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: 52


The solutions are $x=-9$ or $x=1$.

Work Step by Step

$ x^{2}+8x=9\qquad$..add $16$ to both sides to complete the square ($\displaystyle \frac{1}{2}(8)=4$, and $4^{2}=16$.) $ x^{2}+8x+16=9+16\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x+4)^{2}=25$ 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+4=\pm\sqrt{25}\qquad$...add $-4$ to each side. $ x+4-4=\pm\sqrt{25}-4\qquad$...simplify. $x=-4\pm 5$ $x=-4+5$ or $x=-4-5$ $x=1$ or $x=-9$
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