1. **State the problem:** Solve the equation $$\sqrt{x-2} = 8 - x$$ for $x$. 2. **Understand the domain:** The expression under the square root must be non-negative, so: $$x - 2 \geq 0 \implies x \geq 2$$ Also, the right side $8 - x$ must be non-negative because the left side is a square root (which is always non-negative): $$8 - x \geq 0 \implies x \leq 8$$ So the domain is $2 \leq x \leq 8$. 3. **Square both sides to eliminate the square root:** $$\left(\sqrt{x-2}\right)^2 = (8 - x)^2$$ $$x - 2 = (8 - x)^2$$ 4. **Expand the right side:** $$(8 - x)^2 = 64 - 16x + x^2$$ So the equation becomes: $$x - 2 = 64 - 16x + x^2$$ 5. **Bring all terms to one side:** $$0 = 64 - 16x + x^2 - x + 2$$ $$0 = x^2 - 17x + 66$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 1 \cdot 66}}{2} = \frac{17 \pm \sqrt{289 - 264}}{2} = \frac{17 \pm \sqrt{25}}{2}$$ 7. **Calculate the roots:** $$x = \frac{17 + 5}{2} = \frac{22}{2} = 11$$ $$x = \frac{17 - 5}{2} = \frac{12}{2} = 6$$ 8. **Check roots against the domain and original equation:** - For $x=11$, it is outside the domain $2 \leq x \leq 8$, so discard. - For $x=6$, check original equation: $$\sqrt{6 - 2} = \sqrt{4} = 2$$ $$8 - 6 = 2$$ Both sides equal 2, so $x=6$ is a valid solution. **Final answer:** $$\boxed{6}$$