For what value of k does the quadratic equation x^2 - 6x + k = 0 have exactly one distinct real root?

✓ Correct Answer: Option C

A quadratic equation (ax^2 + bx + c = 0) has exactly one distinct real root when its discriminant (b^2 - 4ac) is equal to zero. Here, a=1, b=-6, and c=k. So, (-6)^2 - 4(1)(k) = 0, which means 36 - 4k = 0. Solving for k gives k = 9.