1. **State the problem:** Solve the equation $x=\sqrt{2-x}$ for $x$. 2. **Square both sides to eliminate the square root:** $$x^2 = (\sqrt{2-x})^2$$ $$x^2 = 2 - x$$ 3. **Rearrange the equation to standard quadratic form:** $$x^2 + x - 2 = 0$$ 4. **Factor the quadratic equation:** $$x^2 + x - 2 = (x+2)(x-1) = 0$$ 5. **Find the roots:** $$x+2=0 \Rightarrow x=-2$$ $$x-1=0 \Rightarrow x=1$$ 6. **Check for extraneous solutions by substituting back into the original equation:** - For $x=-2$: $$-2 \stackrel{?}{=} \sqrt{2 - (-2)} = \sqrt{4} = 2$$ This is false, so $x=-2$ is extraneous. - For $x=1$: $$1 \stackrel{?}{=} \sqrt{2 - 1} = \sqrt{1} = 1$$ This is true, so $x=1$ is a valid solution. **Final answer:** $$\boxed{1}$$