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

$s=25$
$s+3\sqrt{s}-40=0\qquad$...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 $-8$. $s=5^{2}=25$