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: 55


The solutions are $x=-8$ or $x=-4$.

Work Step by Step

$ x^{2}+12x+32=0\qquad$..add $-32$ to both sides so we can complete the square on the left side. $ x^{2}+12x=-32\qquad$..add $36$ to both sides to complete the square ($\displaystyle \frac{1}{2}(12)=6$, and $(6)^{2}=36$.) $ x^{2}+12x+36=-32+36\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x+6)^{2}=4$ 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+6=\pm\sqrt{4}\qquad$...add $-6$ to each side. $ x+6-6=\pm\sqrt{4}-6\qquad$...simplify. $x=-6\pm 2$ $x=-6+2$ or $t=-6-2$ $x=-4$ or $x=-8$
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