For what negative value of k does the equation x² + kx + 16 = 0 have exactly one real solution?

A quadratic has exactly one real solution when its discriminant is zero. For $x^2 + kx + 16 = 0$, the discriminant is $k^2 - 4(1)(16) = k^2 - 64$. Setting it to zero gives $k^2 = 64$, so $k = \pm 8$. The question asks for the negative value, so $k = -8$, choice A. Check: $x^2 - 8x + 16 = (x-4)^2$. Choice B (8) is the positive value, which also works but is not what was requested. Choice C (16) is the constant term. Choice D ($-16$) comes from forgetting to take the square root, or from computing $-4 \times 4$.