## Precalculus (6th Edition) Blitzer

$\{-6\}$
For the root to be real, x must be such that $x\geq-13/2.$ The LHS is nonnegative, so the RHS = $x+7 \;$ is also nonnegative. All solutions must satisfy $x\geq-13/2\qquad (*)$ $2x+13=(x+7)^{2}$ $2x+13=x^{2}+14x+49$ $x^{2}+12x+36=0 \quad$... recognize a perfect square $(x+6)^{2}=0$ $x=-6 \;$ is a valid solution (satisfies (*) ) Check: $\sqrt{2(-6)+13}=-6+7$ $\sqrt{-12+13}=1$ $\sqrt{1}=1$ The solution set is $\{-6\}$.