1. **State the problem:** Solve the equation $$\sqrt{7 + x} + 2 = \sqrt{3 - x}$$ for $x$. 2. **Isolate one square root:** Subtract 2 from both sides: $$\sqrt{7 + x} = \sqrt{3 - x} - 2$$ 3. **Square both sides to eliminate the square root:** $$\left(\sqrt{7 + x}\right)^2 = \left(\sqrt{3 - x} - 2\right)^2$$ $$7 + x = (\sqrt{3 - x})^2 - 2 \times 2 \times \sqrt{3 - x} + 2^2$$ $$7 + x = 3 - x - 4\sqrt{3 - x} + 4$$ 4. **Simplify the right side:** $$7 + x = 7 - x - 4\sqrt{3 - x}$$ 5. **Bring like terms to one side:** $$7 + x - 7 + x = -4\sqrt{3 - x}$$ $$2x = -4\sqrt{3 - x}$$ 6. **Divide both sides by -4:** $$\frac{2x}{-4} = \sqrt{3 - x}$$ $$\cancel{\frac{2}{-4}}x = \sqrt{3 - x}$$ $$-\frac{1}{2}x = \sqrt{3 - x}$$ 7. **Square both sides again to eliminate the square root:** $$\left(-\frac{1}{2}x\right)^2 = (\sqrt{3 - x})^2$$ $$\frac{x^2}{4} = 3 - x$$ 8. **Multiply both sides by 4 to clear the denominator:** $$x^2 = 12 - 4x$$ 9. **Bring all terms to one side to form a quadratic equation:** $$x^2 + 4x - 12 = 0$$ 10. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-12)}}{2 \times 1}$$ $$x = \frac{-4 \pm \sqrt{16 + 48}}{2}$$ $$x = \frac{-4 \pm \sqrt{64}}{2}$$ $$x = \frac{-4 \pm 8}{2}$$ 11. **Calculate the two possible solutions:** - $$x = \frac{-4 + 8}{2} = \frac{4}{2} = 2$$ - $$x = \frac{-4 - 8}{2} = \frac{-12}{2} = -6$$ 12. **Check for extraneous solutions by substituting back into the original equation:** - For $x=2$: $$\sqrt{7 + 2} + 2 = \sqrt{9} + 2 = 3 + 2 = 5$$ $$\sqrt{3 - 2} = \sqrt{1} = 1$$ Not equal, so $x=2$ is extraneous. - For $x=-6$: $$\sqrt{7 - 6} + 2 = \sqrt{1} + 2 = 1 + 2 = 3$$ $$\sqrt{3 - (-6)} = \sqrt{9} = 3$$ Equal, so $x=-6$ is a valid solution. **Final answer:** $$x = -6$$