How do you find the roots, real and imaginary, of y= x^2 - 12x -9 -2(x- 3 )^2 using the quadratic formula?

1 Answer
JJ
Aug 26, 2017

x=2isqrt(3), -2isqrt(3)

Explanation:

First, let's simplify the equation.

y=x^2-12x-9-2(x-3)^2 ->

y=x^2-12x-9-2(x^2-6x+9) ->

y=x^2-12x-9-2x^2+12x-18 ->

y=-x^2-27=-1x^2+0x-27

I put the -1 and 0 to show what the a and b terms will be. Using the simplified version of the equation, we can find the roots using the quadratic equation. In the quadratic equation, a=-1, b=0, and c=-27.

x=(-b±sqrt(b^2-4ac))/(2a) ->

x=(-(0)±sqrt((0)^2-4(-1)(-27)))/(2(-1)) ->

x=(±sqrt(-(1)(4)(27)))/-2 ->

x=(±sqrt(-1)sqrt(4)sqrt(27))/-2 ->

x=(±i(2)(3sqrt(3)))/-2 ->

x=±2isqrt(3)
OR
x=2isqrt(3), -2isqrt(3)

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