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 - 11.5 Equations Reducible to Quadratic - 11.5 Exercise Set - Page 731: 18



Work Step by Step

$ s+3\sqrt{s}-40=0\qquad$...substitute $\sqrt{s}$ for $u$ so that $u^{2}=s$ Note that $s$ and $\sqrt{s}$ are positive. . $ u^{2}+3u-40=0\qquad$... solve with the Quadractic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{-3\pm\sqrt{9+160}}{2}$ $u=\displaystyle \frac{-3\pm\sqrt{169}}{2}$ $u=\displaystyle \frac{-3\pm 13}{2}$ $u=\displaystyle \frac{-3+13}{2}=\frac{10}{2}=5$ or $u=\displaystyle \frac{-3-13}{2}=\frac{-16}{2}=-8$ Bring back $\sqrt{s}=u$. $\sqrt{s}=5$ or $\sqrt{s}=-8$ $\sqrt{s}$ is a positive number which is why we discard $-8$. $s=5^{2}=25$
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