Solve. $LaTeX: \displaystyle \sqrt{x + 32}=\sqrt{x + 4} - 2$

Squaring both sides gives $LaTeX: \displaystyle x + 32=x - 4 \sqrt{x + 4} + 8$ . Isolating the radical gives $LaTeX: \displaystyle -6=\sqrt{x + 4}$ Squaring again gives $LaTeX: \displaystyle 36=x + 4$ . Solving for $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x=32$ . Checking the solution, $LaTeX: \displaystyle x = 32$ , in the original equation gives $LaTeX: \displaystyle 8 = 4$ which is false so it is an extraneous solution and there is no solution.