Beginning Algebra, 11th Edition

Using the Squaring Property with a Radical on Each Side

Solve

Squaring property

Subtract x.

Divide by 8.

CHECK Original equation

Let

3 = 3 ✓ True

3112  29

321  21 + 8 x=1.

32 x= 2 x+ 8

x= 1

8 x= 8

9 x=x+ 8 A 2 xB^2 =x; A 2 x+ 8 B^2 =x+ 8

32 A 2 xB 1 ab 22 =a^2 b^2

2

=A 2 x+ 8 B

2

(^) A 32 xB^2 =A 2 x+ (^8) B^2
32 x= 2 x+ 8
32 x= 2 x+ 8.
EXAMPLE 2
532 CHAPTER 8 Roots and Radicals
NOW TRY
EXERCISE 2
Solve 4 2 x= 210 x+ 12.
NOW TRY ANSWERS

2. 526 3. 0

Be careful

here.

This is not

the solution

The solution set of 32 x= 2 x+ 8 is 516. NOW TRY

CAUTION Do not write the final result obtained in the check in the solution set.

In Example 2,the solution set is 516 ,not 536.

OBJECTIVE 2 Identify equations with no solutions.Not all radical equa-

tions have solutions.

NOW TRY

EXERCISE 3

Solve 2 x=-6.

NOTE Because represents the principalor nonnegativesquare root of xin

Example 3,we might have seen immediately that there is no solution.

2 x

Using the Squaring Property When One Side Is Negative

Solve

Squaring property

Proposed solution

CHECK Original equation

Let

False

Because the statement is false, the number 9 is nota solution of the given

equation. Recall from Section 7.6that a proposed solution that is not an actual solu-

tion of the original equation is called an extraneous solutionand must be rejected. In

fact, 2 x=- 3 has no solution. The solution set is 0. NOW TRY

3 =- 3

3 =- 3

29 - 3 x=9.

2 x=- 3

x= 9

A 2 xB

2

= 1 - 322

2 x=- 3

2 x=-3.

EXAMPLE 3

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