Question

How do you find the roots, real and imaginary, of `y=4x^2 + -7x -(x-2)^2` using the quadratic formula?

Answer 1

`x=(3+sqrt57)/6,` `(3-sqrt57)/6`

Explanation:

Solve:

`y=4x^2+(-7x)-(x-2)^2`

Expand `(x-2)^2`.

`y=4x^2+(-7x)-(x-2)(x-2)`

`y=4x^2+(-7x)-(x^2-4x+4)`

Simplify.

`y=4x^2-7x-x^2+4x-4`

Gather like terms.

`y=(4x^2+(-x^2))+(-7x+4x)-4`

Combine like terms.

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

To solve for roots, substitute `0` for `y`. Then solve for `x` using the quadratic formula.

`0=3x^2-3x-4`

Quadratic formula

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

Plug in the given values.

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

Simplify.

`x=(3+-sqrt(9+48))/6`

`x=(3+-sqrt57)/12` `larr` `57` is the product of two prime numbers.

`x=(3+sqrt57)/6,` `(3-sqrt57)/6`