An irrational equation via squaring

Statement

Solve the irrational equation $\sqrt{x + 4} + x = 8$.

Solution

Isolate the root: $\sqrt{x + 4} = 8 - x$.

Conditions. We need $x + 4 \ge 0$ (i.e. $x \ge -4$) and, since the root is non-negative, $8 - x \ge 0$ (i.e. $x \le 8$).

Square both sides: $x + 4 = (8 - x)^2 = 64 - 16x + x^2 \implies x^2 - 17x + 60 = 0.$ Factoring: $(x - 5)(x - 12) = 0$, so $x = 5$ or $x = 12$.

Check. $x = 12$ violates $x \le 8$: substituting, $\sqrt{16} + 12 = 16 \neq 8$, so it is extraneous. Instead $x = 5$: $\sqrt{9} + 5 = 3 + 5 = 8$. Correct. $\boxed{x = 5}$