Solve. $LaTeX: \displaystyle \sqrt{x + 12}=\sqrt{x + 4} + 2$

Squaring both sides gives $LaTeX: \displaystyle x + 12=x + 4 \sqrt{x + 4} + 8$ . Isolating the radical gives $LaTeX: \displaystyle 1=\sqrt{x + 4}$ Squaring again gives $LaTeX: \displaystyle 1=x + 4$ . Solving for $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x=-3$ . Checking the solution, $LaTeX: \displaystyle x = -3$ , in the original equation gives $LaTeX: \displaystyle 3 = 3$ which is true so the solution checks.