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: 36

Answer

The solutions are $x=-6+\sqrt{15}$ or $x=-6-\sqrt{15}$.

Work Step by Step

$f(t)=(t+6)^{2}$ $ f(t)=15\qquad$...substitute $15$ for $f(t)$. $(t+6)^{2}=15$ According to the general principle of square roots: For any real number $k$ and any algebraic expression $x$ : $\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$. $ t+6=\pm\sqrt{15}\qquad$...add $-6$ to each side. $ x+6-6=\pm\sqrt{15}-6\qquad$...simplify. $x=-6\pm\sqrt{15}$ $x=-6+\sqrt{15}$ or $x=-6-\sqrt{15}$
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