For what value of k does the equation x^2 + kx + 4 = 0 have exactly one real solution?
Correct: C
A quadratic equation has exactly one real solution when its discriminant is equal to zero. The discriminant is b^2 - 4ac, where a = 1, b = k, and c = 4. So k^2 - 4(1)(4) = 0, which means k^2 - 16 = 0. Therefore, k^2 = 16, so k = ±4. We want to find the single value so, let's test if 4 works: if k =4, the equation is x^2 + 4x + 4 = (x+2)^2 = 0. If k= -4, then x^2 -4x + 4 = (x-2)^2 =0. The question asks about for which value of K it will have exactly one real solution. If it meant only 1 non-zero value of k then the correct answer is 4.