Section 7.6 Radical Equations and Problem Solving 447

Section 7.6 Radical Equations and Problem Solving 447

Use rational exponents to write each as a single radical expression.

19. 24 x # 2x

20. 25 # 23 2

Simplify. 21. 240

22. 24 16x7y10

23. 23 54x4

Multiply or divide. Then simplify if possible.

25. 25 # 2x

# 26. 23 8x 23 8x2

Perform each indicated operation.

298y6 27.

22y

29. 220 - 275 + 527

30. 23 54y4 - y 23 16y

32. 1 27 + 2322

33. 12x - 25212x + 252

Rationalize each denominator.

7 35. A 3 Rationalize each numerator.

5 36.

23 2x2

7 38. A 3

39.

3 9y A 11

24. 25 - 64b10

24 48a9b3 28.

24 ab3

31. 231 25 - 222 34. 1 2x + 1 - 122

23 - 27 37.

223 + 27 2x - 2 40.

2x

7.6 Radical Equations and Problem Solving

OBJECTIVES

1 Solve Equations That Contain

Radical Expressions.

2 Use the Pythagorean Theorem

to Model Problems.

OBJECTIVE

1 Solving Equations That Contain Radical Expressions In this section, we present techniques to solve equations containing radical expressions such as

22x - 3 = 9 We use the power rule to help us solve these radical equations.

Power Rule If both sides of an equation are raised to the same power, all solutions of the original equation are among the solutions of the new equation.

This property does not say that raising both sides of an equation to a power yields an equivalent equation. A solution of the new equation may or may not be a solution of the original equation. For example, 1 -222 = 22 , but -2 2. Thus, each solution of the new equation must be checked to make sure it is a solution of the original equation. Recall that a proposed solution that is not a solution of the original equation is called an extraneous solution.

E X A M P L E 1 Solve: 22x - 3 = 9.

Solution We use the power rule to square both sides of the equation to eliminate the radical.

22x - 3 = 9 1 22x - 322 = 92

2x - 3 = 81 2x = 84 x = 42

Now we check the solution in the original equation.

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448 CHAPTER 7 Rational Exponents, Radicals, and Complex Numbers

Check:

22x - 3 = 9 221422 - 3 9

284 - 3 9 281 9

9=9

Let x = 42. True

The solution checks, so we conclude that the solution is 42, or the solution set is 5426.

PRACTICE

1 Solve: 23x - 5 = 7.

To solve a radical equation, first isolate a radical on one side of the equation.

E X A M P L E 2 Solve: 2-10x - 1 + 3x = 0. Solution First, isolate the radical on one side of the equation. To do this, we subtract 3x from both sides.

2-10x - 1 + 3x = 0 2-10x - 1 + 3x - 3x = 0 - 3x

2-10x - 1 = -3x

Next we use the power rule to eliminate the radical.

1 2 - 10x - 122 = 1 - 3x22 - 10x - 1 = 9x2

Since this is a quadratic equation, we can set the equation equal to 0 and try to solve by factoring.

9x2 + 10x + 1 = 0 19x + 121x + 12 = 0 9x + 1 = 0 or x + 1 = 0

Factor. Set each factor equal to 0.

x = - 1 or 9

Check: Let x = - 1 . 9

2-10x - 1 + 3x = 0

x = -1

Let x = -1. 2-10x - 1 + 3x = 0

B

-

10

a

-

1 9

b

-

1

+

3a - 1b 9

0

2-101 -12 - 1 + 31 -12 0

10 A9

-

9 9

-

3 9

0

210 - 1 - 3 0

1 A9

-

1 3

0

29 - 3 0

1 - 1 = 0 True

3 - 3 = 0 True

33

Both solutions check. The solutions are - 1 and -1, or the solution set is e - 1 , -1 f .

9

9

PRACTICE

2 Solve: 216x - 3 - 4x = 0.

Section 7.6 Radical Equations and Problem Solving 449

The following steps may be used to solve a radical equation.

Solving a Radical Equation Step 1. Isolate one radical on one side of the equation. Step 2. Raise each side of the equation to a power equal to the index of the

radical and simplify. Step 3. If the equation still contains a radical term, repeat Steps 1 and 2. If not,

solve the equation. Step 4. Check all proposed solutions in the original equation.

E X A M P L E 3 Solve: 23 x + 1 + 5 = 3. Solution First we isolate the radical by subtracting 5 from both sides of the equation.

23 x + 1 + 5 = 3 23 x + 1 = - 2

Next we raise both sides of the equation to the third power to eliminate the radical.

1 23 x + 123 = 1 - 223

x + 1 = -8 x = -9

The solution checks in the original equation, so the solution is - 9 .

PRACTICE

3 Solve: 23 x - 2 + 1 = 3.

E X A M P L E 4 Solve: 24 - x = x - 2.

Solution

24 - x = x - 2

1 24 - x22 = 1x - 222

4 - x = x2 - 4x + 4 x2 - 3x = 0 x1x - 32 = 0 x = 0 or x - 3 = 0

Write the quadratic equation in standard form. Factor. Set each factor equal to 0.

x=3

Check:

24 - x = x - 2

24 - 0 0 - 2 2 = -2

Let x = 0. False

24 - x = x - 2

24 - 3 3 - 2 Let x = 3.

1=1

True

The proposed solution 3 checks, but 0 does not. Since 0 is an extraneous solution, the only solution is 3.

PRACTICE

4 Solve: 216 + x = x - 4.

Helpful Hint In Example 4, notice that 1x - 222 = x2 - 4x + 4 . Make sure binomials are squared correctly.

450 CHAPTER 7 Rational Exponents, Radicals, and Complex Numbers

CONCEPT CHECK

How can you immediately tell that the equation 22y + 3 = -4 has no real solution?

E X A M P L E 5 Solve: 22x + 5 + 22x = 3.

Solution We get one radical alone by subtracting 22x from both sides.

22x + 5 + 22x = 3 22x + 5 = 3 - 22x

Now we use the power rule to begin eliminating the radicals. First we square both sides.

1 22x + 522 = 13 - 22x22

2x + 5 = 9 - 6 22x + 2x Multiply 13 - 22x213 - 22x2.

There is still a radical in the equation, so we get a radical alone again. Then we square both sides.

2x + 5 = 9 - 622x + 2x

622x = 4

Get the radical alone.

3612x2 = 16

Square both sides of the equation to eliminate the radical.

72x = 16

Multiply.

x = 16 72

Solve.

x=2

Simplify.

9

2

2

The proposed solution, , checks in the original equation. The solution is .

9

9

PRACTICE

5 Solve: 28x + 1 + 23x = 2.

Helpful Hint Make sure expressions are squared correctly. In Example 5, we squared 13 - 22x2 as

13 - 22x22 = 13 - 22x213 - 22x2

= 3 # 3 - 3 22x - 3 22x + 22x # 22x

= 9 - 6 22x + 2x

CONCEPT CHECK

What is wrong with the following solution? 22x + 5 + 24 - x = 8

1 22x + 5 + 24 - x22 = 82

12x + 52 + 14 - x2 = 64 x + 9 = 64 x = 55

Answers to Concept Checks: answers may vary

1 22x + 5 + 24 - x22 is not 12x + 52 + 14 - x2.

OBJECTIVE

2 Using the Pythagorean Theorem

Recall that the Pythagorean theorem states that in a right triangle, the length of the hypotenuse squared equals the sum of the lengths of each of the legs squared.

Section 7.6 Radical Equations and Problem Solving 451

Pythagorean Theorem If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then a2 + b2 = c2 .

Hypotenuse c

a

b Legs

E X A M P L E 6 Find the length of the unknown leg of the right triangle.

10 m 4 m

b

Solution In the formula a2 + b2 = c2 , c is the hypotenuse. Here, c = 10, the length of the hypotenuse, and a = 4 . We solve for b. Then a2 + b2 = c2 becomes

42 + b2 = 102 16 + b2 = 100

b2 = 84 Subtract 16 from both sides.

b = {284 = {24 # 21 = {2221

Since b is a length and thus is positive, we will use the positive value only. The unknown leg of the triangle is 2221 meters long.

PRACTICE

6 Find the length of the unknown leg of the right triangle.

6 m

12 m

a

E X A M P L E 7 Calculating Placement of a Wire

A 50-foot supporting wire is to be attached to a 75-foot antenna. Because of surrounding buildings, sidewalks, and roadways, the wire must be anchored exactly 20 feet from the base of the antenna.

75 ft 50 ft

20 ft

a. How high from the base of the antenna is the wire attached?

b. Local regulations require that a supporting wire be attached at a height no less than 3 of the total height of the antenna. From part (a), have local regulations been met? 5

Solution

50 ft

x ft

1. UNDERSTAND. Read and reread the problem. From the diagram, we notice that

a right triangle is formed with hypotenuse 50 feet and one leg 20 feet. Let x be the

height from the base of the antenna to the attached wire.

20 ft

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