## Intermediate Algebra (6th Edition)

$7$
$\sqrt (x-3) + \sqrt (x+2) = 5$ Squaring on both sides to eliminate radicals. $(\sqrt (x-3) + \sqrt (x+2))^{2} = 5^{2}$ $(x-3)+(x+2)+2\sqrt (x-3) \sqrt (x+2) = 25$ $2x-1+2\sqrt (x-3) \sqrt (x+2) = 25$ $2\sqrt (x-3) \sqrt (x+2) = 25+1-2x$ $2\sqrt (x-3) \sqrt (x+2) = 26-2x$ $2\sqrt (x-3) \sqrt (x+2) = 2(13-x)$ $\sqrt (x-3) \sqrt (x+2) = (13-x)$ Squaring on both sides to eliminate radicals. $(\sqrt (x-3) \sqrt (x+2))^{2} = (13-x)^{2}$ $(x-3) (x+2) =169-26x+x^{2}$ $x^{2}-x-6 = 169-26x+x^{2}$ $x^{2}-x-6 - 169+26x-x^{2}=0$ $25x-175=0$ $25x=175$ $x = 7$