Solve. $LaTeX: \displaystyle \sqrt{x + 32}=\sqrt{x + 6} + 1$

Squaring both sides gives $LaTeX: \displaystyle x + 32=x + 2 \sqrt{x + 6} + 7$ . Isolating the radical gives $LaTeX: \displaystyle \frac{25}{2}=\sqrt{x + 6}$ Squaring again gives $LaTeX: \displaystyle \frac{625}{4}=x + 6$ . Solving for $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x=\frac{601}{4}$ . Checking the solution, $LaTeX: \displaystyle x = \frac{601}{4}$ , in the original equation gives $LaTeX: \displaystyle \frac{27}{2} = \frac{27}{2}$ which is true so the solution checks.