## Precalculus (6th Edition) Blitzer

$\{6\}$
The LHS is nonnegative, so the RHS = $x-3$ is also nonnegative. All solutions must satisfy $x\geq 3\qquad (*)$ $\sqrt{x+3}=x-3 \quad$... Square both sides $x+3=x^{2}-6x+9$ $x^{2}-7x+6=0$ We factor this trinomial by finding two factors of $+6$ that add to $-7$; they are $-6$ and $-1$. $(x-6)(x-1)=0$ $x=6,\quad x=1$ $x=1 \;$ is extraneous (does not satisfy (*) ), so we discard it. $x=6$ is a valid solution Check: $\sqrt{6+3}=6-3$ $\sqrt{9}=3$ The solution set is $\{6\}$.