1. **State the problem:** Solve the equation $$\sqrt{64x} = x + 12$$ for $x$. 2. **Recall the formula and rules:** The square root function $\sqrt{y}$ returns the non-negative root of $y$. To solve equations involving square roots, we often square both sides to eliminate the root, but we must check for extraneous solutions afterward. 3. **Square both sides:** $$\left(\sqrt{64x}\right)^2 = (x + 12)^2$$ which simplifies to $$64x = (x + 12)^2$$ 4. **Expand the right side:** $$(x + 12)^2 = x^2 + 24x + 144$$ 5. **Rewrite the equation:** $$64x = x^2 + 24x + 144$$ 6. **Bring all terms to one side:** $$0 = x^2 + 24x + 144 - 64x$$ which simplifies to $$0 = x^2 - 40x + 144$$ 7. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-40$, and $c=144$. Calculate the discriminant: $$\Delta = (-40)^2 - 4 \times 1 \times 144 = 1600 - 576 = 1024$$ Calculate the roots: $$x = \frac{40 \pm \sqrt{1024}}{2} = \frac{40 \pm 32}{2}$$ 8. **Find the two possible solutions:** - $$x = \frac{40 + 32}{2} = \frac{72}{2} = 36$$ - $$x = \frac{40 - 32}{2} = \frac{8}{2} = 4$$ 9. **Check for extraneous solutions:** Substitute $x=36$ into the original equation: $$\sqrt{64 \times 36} = \sqrt{2304} = 48$$ and $$36 + 12 = 48$$ True. Substitute $x=4$: $$\sqrt{64 \times 4} = \sqrt{256} = 16$$ and $$4 + 12 = 16$$ True. Both solutions satisfy the original equation. **Final answer:** $$x = 4, 36$$