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: 25



Work Step by Step

$ r-2\sqrt{r}-6=0\qquad$...substitute $\sqrt{r}$ for $u$ so that $u^{2}=r$. $ u^{2}+2u-6=0\qquad$... solve with the Quadratic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{-2\pm\sqrt{4+24}}{2}$ $u=\displaystyle \frac{-2\pm\sqrt{28}}{2}$ $ u=\displaystyle \frac{-2\pm 2\sqrt{7}}{2}\qquad$... the symbol $\pm$ indicates two solutions. $u=\displaystyle \frac{-2+2\sqrt{7}}{2}=-1+\sqrt{7}$ or $u=\displaystyle \frac{-2-2\sqrt{7}}{2}=-1-\sqrt{7}$ Bring back $\sqrt{r}=u$. $\sqrt{r}=-1+\sqrt{7}$ or $\sqrt{r}=-1-\sqrt{7}$ $\sqrt{r}$ is a positive number which is why we discard $-1-\sqrt{7}$. $r=(-1+\sqrt{7})^{2}=7-2\sqrt{7}+1=8+2\sqrt{7}$ $r=8+2\sqrt{7}$
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