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: 26

Answer

$s=9+4\sqrt{5}$

Work Step by Step

$ s-4\sqrt{s}-1=0\qquad$...substitute $\sqrt{s}$ for $u$ so that $u^{2}=s$ Note that $s$ and $\sqrt{s}$ are positive. . $ u^{2}-4u-1=0\qquad$... solve with the Quadractic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{4\pm\sqrt{16+4}}{2}$ $u=\displaystyle \frac{4\pm\sqrt{20}}{2}$ $u=\displaystyle \frac{4\pm 2\sqrt{5}}{2}$ $u=\displaystyle \frac{4+2\sqrt{5}}{2}=2+\sqrt{5}$ or $u=\displaystyle \frac{4-2\sqrt{5}}{2}=2-2\sqrt{5}$ Bring back $\sqrt{s}=u$. $\sqrt{s}=2+\sqrt{5}$ or $\sqrt{s}=2-2\sqrt{5}$ $\sqrt{s}$ is a positive number which is why we discard $2-2\sqrt{5}$. $s=(2+\sqrt{5})^{2}=4+4\sqrt{5}+5=9+4\sqrt{5}$
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