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: 33


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

Work Step by Step

$f(x)=(x-5)^{2}$ $f(x)=16\qquad$...substitute $16$ for $f(x)$ $(x-5)^{2}=16$ 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-5=\pm\sqrt{16}\qquad$...add $5$ to each side. $ x-5+5=\pm\sqrt{16}+5\qquad$...simplify. $x=5\pm 4$ $x=5+4$ or $x=5-4$ $x=9$ or $x=1$
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