Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 706: 37


The solutions are $x=-14$ or $x=0$.

Work Step by Step

$g(x)=x^{2}+14x+49$ $ g(x)=49\qquad$...substitute $49$ for $g(x)$. $49=x^{2}+14x+49$ Apply Perfect square formula: $(x+a)^{2}=x^{2}+2ax+a^{2}$, $49=(x+7)^{2}$ According to the general principle of square roots: For any real number $k$ and any algebraic expression $x$ : $\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$. $ x+7=\pm\sqrt{49}\qquad$...add $-7$ to each side. $ x+7-7=\pm\sqrt{49}-7\qquad$...simplify. $x=-7\pm 7$ $x=0$ or $x=-14$
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