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
Aug 26, 2017

Answer:

#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)#