1. **State the problem:** Solve the equation $$\sqrt{3n + 11} = \sqrt{n + 5}$$ and check for extraneous solutions. 2. **Formula and rules:** To solve equations involving square roots, we square both sides to eliminate the radicals, but we must check for extraneous solutions because squaring can introduce invalid answers. 3. **Square both sides:** $$\left(\sqrt{3n + 11}\right)^2 = \left(\sqrt{n + 5}\right)^2$$ $$3n + 11 = n + 5$$ 4. **Simplify the equation:** $$3n + 11 = n + 5$$ Subtract $n$ from both sides: $$3n - n + 11 = 5$$ $$2n + 11 = 5$$ 5. **Isolate $n$:** $$2n = 5 - 11$$ $$2n = -6$$ Divide both sides by 2: $$n = \frac{\cancel{2}n}{\cancel{2}} = \frac{-6}{2}$$ $$n = -3$$ 6. **Check for extraneous solutions:** Substitute $n = -3$ back into the original equation: $$\sqrt{3(-3) + 11} = \sqrt{-3 + 5}$$ $$\sqrt{-9 + 11} = \sqrt{2}$$ $$\sqrt{2} = \sqrt{2}$$ This is true, so $n = -3$ is a valid solution. **Final answer:** $$n = -3$$