Beginning Algebra, 11th Edition

Step 4 Solve by using the square root property.

or Square root property

or Add

or Simplify.

or Lowest terms

A check confirms that the solution set is E-1,^92 F. NOW TRY

x= x=- 1

9

2

x=-

4

4

x=

18

4

7

4 ; 2

121

16 =

11

x= 4.

7

4

-

11

4

x=

7

4

+

11

4

x-

7

4

=-

B

121

16

x-

7

4

=

B

121

16

564 CHAPTER 9 Quadratic Equations

Factor on the left.

Add on the right.

NOW TRY
EXERCISE 6
Solve 3t^2 - 12 t+ 15 =0.

NOW TRY ANSWERS

5. E-3,^34 F 6. 0

Solving a Quadratic Equation by Completing the Square

Solve

Divide by 4.

p 2 + 2 p=- Add - 45 to each side.

5

4

p^2 + 2 p+

5

4

= 0

4 p^2 + 8 p+ 5 = 0

4 p^2 + 8 p+ 5 =0.

EXAMPLE 6

The coefficient of

the second-degree

term must be 1.

The coefficient of pis 2. Take half of 2, square the result, and add it to each side.

; Add 1.

If we apply the square root property to solve this equation, we get the square root of

• , which is not a real number. The solution set is. 0 NOW TRY
  1
  4

1 p+ 122 =-

1

4

p 2 + 2 p+ 1 =- C 21 122 D^2 = 12 = 1

5

4

+ 1

OBJECTIVE 3 Simplify the terms of an equation before solving.

Simplifying the Terms of an Equation before Solving

Solve

Multiply by using the FOIL method.

Add 3.

Add

Factor on the left. Add on the right.

or Square root property

or Subtract 1.

The solution set is E- 1  (^26) F. NOW TRY
x=- 1 + 26 x =- 1 - 26
x+ 1 = 26 x + 1 =- 26
1 x+ 122 = 6
x^2 + 2 x+ 1 = 5 + 1 C^12 122 D^2 = 12 =1.
x^2 + 2 x= 5
x^2 + 2 x- 3 = 2
1 x+ 321 x- 12 = 2
1 x+ 321 x- 12 =2.
NOW TRY EXAMPLE 7
EXERCISE 7
Solve 1 x- 521 x+ 12 =2.
7. E 2  (^211) F
NOW TRY
EXERCISE 5
Solve 4x^2 + 9 x- 9 =0.
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