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: 35


The solutions are $x=-4+\sqrt{13}$ or $x=-4-\sqrt{13}$.

Work Step by Step

$F(t)=(t+4)^{2}$ $ F(t)=13\qquad$...substitute $13$ for $F(t)$. $(t+4)^{2}=13$ 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+4=\pm\sqrt{13}\qquad$...add $-4$ to each side. $ x+4-4=\pm\sqrt{13}-4\qquad$...simplify. $x=-4\pm\sqrt{13}$ $x=-4+\sqrt{13}$ or $x=-4-\sqrt{13}$
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