Question

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

Answer 1

The root is `x=–2,` with multiplicity two.

Explanation:

Convert the right-hand side to the form `ax^2+bx+c`.

`y="   "(–2x+4)(x+1)"  " + 3x^2 + 2x`
`color(white)y=–2x^2-2x+4x+4 + 3x^2 + 2x`
`color(white)y=x^2+4x+4`

The roots of this function occur when `y=0`. We replace `y` with `0` and solve for `x:`

`0=x^2+4x+4`

Using the quadratic formula, the solutions for a quadratic `ax^2 + bx + c=0` are

`x=(–b+-sqrt(b^2-4ac))/(2a)`

For the given quadratic, `a = 1, b=4, and c = 4`. We plug these values into the quadratic formula as follows:

`x=(–4+-sqrt(4^2-4(1)(4)))/(2(1))`

`x=(–4+-sqrt(16-16))/2`

`x=(–4+-sqrt(0))/2`

`x=(–4)/2" " = –2`

But wait—shouldn't we get two roots?
Yes, we should, and we did! Since the discriminant `b^2-4ac` was equal to 0, the "one" root we got actually has a multiplicity of two. Meaning, the factor `x+2` appears twice in the factorization of `x^2+4x+4.`

Indeed, we can confirm this if we factor `x^2+4x+4` the shorter way, where we find two numbers that add to 4 and multiply to 4; those two numbers are 2 and 2, giving us

`x^2+4x+4 = (x+2)(x+2).`