Algebra and Trigonometry question. Please show work!! Serious Replies only!! Thank you so much!!?

Question

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

a). x^2 + 3 square root 3 x - 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

The equation is (x - a)(x - b) where a and b are roots:

a=sqrt(3)
b = -4sqrt(3)

(x - a)(x - b) = (x - sqrt(3))(x + 4sqrt(3)) =
= x^2 - sqrt(3)x + 4 sqrt(3)x - 12 = x^2 + 3sqrt(3)x - 12

Choose a for your answer.

Answer 2

The equation would be:

(x - √3)(x + 4√3)

Multiply this out for your answer:

x² + 3x√3 - 12 = 0

So, "a)" is your answer.

Answer 3

x = √3 → x - √3 = 0
x = -4√3 → x + 4√3 = 0

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

and that's answer a).

Answer 4

[x-sqr3][x+4rt3]=0
x^2+[-sqrt3+4sqrt3]x-4*3=0
x^2+3[sqrt3]x-12=0
so it is first option

rt 3*rt3=3