Question

How do you find the roots, real and imaginary, of `y=-(2x-1)^2 -2x^2 - 3x + 4` using the quadratic formula?

Answer 1

`x=(1+-sqrt73)/12`

Explanation:

First, get the equation into the standard form of a quadratic equation

`y=ax^2+bx+c`

which can be very easily applied to the quadratic equation

`x=(-b+-sqrt(b^2-4ac))/(2a`

To get the equation into standard form, simplify by combining like terms.

First, square and expand the `2x-1` term.

`y=-(2x-1)(2x-1)-2x^2-3x+4`

`y=-(4x^2-4x+1)-2x^2-3x+4`

`y=-4x^2+4x-1-2x^2-3x+4`

`y=-6x^2+x+3`

Thus,

`a=-6`
`b=1`
`c=3`

so,

`x=(-1+-sqrt(1^2-4(-6)(3)))/(2(-6))`

`x=(-1+-sqrt(1+72))/(-12`

`x=(1+-sqrt73)/12`