Beginning Algebra, 11th Edition

The equation is quadratic because of the -term. To solve it,

as shown in Section 6.5,we must write the equation in standard form.

Standard form

Factor.

or Zero-factor property

or Proposed solutions

CHECK

Let Let

✓ True False

Only 6 is a valid solution. (2 is extraneous.) The solution set is 566. NOW TRY

3 = 3 1 - 1

29  3 21 - 1

212 - 3  3 24 - 3 - 1

22162 - 3  6 - 3 x=6. 22122 - 3  2 - 3 x=2.

22 x- 3 =x- 3 22 x- 3 =x- 3

x= 6 x= 2

x- 6 = 0 x - 2 = 0

1 x- 621 x- 22 = 0

x^2 - 8 x+ 12 = 0

2 x- 3 =x^2 - 6 x+ 9 x^2

534 CHAPTER 8 Roots and Radicals

NOW TRY
EXERCISE 5
Solve.

24 x+ 1 =x- 5

NOW TRY ANSWERS

5. 5126 6.E 43 , 3F

NOW TRY

EXERCISE 6

Solve 227 x- 3 = 2 x.

Subtract 2x, add 3, and

interchange sides.

Rewriting an Equation before Using the Squaring Property

Solve

We must apply Step 1 here, and isolate the radical beforesquaring each side. If

we skip Step 1 and begin by squaring instead, we obtain

a more complicated equation that still contains a radical. Follow the steps below.

Add 1 to isolate the radical. (Step 1)

Square both sides. (Step 2)

No terms contain radicals.

Standard form (Step 3)

Factor. (Step 5; Step 4 is not needed.)

or Zero-factor property

or Proposed solutions

CHECK (Step 6) (Step 6)

Let Let

✓ True ✓ True

Both proposed solutions check, so the solution set is E^14 , 1F. NOW TRY

2 = 2

1

2

=

1

2

x= 41. 29112 - 1  2112 x=1.

B

9 a

1

4

b - 1  2 a

1

4

b

29 x- 1 = 2 x 29 x- 1 = 2 x

x= x= 1

1

4

4 x- 1 = 0 x - 1 = 0

14 x- 121 x- 12 = 0

4 x^2 - 5 x+ 1 = 0

9 x= 4 x^2 + 4 x+ 1 1 x+y 22 =x^2 + 2 xy+y^2

(^) A 29 xB^2 = 12 x+ 122
29 x= 2 x+ 1
29 x- 1 = 2 x
9 x- 229 x+ 1 = 4 x^2 ,
A 29 x- 1 B
2
= 12 x 22
29 x- 1 = 2 x.
EXAMPLE 6
This is a
key step.
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