Question: Solve for $x$: $x^2 = -2(3 - 2)$
A. $-3 \pm \sqrt{26}$
B. $3 \pm 2\sqrt{13}$
C. $-3 \pm \sqrt{13}$
D. $3 \pm \sqrt{5}$
1. **State the problem:** Solve for $x$ in the equation $$x^2 = -2(3 - 2)$$.
2. **Simplify the right side:** Calculate inside the parentheses first:
$$3 - 2 = 1$$
So the equation becomes:
$$x^2 = -2 \times 1$$
$$x^2 = -2$$
3. **Analyze the equation:** We have $$x^2 = -2$$.
4. **Recall the rule:** The square of a real number cannot be negative. Since $$-2$$ is negative, there are no real solutions.
5. **Consider complex solutions:** If we allow complex numbers, then:
$$x = \pm \sqrt{-2} = \pm \sqrt{-1 \times 2} = \pm \sqrt{-1} \times \sqrt{2} = \pm i \sqrt{2}$$
6. **Check the answer choices:** None of the options A, B, C, or D match $$\pm i \sqrt{2}$$ or any complex number.
7. **Conclusion:** There is no real solution matching the given options. The problem as stated has no real solution.
**Final answer:** No real solution matches the options provided.