What is the sum of all real solutions to x^4 - 13x² + 36 = 0?

Treat the equation as a quadratic in $x^2$: $x^4 - 13x^2 + 36 = (x^2 - 4)(x^2 - 9) = 0$. So $x^2 = 4$ or $x^2 = 9$, which gives $x = \pm 2$ or $x = \pm 3$. The four real solutions are $-3, -2, 2, 3$, and they cancel in pairs, so the sum is 0, choice B. Choice A (13) is the coefficient in the equation, and choice C (36) is the constant term; neither is a sum of solutions. Choice D (6) comes from adding only the positive roots, $2 + 3 = 5$, with an error, or from counting $x^2$ values rather than $x$ values. Each positive $x^2$ gives two opposite roots.