
2.4.10 5a −5p
ab
2.4.11 −10abc
2.4.12 1−6x
2.4.13 x −10p
x +25
2.4.14 3x −p
3x −6
2.4.15 x −4y 2
2.4.16 3a −3b
2.4.17 x −1
2.4.18 ap
2−a
2.4.19 3x +6p
3x +9
2.4.20 5x −10p
x y +5y
2.4.21 8+2p
7
2.4.22 4−2√
4x −x2
2.5.1
p2
2
2.5.2p
3
2.5.3
px
3
2.5.4
px + y
2
2.5.5 −2p
a +4
3
2.5.6p
2−1
2.5.7 −p
3−2
2.5.8p
5+2
2.5.9p
3−p
2
2.5.10 7−4p
3
2.5.11 5p
3+3p
5+p
15+3
2.5.12p
x −py
2.5.13p
2a +1
2.5.145(p
2x +5
2x −25
2.5.153p
2
2
2.5.16
3√
5t 2
5t
2.5.173p
2x
2x
2.5.18p
x −4
2.5.19 10−3p
10
201
Answers and Hints
2.5.204
x −4
2.6.1 2
2.6.2 5
2.6.3 8
2.6.4 No solution.
2.6.5 4
2.6.6 34
2.6.7 9/4
2.6.8 3
2.6.9 −1
2.6.10 −1
2.6.11 2,3
2.6.12 1
2.6.14 4
2.6.16 1,−6
2.6.17 8,16
2.6.18 −9
2.6.19 1,13
2.6.20 −1,−9/4
2.6.21 No solution.
2.7.1 −3
2.7.2 1/7
2.7.3 16
2.7.4 −1/125
2.7.5 −27
2.7.6 64
2.7.7 −10
2.7.8 81/16
2.7.9 2
2.7.10 4p
3x
2.7.113√
(4a +5b)2
2.7.12√
x2 + y 2
2.7.13 −4(x2 + y 2)1/3
2.7.14 (a2 −b2)1/2
2.7.15 (6x2 y 3)1/6
2.7.16 −3x2
2.7.17 6x2 y 3
2.7.18 x +64
2.7.19 1/6
2.7.20 4
2.7.21 −2
2.7.22 2x5/6
2.8.1 (3.5)(10)6
2.8.2 (7.803021)(10)3
2.8.3 (3.5)(10)−4
2.8.4 (5.37)(10)7
2.8.5 2670
2.8.6 50.9
2.8.7 0.000300601
2.8.8 0.000300601
2.9 (−1)9 +18 = 0
2.9
√
9x2 +16x2
9 ·16=
5
12x
2.9 (1
4)−
1
4 = (1
22)−1/4 = (2−2)−1/4) = 21/2 =
p2
2.9 (1
4−
1
3)−1 = (−
1
12)−1 =−12
2.9 = x7/8
2.9 6(66) = 67
2.9 = 515
2.9 =
√
230 +220
212 +222=
√
28 = 16
2.9 = 3613
2.9 m =m −2
1−m, m −m2 = m −2 thus m =
p2
2.9 Multiply by the denominator to get 1 + 1 = 20(x −√
x2 −1) That is (x −√
x2 −1) =1
10. Observe that (x −
√
x2 −1)(x +√
x2 −1) = 1 so they are reciprocals thus (x +√
x2 −1) = 10. Add both expressions 2x = (x −√
x2 −1)+
(x +√
x2 −1) =1
10+10 = 10.1. Substitute x2 for x to get
2x2 = 2(10.1/2)2 = 51.005.
3.1.1 4p
3i
202
Appendix A
3.1.2 2p
5i
3.1.3 −10p
10i
3.1.4 8i
3.1.5 ip
22
3.1.7 9+8i
3.1.8 −8+6i
3.1.9 17−16i
3.1.10 −5−4i
3.1.11 41−13i
3.1.12 12+18i
3.1.13 5
3.1.14 25
3.1.15 −5−12i
3.1.16 −5i
3.1.173
2i
3.1.181
3−
2
3i
3.1.19 2+4i
3.1.20−20
25(3i +4) =−
16
5−
12
5i
3.1.21p
3−2i
3.1.22 15/34+9/34i
3.1.23 1
3.1.24 i
3.2.1 −5,4
3.2.2 −3/2,1/6
3.2.3 0,−4k
3.2.4 0,25
3.2.5 3,−3/2
3.2.6 1/2,1/5
3.2.7 −4/3,5/4
3.2.8 0,−5,5
3.2.9 0,1,3
3.2.10 ∓6
3.2.11 ∓2p
11
3.2.12 ∓p
15/3
3.2.13 (2+∓4i )/3
3.2.14 −2i ,2i
3.2.15−1∓
p5
2
3.2.16 ∓p
2
3.2.17 ∓2p
3i
3.2.18 4,−2
3.2.19 −3+ i ,−3− i
3.2.20 −1/3,1/3
3.2.21 8p
2
3.2.22 2p
21
3.3.1 6,−3
3.3.2 5,−1/3
3.3.3 2∓p
5
3.3.4 x = 4∓2p
3
3.3.5 x =−1/2∓p
17/2
3.3.6 x = 3/2∓p
5/2
3.3.7 4∓ i
3.3.8 −1∓ ip
11
3.3.97∓
p53
2
3.3.10 1,−7
3.3.11 2/3,−9/2
3.3.12 −5+∓3p
2
3.3.13−2∓
p10
2
3.3.14 x = 1∓2p
2
3.3.15 x =−1/2∓p
2/2
3.3.16 x = 4∓p
6
3.3.17 x =−1∓p
3/3
3.3.18 x = 5/2∓3p
5/2
3.4.1 1/2
3.4.2−2∓
p10
3
3.4.3 3∓2p
3
3.4.4 6,10
203
Answers and Hints
3.4.51∓
p33
4
3.4.63∓
p91i
10
3.4.71∓
p65
8
3.4.8∓p
15i
5
3.4.95∓
p13
2
3.4.102∓
p133
3
3.4.11 2,−10/7
3.4.12 4∓p
14
3.4.13 −6∓2p
10
3.4.141∓
p2
2
3.4.152∓2
p14
13
3.4.16−1∓
p3i
2
3.4.171∓
p7i
2
3.4.18 0,10
3.4.191∓
p31i
4
3.4.20 3∓p
19
3.4.21 2∓p
7
3.5.1 ∓2,∓ip
3
3.5.2 ∓p
5
3.5.3 0,6
3.5.4 27,−8
3.5.6 ∓2,∓2i
3.5.7 ∓3,∓p
10i
3.5.8 T w o
3.5.9 None
3.5.10 T w o
3.6.1 x ≤−3, or x ≥ 7/2
3.6.2 −4 < x < 2
3.6.3 −5 ≤ x ≤ 0
3.6.4 (−3,1)
3.6.5 −5 < x < 6
3.6.6 −3 < x < 3
3.6.7 −4 ≤ x ≤ 4
3.6.8 All real numbers
3.6.9 x <−2, or x > 5/2
3.6.10 −6 ≤ x ≤ 0
3.6.11 1 < x < 5
3.6.12 −2 < x < 2.
3.6.13 0 < x < 2
3.6.14 0 < x < 1/2
3.6.15 x <−4, x > 2
3.6.16 x ≤−10/3, or x > 0
3.6.17 x <−6, or x >≥ 1
3.6.18 x ≤−9, or x ≥−6
3.6.19 y >−1
3.6.20 −5 < x < 1/5 or x > 1
3.6.21 x ≥ 0
3.6.22 x < 0, or x > 6
3.6.23 x < 0
3.6.24 x > 3, or −3 < x −1
4.1.1 d =√
(1−3)2 + (0−1)2 =p
5
4.1.2 d =√
(−8+2)2 + (−2−5)2 =p
36+49 =p
85
4.1.3 d =√
(1)2 + (1)2 =p
2
4.1.4 (−1,1)
4.1.5 (−2,0)
4.1.6 (4,3)
4.1.7 (x +2)2 + (y +3)2 = 9
4.1.8 (x −5)2 + (y +4)2 = 18
4.1.9 x2 + y 2 = 64
4.1.10 x2 + y 2 = 12
4.1.11 (x +6)2 + (y −6)2 = 36
4.1.12 (0,0),r = 7
4.1.13 (0,0),r = 4
204
Appendix A
4.1.14 (−1,0),r = 3
4.1.15 (−3,0),r = 3
4.1.16 (4,7),r = 1
4.1.17 (3,−4),r = 4p
5
4.1.18 (0,6),r = 6
4.1.19 (0,0),r = 5
4.1.20 (2,0),r = 2
5.1.1 Not a function
5.1.2 Function
5.1.3 Function
5.1.4 Not a function
5.1.5 1. h(2) = 0
2. h(3) = 1/2
3. h(1
3) =−5/9
4. h(a) =1
3a −
2
3
5.1.6 1. f (a) =−3a +1
2. f (a +3) =−3a −8
3. f (a +h) =−3a −3h +1
5.1.7 1. f (a) = a2 −5a
2. f (a −1) = a2 −7a +6
3. f (a +h) = (a +h)2 −5(a +h)
5.1.8 1. f (−a) = a2 +6a +9
2. f (3) = 0
3. f (a −3) = a2 −12a +36
4. f (3x) = 9x2 −18x +9
5. f (x2) = x4 −6x2 +9
5.1.9 1. f (−2)+ f (5) = 38
2. f (−2+5) = 21
3. f (a)+3 = a2 +5a
4. f (x)− f (3) = x2 +5x −24
5. 4f (x) = 4x2 +20x −12
5.1.10 x ≥ 0
5.1.11 x 6= 0
5.1.12 All real numbers
5.1.13 x ≥ 9/4
5.1.14 All real numbers
5.1.15 x ≤ 2
5.1.16 x 6= −2/5
5.1.17 x 6= −2,1/3
5.1.18 D = {0,1,2,3}
5.1.19 Set of natural numbers N
5.1.20 All real numbers
5.2.12 (0,−5), x = 0
5.2.13 (−4,−2), x =−4
5.2.14 (3,9), x = 3
5.2.15 (5,5/6), x = 5
5.2.16 (−1,4), x =−1
5.2.17 (6,−7), x = 6
5.2.18 (−2,0), x =−2
5.3.1 y = (x −2)2 −4
5.3.2 y = (x +1)2 −1
5.3.3 y = (x −3)2
5.3.4 y = (x +7)2 +1
5.3.5 f (x) =−(x +5/2)2 +17/4
5.3.6 f (x) = 2(x +2)2 −13
5.3.7 f (x) =−3(x +5)2 +5
5.3.8 V = (0,−4), x = 0, i nt er cep t s : (0,−4),(2,0),(−2, 0)
5.3.9 V = (0,−1), x = 0, i nt er cep t s : (0,−1)
5.3.10 V = (0,9), x = 0, i nt er cep t s : (0,9),(3,0),(−3,0)
5.3.11 V = (1,2), x = 1, i nt er cep t s : (0,3)
5.3.12 V = (−5,0), x =−5, i nt er cep t s : (0,25),(−5,0)
5.3.13 V = (4,−4), x = 4, i nt er cep t s : (0,12),(6,0),(2, 0)
5.3.14 (1,0), x = 1, i nt er cep t s : (0,3),(1,0)
5.3.15 V = (2,5), x = 2, i nt er cep t s : (0,1),(2 +p
5,0),(2 −p5,0)
5.3.16 V = (0,−6), x = 0
5.3.17 V = (−5/2,25/4), x =−5/2, i nt er cep t s : (0,0),(−5,0)
5.3.18 V = (4,−1), x = 4, i nt er cep t s : (0,15),(5,0),(3, 0)
5.4.1 y = x2, 3 units to the right, 5 units upward
5.4.2 y = x3, 4 units to the left, reflection
5.4.3 y = |x|, 5 units to the right, 3 units downward
5.4.4 y =p
x, 2 units to the right, 4 units downward,reflection
205
Answers and Hints
5.4.5 y = |x|−2
5.4.6 y = |x +5|+1
5.4.7 y = (x −3)3 +3
5.4.8 y =−x3 −4
5.4.9 y =−p
x −7+2
5.5.1 f +g = 4x, f −g = 6, f ·g = 4x2 −9, f /g =2x +3
2x −3
5.5.2 2x2 −x −6,−x +6, x4 −x3 −6x2 −6x,
(x2 −x)/(x2 −6)
5.5.3p
x +1+p
x,p
x +1p
x,p
x +1/p
x
5.5.4 −2x2 +10x −12,0,(−x +5x −6)2,1
5.5.5 f ◦g (x) =−6x +3, g ◦ f (x) =−6x −6
5.5.6 f ◦g = (−x2 +1)2 +1, g ◦ f =−(x2 +1)2 +1
5.5.7p
x −4+4,p
2x
5.5.8 x2 +3, x2 +8x +15
5.5.9 1/(2x +3),2/x +5
5.5.10 f ◦g =−x2, g ◦ f =−x2
5.5.11 f ◦g (−1) =−2, g ◦ f (3) = 98
5.5.12 f ◦g (2) = 1/5, g ◦ f (1) = 2
5.5.13 f ◦g (−4) = 8, g ◦ f (−4) = 4
5.5.14 1. f (0) =−3
2. f (x +h) = 2x2 +2xh +2h2 −3
3.f (x +h)− f (x)
h= 2(x +h)
5.5.15 1. f (h) = 3h −6
2. f (a) = 3a −6
3.f (a)− f (h)
a −h= 3
5.5.16 −2/x(x +h)
5.6.1 F−1 = {(0,1),(3,2),(8,3),(15, 4)}
5.6.2 G−1 = {(a,−1),(b,0),(c ,1))}
5.6.3 f −1 =1
4x
5.6.4 f −1 =x +2
3
5.6.5 f −1 = x
5.6.6 f −1 = 3 3p
x
5.6.7 f −1 = (x +4)2
5.6.8 f −1 =5
3(x −10)
5.6.9 f −1 =2
x
5.6.10 f −1 =2x +1
3x
5.6.11 f −1 =p
x −16
6.3.1 x = 3, y =−2
6.3.2 Infinite many solutions
6.3.3 x = 2, y = 5
6.3.4 No solution
6.3.5 x = 9/5, y = 19/5
6.3.6 x = 50/7, y =−2/7
6.3.7 x = 12, y = 9
6.3.8 x = 3, y = 1/3
6.3.9 x =−22/5, y =−15
6.3.10 x =−1, y =−1
6.3.11 x = 1/44, y = 17/44
6.3.12 x =−9/8, y =−15/4
6.3.13 x =−3/2, y = 19/2
6.3.14 No solution
6.3.15 x = 37, y = 17
206