Step 3 The new equation is quadratic, so write it in standard form.
Equation from Step 2
Subtract 4. Add x.
Factor.or Zero-factor property
x=- 5 Solve for x.
x= 0 x+ 5 = 0
x 1 x+ 52 = 0
x^2 + 5 x= 0
4 - x=x^2 + 4 x+ 4
470 CHAPTER 8 Roots, Radicals, and Root Functions
NOW TRY
EXERCISE 3Solve. 216 - x=x+ 4
Step 4 Check each proposed solution in the original equation.
CHECK
Let.2 = 2 ✓ True
24 2
24 - 0 0 + 2 x= 0
24 - x= x+ 2
Set eachfactor
equal to 0.Let.3 =- 3 False
29 - 3
24 - 1 - 52 - 5 + 2 x=- 5
24 - x=x+ 2
The solution set is. The other proposed solution, is extraneous. NOW TRY
Using the Power Rule (Squaring a Binomial)Solve
Squaring gives on the right.
Twice the product of xand
Subtract and 9. Add 2x.
Divide by.CHECK Original equation
Let.3 = 3 ✓ True
242 - 4 # 4 + 9 4 - 1 x= 4
2 x^2 - 4 x+ 9 =x- 1
x= 4 - 2
- 2 x=- 8 x^2
- 1
x^2 - 4 x+ 9 =x^2 - 2 x+ 1
A 2 x^2 - 4 x+ 9 B
2= 1 x- 122
1 x- 122 = x^2 - 21 x 2112 + 12
2 x^2 - 4 x+ 9 = x- 1.
EXAMPLE 4
506 - 5,
NOW TRY
EXERCISE 4Solve
2 x^2 - 3 x+ 18 =x+ 3.The solution set is. NOW TRY
Using the Power Rule (Squaring Twice)Solve
Isolate one radical on one side of the equation by subtracting from
each side.
SubtractSquare each side.Twice the product of 2 and - 23 x+ 45 x+ 6 = 4 - 423 x+ 4 + 13 x+ 42
(^) A 25 x+ (^6) B
2
= A 2 - 23 x+ (^4) B
2
25 x+ 6 = 2 - 23 x+ 4 23 x+ 4.
23 x+ 4
25 x+ 6 + 23 x+ 4 =2.
EXAMPLE 5
546
Remember the
middle term.Remember the
middle term.Be careful here.NOW TRY ANSWERS
- 506 4. 516