1. **State the problem:** Solve the equation $$\sqrt{x + 2} + 1 = x - 3$$ for $x$. 2. **Isolate the square root:** Subtract 1 from both sides: $$\sqrt{x + 2} = x - 4$$ 3. **Check domain:** Since the left side is a square root, $x + 2 \geq 0 \Rightarrow x \geq -2$. Also, the right side must be non-negative because the square root is always non-negative, so: $$x - 4 \geq 0 \Rightarrow x \geq 4$$ 4. **Square both sides:** To eliminate the square root, square both sides: $$\left(\sqrt{x + 2}\right)^2 = (x - 4)^2$$ $$x + 2 = (x - 4)^2$$ 5. **Expand the right side:** $$x + 2 = x^2 - 8x + 16$$ 6. **Bring all terms to one side:** $$0 = x^2 - 8x + 16 - x - 2$$ $$0 = x^2 - 9x + 14$$ 7. **Factor the quadratic:** $$0 = (x - 7)(x - 2)$$ 8. **Solve for $x$:** $$x = 7 \quad \text{or} \quad x = 2$$ 9. **Check solutions in original equation:** - For $x=7$: $$\sqrt{7 + 2} + 1 = \sqrt{9} + 1 = 3 + 1 = 4$$ $$7 - 3 = 4$$ Both sides equal 4, so $x=7$ is a solution. - For $x=2$: $$\sqrt{2 + 2} + 1 = \sqrt{4} + 1 = 2 + 1 = 3$$ $$2 - 3 = -1$$ Left side is 3, right side is -1, not equal, so $x=2$ is extraneous. **Final answer:** $x = 7$