1. **State the problem:** Solve the equation $x + 5 = \sqrt{5x + 21}$ for $x$. 2. **Recall the formula and rules:** To solve equations involving square roots, we isolate the square root and then square both sides to eliminate it. However, squaring can introduce extraneous solutions, so we must check all solutions in the original equation. 3. **Isolate and square:** $$x + 5 = \sqrt{5x + 21}$$ Square both sides: $$(x + 5)^2 = 5x + 21$$ 4. **Expand and simplify:** $$(x + 5)^2 = x^2 + 10x + 25$$ So, $$x^2 + 10x + 25 = 5x + 21$$ Bring all terms to one side: $$x^2 + 10x + 25 - 5x - 21 = 0$$ $$x^2 + 5x + 4 = 0$$ 5. **Factor the quadratic:** $$x^2 + 5x + 4 = (x + 4)(x + 1) = 0$$ So, $$x = -4 \text{ or } x = -1$$ 6. **Check for extraneous solutions:** - For $x = -4$: $$LHS = -4 + 5 = 1$$ $$RHS = \sqrt{5(-4) + 21} = \sqrt{-20 + 21} = \sqrt{1} = 1$$ Valid solution. - For $x = -1$: $$LHS = -1 + 5 = 4$$ $$RHS = \sqrt{5(-1) + 21} = \sqrt{-5 + 21} = \sqrt{16} = 4$$ Valid solution. **Final answer:** $-4, -1$