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

$4$
$w+4\sqrt{w}-12=0\qquad$...substitute $\sqrt{w}$ for $u$ so that $u^{2}=w$. $u^{2}+4u-12=0\qquad$... solve with the Quadractic formula. $a=1,\ b=4,\ c=-12$ $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $4,\ a$ for $1$ and $c$ for $-12$. $u=\displaystyle \frac{-4\pm\sqrt{4^{2}-4\cdot(-12)\cdot 1}}{2\cdot 1}\qquad$... simplify. $u=\displaystyle \frac{-4\pm\sqrt{16+48}}{2}$ $u=\displaystyle \frac{-4\pm\sqrt{64}}{2}$ $u=\displaystyle \frac{-4\pm 8}{2}\qquad$... the symbol $\pm$ indicates two solutions. $u=\displaystyle \frac{-4+8}{2}=\frac{4}{2}=2$ or $u=\displaystyle \frac{-4-8}{2}=\frac{-12}{2}=-6$ Bring back $\sqrt{w}=u$. $\sqrt{w}=2$ or $\sqrt{w}=-6$ $w=4$ or $w=36$ Check to see that $w=36$ can not be a solution. $36+4\sqrt{36}-12=0$ $36+4\cdot 6-12=0$ $36+24-12=0$ $60-12=0$ $48\neq 0$ Therefore, the solution is $4$.