Question

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

Answer 1

`4 +- [ (15)^(1/2) ] / 2`

Explanation:

If you expand the right side you should get
`x^2 - 8x +1`
Then just plug in to
`x = ( -b +- [b^2 - 4ac]^(1/2) ) / (2a)`
where a = 1 b = -8 c = 1

Answer 2

`x=4+sqrt15,` `4-sqrt15`

Refer to the explanation for the process.

Explanation:

Given:

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

FOIL `(2x-1)^`

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

Gather like terms.

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

Combine like terms.

`y=x^2-8x+1` `larr` standard form: `y=ax^2+bx+c`,

where:

`a=1`, `b=-8`, and `c=1`

Substitute `0` for `y` and solve for `x` using the quadratic equation.

`0=x^2-8x+1`

Quadratic Formula

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

Plug in the given values.

`x=(-(-8)+-sqrt((-8)^2-4*1*1))/(2*1)`

Simplify.

`x=(8+-sqrt(64-4))/2`

`x=(8+-sqrt60)/2`

Prime factorize `60`.

`x=(8+-sqrt(2xx2xx3xx15))/2`

Simplify.

`x=(8+-2sqrt15)/2`

Simplify.

`x=(color(red)cancel(color(black)(8^4))+-color(red)cancel(color(black)(2^1))sqrt15)/color(red)cancel(color(black)(2^1))`

`x=4+-sqrt15`

Roots

`x=4+sqrt15,` `4-sqrt15`