copyright © 2009 pearson education, inc. chapter 3: quadratic functions and equations; inequalities...

Copyright © 2009 Pearson Education, Inc.

CHAPTER 3: Quadratic Functions and Equations; Inequalities

3.1 The Complex Numbers

3.2 Quadratic Equations, Functions, Zeros, and Models

3.3 Analyzing Graphs of Quadratic Functions

3.4 Solving Rational Equations and Radical Equations

3.5 Solving Equations and Inequalities with Absolute Value

Copyright © 2009 Pearson Education, Inc.

3.2 Quadratic Equations, Functions,

Zeros, and Models

Find zeros of quadratic functions and solve quadratic equations by using the principle of zero products, by using the principle of square roots, by completing the square, and by using the quadratic formula.

Solve equations that are reducible to quadratic. Solve applied problems using quadratic equations.

2 - 4Copyright © 2009 Pearson Education, Inc.

Quadratic Equations

A quadratic equation is an equation that can be written in the form

ax2 + bx + c = 0, a 0,

where a, b, and c are real numbers.

A quadratic equation written in this form is said to be in standard form.


Quadratic Functions

A quadratic function f is a function that can be written in the form

f (x) = ax2 + bx + c, a 0, where a, b, and c are real numbers.

The zeros of a quadratic function f (x) = ax2 + bx + c are the solutions of the associated quadratic equation ax2 + bx + c = 0. Quadratic functions can have real-number or imaginary-number zeros and quadratic equations can have real-number or imaginary-number solutions.


Equation-Solving Principles

The Principle of Zero Products:

If ab = 0 is true, then

a = 0 or b = 0,

and if a = 0 or b = 0,

then ab = 0.


Example

Solution

Solve 2x2 x = 3.

2x2 −x=3

2x2 −x−3=0

x +1( ) 2x−3( ) =0

x +1=0 or 2x−3=0

x =−1 or 2x=3

x =−1 or x=32


Example - Checking the Solutions

Check: x = – 1

2x2 −x=32 −1( )2 − −1( ) ? 3

2⋅1+1 3

2 +1 3

3 3

23

2⎛⎝⎜

⎞⎠⎟2

−32

⎛⎝⎜

⎞⎠⎟

? 3

2⋅94−32

3

9

2−32

3

6

23

3 3

Check:

2x2 −x=3

x =32

3

2.The solutions are –1 and

TRUE

TRUE


Equation-Solving Principles

The Principle of Square Roots: If x2 = k, then

x = k or x=− k.


Example

Solve 2x2 10 = 0.

Solution

2x2 −10 =02x2 =10

x2 =5x = 5 or x=− 5

2 ± 5( )2−10 ? 0

2⋅5−10 0

10 −10 0

0 0 TRUE

− 5 .The solutions are and5

2x2 −10 =0

Check:


Completing the Square

1. Isolate the terms with variables on one side of the equation and arrange them in descending order.

2. Divide by the coefficient of the squared term if that coefficient is not 1.

3. Complete the square by taking half the coefficient of the first-degree term and adding its square on both sides of the equation.

4. Express one side of the equation as the square of a binomial.

5. Use the principle of square roots.6. Solve for the variable.

To solve a quadratic equation by completing the square:


Example

Solve 2x2 1 = 3x.

2x2 −1=3x

2x2 −3x−1=0

2x2 −3x =1

x2 −32x =

12

x2 −32x+

916

=12+

916

x −34

⎛⎝⎜

⎞⎠⎟2

=1716

x −34=±

174

x =34±

174

The solutions are 3± 174

.

Solution


Quadratic Formula

The solutions of ax2 + bx + c = 0, a 0, are given by

This formula can be used to solve any quadratic equation.

x =−b± b2 −4ac

2a.


Example

Solution:

3x2 + 2x 7 = 0 a = 3, b = 2, c = 7The exact solutions are:

Solve 3x2 + 2x = 7. Find exact solutions and approximate solutions rounded to the thousandths.

The approximate solutions are –1.897 and 1.230.

x =−2± 22 −4 3( ) −7( )

2 3( )=−2 ± 4 + 84

6=−2 ± 88

6

=−2 ± 2 22

6=−1− 22

3 or

−1+ 22

3=

2 −1 ± 22( )

2 ⋅3


Discriminant

When you apply the quadratic formula to any quadratic equation, you find the value of b2 4ac, which can be positive, negative, or zero. This expression is called the discriminant.

For ax2 + bx + c = 0, where a, b, and c are real numbers: b2 4ac = 0 One real-number solution; b2 4ac > 0 Two different real-number solutions; b2 4ac < 0 Two different imaginary-number solutions, complex conjugates.


Equations Reducible to Quadratic

Some equations can be treated as quadratic, provided that we make a suitable substitution.

Example: x4 5x2 + 4 = 0 Knowing that x4 = (x2)2, we can substitute u for x2 and the resulting equation is then u2 5u + 4 = 0. This equation can then be solved for u by factoring or using the quadratic formula. Then the substitution can be reversed by replacing u with x2, and solving for x. Equations like this are said to be reducible to quadratic, or quadratic in form.


Solving an Equation Reducible to Quadratic - Example

Solution

x4 5x2 + 4 = 0

u2 5u + 4 = 0 (substituting u for x2)

(u 1)(u 4) = 0

u 1 = 0 or u 4 = 0

u = 1 or u = 4

x2 = 1 or x2 = 4 (substitute x2 for u

x = ±1 or x = ±2 and solve for x)

The solutions are 1, 1, 2, and 2.

Solve: x4 5x2 + 4 = 0.


Applications

Some applied problems can be translated to quadratic equations.

ExampleTime of Free Fall. The Petronas Towers in Kuala Lumpur, Malaysia are 1482 ft tall. How long would it take an object dropped from the top reach the ground?


Example (continued)

1. Familiarize. The formula s = 16t2 is used to approximate the distance s, in feet, that an object falls freely from rest in t seconds.

1482 =16t2

1482

16=t2

1482

16=t

9.624 ≈t

2. Translate. Substitute 1482 for s in the formula: 1482 = 16t2.

3. Carry out. Use the principle of square roots.


Example (continued)

4. Check. In 9.624 seconds, a dropped object would travel a distance of 16(9.624)2, or about 1482 ft. The answer checks.

5. State. It would take about 9.624 sec for an object dropped from the top of the Petronas Towers to reach the ground.

copyright © 2009 pearson education, inc. chapter 3: quadratic functions and equations; inequalities...

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