slide 1.5- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley
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Slide 1.5- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
OBJECTIVES
Solving Other Types of Equations
Learn to solve equations by factoring.
Learn to solve fractional equations.
Learn to solve equations involving radicals.
Learn to solve equations that are quadratic in form.
SECTION 1.5
1
2
3
4
Slide 1.5- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
PROCEDURE FOR SOLVING EQUATIONS BY FACTORING
Step 1 Make one side zero. Move all nonzero terms in the equation to one side (say the left side), so that the other side (right side) is 0.
Step 2 Factor the left side.
Step 3 Use the zero-product property. Set each factor in Step 2 equal to 0, and then solve the resulting equations.
Step 4 Check your solutions.
Slide 1.5- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1 Solving an Equation by Factoring
Solve by factoring:
The solution set is {–3, 0, 3}.
x4 9x2
Solution
Step 1 x4 9x2 0
Step 2 x2 x2 9 0
x2 x 3 x 3 0
Step 3 x2 0 or x 3 0 or x 3 0
x 0 or x 3 or x 3Step 4 04 9 0 2 3 4 9 3 234 9 3 2
Slide 1.5- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2 Solving an Equation by Factoring
Solve by factoring:
The solution set is {2,i,–i}.
x3 2x2 x 2
Solution
Step 1 x3 2x2 x 2 0
Step 2 x2 x 2 x 2 0
x 2 x2 1 0
Step 3 x 2 0
x 2
Step 4 23 2 2 2 2 2
x2 1 0
x2 1 i
i3 2 i 2 i 2 i 3 2 i 2 i 2
Slide 1.5- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3 Solving a Rational Equation
Solve: 1
6
1
x 1
1
x
Solution
Step 1 Find the LCD: 6x(x + 1)
Step 2 6x x 1 1
6
1
x 1
6x x 1 1
x
6x x 1 6
6x x 1
x 1
6x x 1 x
x x 1 6x 6 x 1
Slide 1.5- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3 Solving a Rational Equation
The solution set is {–3,2}.
Solution continuedx2 x 6x 6x 6
x2 x 6 0
x 3 0
x 3Step 4
Step 3 x 3 x 2 0
or x 2 0
or x 2
Step 5 Both solutions check in the original equation.
Slide 1.5- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Solving Equations Involving Radicals
Solve: x 6x x3
x2 6x x3 2x2 6x x3
x3 x2 6x 0
x x2 x 6 0
x x 3 x 2 0
Solution
Since we raise both sides to power 2. a 2 a,
Slide 1.5- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4 Solving Equations Involving Radicals
Solution continued
x 0 or x 3 0 or x 2 0
x 0 or x 3 or x 2
–3 is an extraneous solution.The solution set is {0, 2}.
Check each solution.
3 6 3 3 3
3 3
? 0 6 0 0 3
0 0 ?
2 6 2 2 3
2 2 ?
Slide 1.5- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5 Solving Equations Involving Radicals
Solve: 2x 1 1 x
Solution
Step 1 Isolate the radical on one side.
2x 1 x 1
Step 2 Square both sides and simplify.2x 1 2 x 1 2
2x 1 x2 2x 1
x2 4x 0
x x 4 0
Slide 1.5- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5 Solving Equations Involving Radicals
Solution continued
Step 3 Set each factor = 0.
x 0 or x 4 0
x 0 or x 4
0 is an extraneous solution.The solution set is {4}.
Step 4 Check.
2 0 1 1 0
2 0
? 2 4 1 1 4
4 4 ?
Slide 1.5- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Solving an Equation Involving Two Radicals
Solve: 2x 1 x 1 1
Solution
Step 1 Isolate one of the radicals.2x 1 1 x 1
Step 2 Square both sides and simplify.
2x 1 2 1 x 1 22x 1 1 2 x 1 x 1
2x 1 2 x 1 x
Slide 1.5- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Solving an Equation Involving Two Radicals
Solution continued
Step 3 Repeat the process - isolate the radical, square both sides, simplify and factor.
x 1 2 x 1
x 1 2 2 x 1 2x2 2x 1 4 x 1 x2 6x 5 0
x 5 x 1 0
Slide 1.5- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6 Solving an Equation Involving Two Radicals
Solution continued
Step 4 Set each factor = 0.x 5 0 or x 1 0
x 5 or x 1
The solution set is {1,5}.
Step 5 Check.
2 5 1 5 1 1
3 2 1? 2 1 1 1 1 1
1 0 1?
Slide 1.5- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
SOLVING EQUATIONS CONTAINING SQUARE ROOTS
Step 1 Isolate one radical to one side of the equation.
Step 2 Square both sides of the equation in Step 1 and simplify.
Step 3 If the equation in Step 2 contains a radical, repeat Steps 1 and 2 to get an equation that is free of radicals.
Step 5 Check the solutions in the original equation.
Step 4 Solve the equation obtained in Steps 1 - 3.
Slide 1.5- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
An equation in a variable x is quadratic in form if it can be written as
EQUATIONS THAT ARE QUADRATIC IN FORM
au2 bu c 0 a 0 ,where u is an expression in the variable x . We solve the equation au2 + bu + c = 0 for u. Then the solutions of the original equation can be obtained by replacing u by the expression in x that u represents.
Slide 1.5- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7Solving an Equation That Is Quadratic in Form by Substitution
Solve: x2 1 2 6 x2 1 16 0
Solution
Let u = x2 – 1, then u2 = (x2 – 1)2
x2 1 2 6 x2 1 16 0
u2 6u 16 0
u 2 u 8 0
u 8 0 or u 2 0
u 8 or u 2
Slide 1.5- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7Solving an Equation That Is Quadratic in Form by Substitution
Solution continued
Replace u with x2 – 1, and solve for x.
x2 1 2
x2 1
x i
x2 1 8
x2 9
x 3
All four solutions check in the original equation.
The solution set is {i, –i, 3, –3}.
Slide 1.5- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8Solving an Equation That Is Quadratic in Form by Substitution
Solve: x 1
x
2
6 x 1
x
8 0
x 1
x
2
6 x 1
x
8 0
u2 6u 8 0
u 2 u 4 0
Solution
Let thenu x 1
x, u2 x
1
x
2
.
Slide 1.5- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8Solving an Equation That Is Quadratic in Form by Substitution
Solution continued
u 2 0 or u 4 0
u 2 or u 4
Replace u with and solve for x.x 1
x,
x 1
x2
x2 x 2x
x2 2x 1 0
x 1 2 0
x 1 2 0
x 1 0
x 1
x = 1 checks in the original equation
Slide 1.5- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8Solving an Equation That Is Quadratic in Form by Substitution
Solution continued
x 1
x4
x2 1 4x
x2 4x 1 0
x 4 4 2 4 1 1
2 1
x 4 12
2
Both solutions check in the original equation. The solution set is 1,2 3,2 3 .
x 4 2 3
2
x 2 2 3
2
x 2 3
Slide 1.5- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 9 Investigating Space Travel
Your sister is 5 years older than you are. She decides she has had enough of Earth and needs a vacation. She takes a trip to the Omega-One star system. Her trip to Omega-One and back in a spacecraft traveling at an average speed v took 15 years, according to the clock and calendar on the spacecraft. But on landing back on Earth, she discovers that her voyage took 25 years, according to the time on Earth. This means that, although you were 5 years younger than your
Slide 1.5- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 9 Investigating Space Travel
sister before her vacation, you are 5 years older than her after her vacation! Use the time-dilation equation
from the introduction to this section to calculate the speed of the spacecraft.
t0 t 1 v2
c2
Slide 1.5- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 9 Investigating Space Travel
Substitute t0 = 15 (moving-frame time) and t = 25 (fixed-frame time) to obtain
15 25 1 v2
c2
3
5 1
v2
c2
9
251
v2
c2
Solution
v2
c2 1 9
25
v
c
2
16
25
v
c
4
5
v 4
5c 0.8c
So the spacecraft was moving at 80% (0.8c) the speed of light.
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