## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$r=8+2\sqrt{7}$
$r-2\sqrt{r}-6=0\qquad$...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 $-1-\sqrt{7}$. $r=(-1+\sqrt{7})^{2}=7-2\sqrt{7}+1=8+2\sqrt{7}$ $r=8+2\sqrt{7}$