Algebra and Trigonometry question. Please show work!! Serious Replies only..Thankyou very much!!!?

Question

Write a quadratic equation having solutions (square root 3) and (-4 square root 3)

a).x^2 + 3 square root 3x - 12 = 0
b). x^2 -12 =0
c). x^2 - 4 square root 3x + 12 =0
d).x^2 - 4 square root 3x -36 =0

Answer 1

(x - √3)(x + 4√3) = 0
x² + (3√3)x - 12 = 0

The answer is a).

Answer 2

If the solutions of a quadratic are a and b then the quadratic must be (x - a)(x - b) = x^2 -(a + b)x + ab.
So your quadratic is (x - sqrt(3))(x + 4sqrt(3)) =
x^2 + (4sqrt(3) - sqrt(3))x - (4sqrt(3))(sqrt(3)) =
x^2 + 3sqrt(3)x - (4)(3) = x^2 + 3sqrt(3) - 12

(a) is the correct answer.