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: 51


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Work Step by Step

$ x^{2}+6x=7\qquad$..add $9$ to both sides to complete the square ($\displaystyle \frac{1}{2}(6)=3$, and $3^{2}=9$.) $ x^{2}+6x+9=7+9\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x+3)^{2}=16$ 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+3=\pm\sqrt{16}\qquad$...add $-3$ to each side. $ x+3-3=\pm\sqrt{16}-3\qquad$...simplify. $x=-3\pm 4$ $x=-3+4$ or $x=-3-4$ $x=1$ or $x=-7$
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