Set 14: Quadratic Equations

Choice D is correct. Choice D is the correct answer. 1. Subtract 9 from both sides: $x^2 = -9$. 2. Take the square root: $x = \pm \sqrt{-9}$. Since the square of any real number is non-negative, there is no real number that squares to -9. The solutions are imaginary ($\pm 3 i$). Choice A is incorrect because $3^2 = 9 \neq -9$. Choice B is incorrect because $(-3)^2 = 9 \neq -9$. Choice C is incorrect because $\pm 3$ are solutions to $x^2 - 9 = 0$.