Solve. $LaTeX: \displaystyle \sqrt{x + 28}=\sqrt{x - 8} + 1$

Squaring both sides gives $LaTeX: \displaystyle x + 28=x + 2 \sqrt{x - 8} - 7$ . Isolating the radical gives $LaTeX: \displaystyle \frac{35}{2}=\sqrt{x - 8}$ Squaring again gives $LaTeX: \displaystyle \frac{1225}{4}=x - 8$ . Solving for $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x=\frac{1257}{4}$ . Checking the solution, $LaTeX: \displaystyle x = \frac{1257}{4}$ , in the original equation gives $LaTeX: \displaystyle \frac{37}{2} = \frac{37}{2}$ which is true so the solution checks.