Solve. $LaTeX: \displaystyle \sqrt{x + 28}=\sqrt{x - 8} + 2$

Squaring both sides gives $LaTeX: \displaystyle x + 28=x + 4 \sqrt{x - 8} - 4$ . Isolating the radical gives $LaTeX: \displaystyle 8=\sqrt{x - 8}$ Squaring again gives $LaTeX: \displaystyle 64=x - 8$ . Solving for $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x=72$ . Checking the solution, $LaTeX: \displaystyle x = 72$ , in the original equation gives $LaTeX: \displaystyle 10 = 10$ which is true so the solution checks.