Intermediate Algebra (11th edition)

SECTION 8.6 Solving Equations with Radicals 471

The equation still contains a radical, so isolate the radical term on the right and

square both sides again.

Result after squaring

Combine like terms.

Subtract 8 and 3x.

Divide by 2.

Square each side again.

On the right,

Apply the exponents.

Distributive property

Standard form

Factor.

or Zero-factor property

or Solve each equation.

CHECK Original equation

Let.

Simplify.

Take square roots.

16 = 2 False

9 + 7  2

281 + 249  2

251152 + 6 + 231152 + 4  2 x= 15

25 x+ 6 + 23 x+ 4 = 2

x= 15 x=- 1

x- 15 = 0 x+ 1 = 0

1 x- 1521 x+ 12 = 0

x^2 - 14 x- 15 = 0

x^2 - 2 x+ 1 = 12 x+ 16

x^2 - 2 x+ 1 = 413 x+ 42

x 2 - 2 x+ 1 = 1 - 222 A 23 x+ 4 B 1 ab 22 =a^2 b^2.

2

1 x- 122 = A- 223 x+ 4 B

2

x- 1 =- 223 x+ 4

2 x- 2 =- 423 x+ 4

5 x+ 6 = 8 - 423 x+ 4 + 3 x

5 x+ 6 = 4 - 423 x+ 4 + 3 x+ 4

NOW TRY

EXERCISE 6

Solve. 234 x- 5 = 233 x+ 2

Thus, 15 is an extraneous solution and must be rejected. Confirm that the proposed

solution - 1 checks, so the solution set is 5 - 16.

Divide each

term by 2.

NOW TRY

OBJECTIVE 3 Solve radical equations with indexes greater than 2.

Using the Power Rule for a Power Greater Than 2

Solve

Cube each side.

Subtract z. Add 6.

CHECK Original equation

Let.

✓ True

The solution set is 5116. NOW TRY

2316 = 2316

2311 + 5  232 # 11 - 6 z= 11

23 z+ 5 = 232 z - 6

11 = z

z+ 5 = 2 z- 6

(^) A 23 z + (^5) B
3
= A 232 z - (^6) B
3

23 z+ 5 = 232 z- 6.

EXAMPLE 6

NOW TRY
EXERCISE 5
Solve

23 x+ 1 - 2 x+ 4 = 1.

NOW TRY ANSWERS

5. 556 6. 576