## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.7 - Equations - Exercise Set - Page 108: 152

#### Answer

Observe the first two terms on the LHS, $x^{2}+6x=x^{2}+2(x)(3)$ We wrote the second term in this form to be in line with the formula $(a+b)^{2}=a^{2}+\fbox{$2ab $}+b^{2}$ So, our b is 3, and the missing $b^{2}$ term equals $3^{2}=9$. We now rewrite the equation by adding $+9$ to both sides $x^{2}+6x+9+8= 9\quad$... recognize square, add $-8$ $(x+3)^{2}=9-8$ $(x+3)^{2}=1\qquad$... take the square root, $x+3$ can be -1 or 1 $x+3=\pm 1\qquad$... isolate x, subtract 3 $x=-3\pm 1$ $x=-4$ or $x=-2$ Solution set =$\{-4,-2\}$

#### Work Step by Step

Given above.

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