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 - Mid-Chapter Review - Mixed Review - Page 733: 3


The solutions are $x=3$ or $x=-7$.

Work Step by Step

$ x^{2}+4x=21\qquad$..add $4$ to both sides to complete the square ($\displaystyle \frac{1}{2}(4)=2$, and $2^{2}=4$.) $ x^{2}+4x+4=21+4\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x+2)^{2}=25$ 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+2=\pm\sqrt{25}\qquad$...add $-2$ to each side. $ x+2-2=\pm\sqrt{25}-2\qquad$...simplify. $x=-2\pm 5$ $x=-2+5$ or $x=-2-5$ $x=3$ or $x=-7$ The solutions are $x=3$ or $x=-7$.
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