Explanation Given, x 2 − 6 x = − 1 Step-1: Not necessary. Step-2: Not necessary. Step-3: Make the expression of the left side as a perfect square, Add the square of the half of the coefficient of x on both sides, x 2 − 6 x + ( − 3 ) 2 = − 1 + ( − 3 ) 2 x 2 − 6 x + 3 2 = − 1 + 3 2 x 2 − 6 x + 9 = − 1 + 9 x 2 − 6 x + 9 = 8 Now, we express x 2 − 6 x + 9 in the form of ( a − b ) 2 with a = x and b = 3 . ( x − 3 ) 2 = 8 Step-4: By using the generalized square root property

Prealgebra & Introductory Algebra (5th Edition) (What's New in Developmental Math)

5th Edition

ISBN: 9780134707631

Author: Elayn Martin-Gay

Publisher: PEARSON

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Question

Chapter 16.GRT, Problem 1GRT

To determine

To Choose:

Solve x2−6x=−1 by completing the square, choose the next correct step.

a. x2−6x+9=−1

b. x2−6x+9=−1+9

c. x2−6x−9=−1−9

d. (x−6)2=(−1)2

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Chapter 16.CR, Problem 72CR

Chapter 16.GRT, Problem 2GRT

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Chapter 16 Solutions

Prealgebra & Introductory Algebra (5th Edition) (What's New in Developmental Math)

Ch. 16.1 - Prob. 1P Ch. 16.1 - Prob. 2P Ch. 16.1 - Prob. 3P Ch. 16.1 - Prob. 4P Ch. 16.1 - Prob. 5P Ch. 16.1 - Prob. 6P Ch. 16.1 - Prob. 7P Ch. 16.1 - Prob. 8P Ch. 16.1 - Prob. 9P Ch. 16.1 - Prob. 1E

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To Choose: Solve x 2 − 6 x = − 1 by completing the square, choose the next correct step. a. x 2 − 6 x + 9 = − 1 b. x 2 − 6 x + 9 = − 1 + 9 c. x 2 − 6 x − 9 = − 1 − 9 d. ( x − 6 ) 2 = ( − 1 ) 2

To Choose: Solve x2−6x=−1 by completing the square, choose the next correct step. a. x2−6x+9=−1 b. x2−6x+9=−1+9 c. x2−6x−9=−1−9 d. (x−6)2=(−1)2

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Chapter 16 Solutions

To Choose:

Solve x2−6x=−1 by completing the square, choose the next correct step.

a. x2−6x+9=−1

b. x2−6x+9=−1+9

c. x2−6x−9=−1−9

d. (x−6)2=(−1)2