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


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.


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


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


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


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.


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,



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 ?



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



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 ?


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



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



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?


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.


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.


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



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}.



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

.



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



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


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



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



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.

slide 1.5- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

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