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

Squaring both sides gives $LaTeX: \displaystyle x + 28=x - 4 \sqrt{x + 8} + 12$ . Isolating the radical gives $LaTeX: \displaystyle -4=\sqrt{x + 8}$ Squaring again gives $LaTeX: \displaystyle 16=x + 8$ . Solving for $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x=8$ . Checking the solution, $LaTeX: \displaystyle x = 8$ , in the original equation gives $LaTeX: \displaystyle 6 = 2$ which is false so it is an extraneous solution and there is no solution.