Complex number roots?!?! Math Help please!!!?

Question

if im given 3-2i radical5 and need to find the other root and the quadratic equation for it what do i do??

know i know that the other root must be 3 +2i radical5 because complex roots come in conjugate pairs , but hoow do i find thequadratic equation???

Answer 1

one root is 3 - 2√5 i
then the other root is its conjugate
3 + 2√5 i
and the quadratic equation is

(x - 3 + 2√5 i ) (x - 3 - 2√5 i ) = 0
(x - 3)² - (2√5 i)² = 0
(x - 3)² + 4(5) = 0
x² - 6x + 9 + 20 = 0
x² - 6x + 29 = 0

Answer 2

If your equation is x²+px+q = 0,
then the sum of the roots is -p = -6
and the product of the roots is q = (3-2i√5)(3+2i√5) = 29
Your equation is x² -6x+29 = 0.

Answer 3

Let C be one root and C' be the conjugate root.

The equation is (x - C)(x - C') = 0
or x^2 - Cx - C'x + CC' = 0
or x^2 - (C + C')x + CC' = 0

CC' is the norm of C: If C = a + bi, then this is (a + bi)(a - bi) or (a^2 + b^2)
-(C' + C): if C = a + bi, then this is -((a - bi) + (a + bi)) or -2a

So, your equation is x^2 - (2a)*x + (a^2 + b^2) = 0

Here, your a is 3 and your b is 2*sqrt(5). Plug those values in and you have your equation.

x^2 - 6x + (9 + 4*5) = 0
x^2 - 6x + 29 = 0

Answer 4

(x - first root)(x - second root)
foil them for the equation.