## College Algebra (6th Edition)

$x = 9$
$$x - 5 = \sqrt {x + 7}$$ $(x - 5)^{2} = (\sqrt {x + 7})^{2}$ $(x^{2} - 10x + 25) = (x + 7)$ $x^{2} - 10x - x + 25 - 7 = 0$ $x^{2} - 11x + 18 = 0$; which can be easily factorized, since two factors of $18$ that, when added, yield $-11$, are ($-9\times-2$): $(x - 9)(x - 2) = 0$ $x = 9$ or $x = 2$ By substituting in the original equation to corroborate, we find that only $x=9$ satisfies the conditions since $x=2$ would produce $\sqrt{9} = -3$.