For what positive value of k does the equation x² - kx + 9 = 0 have exactly one real solution?

A quadratic has exactly one real solution when its discriminant is zero. For $x^2 - kx + 9 = 0$ the discriminant is $b^2 - 4ac = k^2 - 4(1)(9) = k^2 - 36$. Setting this equal to 0 gives $k^2 = 36$, so $k = \pm 6$. The question asks for the positive value, so $k = 6$, choice D. Check: $x^2 - 6x + 9 = (x-3)^2$, which has the single root $x = 3$. Choice A (9) uses the constant term. Choice B (3) is the root itself rather than $k$. Choice C (18) comes from forgetting to take the square root, or from computing $2 \cdot 9$.