Solve $LaTeX: \displaystyle x + 4 = \sqrt{6 x + 64}$ .

Squaring both sides gives $LaTeX: \displaystyle x^{2} + 8 x + 16 = 6 x + 64$ . The equation is quadratic setting it equal to zero gives $LaTeX: \displaystyle x^{2} + 2 x - 48 = 0$ . Factoring gives $LaTeX: \displaystyle (x - 6)(x + 8)=0$ so the possible solutions are $LaTeX: \displaystyle x = 6$ and $LaTeX: \displaystyle x = -8$ . Checking the solution $LaTeX: \displaystyle x = 6$ in the original equation gives $LaTeX: \displaystyle 10 = 10$ . The solution checks, so $LaTeX: \displaystyle x = 6$ 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.