For what values of k does the equation x² + kx + 16 = 0 have two distinct real solutions?

A quadratic has two distinct real solutions when its discriminant is positive. For $x^2 + kx + 16 = 0$, the discriminant is $k^2 - 4(1)(16) = k^2 - 64$. Requiring $k^2 - 64 > 0$ gives $k^2 > 64$, so $k > 8$ or $k < -8$, choice D. Choice A ($-8 < k < 8$) is the opposite case, with no real solutions. Choice B ($k = \pm 8$) makes the discriminant exactly zero, which gives one repeated solution. Choice C ($k > 16$) uses the constant term instead of 8 and misses all the negative values of $k$ below $-8$.