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

Squaring both sides gives $LaTeX: \displaystyle x^{2} + 8 x + 16 = 2 x + 32$ . The equation is quadratic setting it equal to zero gives $LaTeX: \displaystyle x^{2} + 6 x - 16 = 0$ . Factoring gives $LaTeX: \displaystyle (x - 2)(x + 8)=0$ so the possible solutions are $LaTeX: \displaystyle x = 2$ and $LaTeX: \displaystyle x = -8$ . Checking the solution $LaTeX: \displaystyle x = 2$ in the original equation gives $LaTeX: \displaystyle 6 = 6$ . The solution checks, so $LaTeX: \displaystyle x = 2$ is a true solution. Checking the solution $LaTeX: \displaystyle x = -8$ in the original equation gives $LaTeX: \displaystyle -4 = 4$ . The solution does no check, so $LaTeX: \displaystyle x = -8$ is an extraneous solution.