Solve $LaTeX: \displaystyle x + 3 = \sqrt{3 x + 27}$ .

Squaring both sides gives $LaTeX: \displaystyle x^{2} + 6 x + 9 = 3 x + 27$ . The equation is quadratic setting it equal to zero gives $LaTeX: \displaystyle x^{2} + 3 x - 18 = 0$ . Factoring gives $LaTeX: \displaystyle (x - 3)(x + 6)=0$ so the possible solutions are $LaTeX: \displaystyle x = 3$ and $LaTeX: \displaystyle x = -6$ . Checking the solution $LaTeX: \displaystyle x = 3$ in the original equation gives $LaTeX: \displaystyle 6 = 6$ . The solution checks, so $LaTeX: \displaystyle x = 3$ is a true solution. Checking the solution $LaTeX: \displaystyle x = -6$ in the original equation gives $LaTeX: \displaystyle -3 = 3$ . The solution does no check, so $LaTeX: \displaystyle x = -6$ is an extraneous solution.