Solve $LaTeX: \displaystyle x + 1 = \sqrt{4 x + 9}$ .

Squaring both sides gives $LaTeX: \displaystyle x^{2} + 2 x + 1 = 4 x + 9$ . The equation is quadratic setting it equal to zero gives $LaTeX: \displaystyle x^{2} - 2 x - 8 = 0$ . Factoring gives $LaTeX: \displaystyle (x - 4)(x + 2)=0$ so the possible solutions are $LaTeX: \displaystyle x = 4$ and $LaTeX: \displaystyle x = -2$ . Checking the solution $LaTeX: \displaystyle x = 4$ in the original equation gives $LaTeX: \displaystyle 5 = 5$ . The solution checks, so $LaTeX: \displaystyle x = 4$ is a true solution. Checking the solution $LaTeX: \displaystyle x = -2$ in the original equation gives $LaTeX: \displaystyle -1 = 1$ . The solution does no check, so $LaTeX: \displaystyle x = -2$ is an extraneous solution.