Question

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

Answer 1

`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)`