Solve. $LaTeX: \displaystyle \sqrt{x + 28}=\sqrt{x + 8} + 3$

Squaring both sides gives $LaTeX: \displaystyle x + 28=x + 6 \sqrt{x + 8} + 17$ . Isolating the radical gives $LaTeX: \displaystyle \frac{11}{6}=\sqrt{x + 8}$ Squaring again gives $LaTeX: \displaystyle \frac{121}{36}=x + 8$ . Solving for $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x=- \frac{167}{36}$ . Checking the solution, $LaTeX: \displaystyle x = - \frac{167}{36}$ , in the original equation gives $LaTeX: \displaystyle \frac{29}{6} = \frac{29}{6}$ which is true so the solution checks.