Solve the equation $\sqrt{x + 5} = x - 1$.

Squaring both sides gives $x + 5 = x^2 - 2x + 1$, so $x^2 - 3x - 4 = 0$ and $(x-4)(x+1) = 0$. Checking in the original equation: $x = 4$ gives $\sqrt{9} = 3 = 4 - 1$, which holds; $x = -1$ gives $\sqrt{4} = 2$ but $x - 1 = -2$, so it is extraneous. Only $x = 4$ is a solution.