Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

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


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\}$

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