Question

How do you find the roots, real and imaginary, of `y=3x^2 - 6x + 3` using the quadratic formula?

Answer 1

Single root at `x=1` (multiplicity `2`)

Explanation:

For a quadratic in the general form: `y=ax^2+bx+c`
the roots are given by the quadratic formula as
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`

Given: `y=3x^2-6x+3`
`color(white)("XXX")a=3`
`color(white)("XXX")b=-6`
`color(white)("XXX")c=3`

So the roots are
`color(white)("XXX")x=(6+-sqrt((-6)^2-4(3)(3)))/(2(3)`

`color(white)("XXXX")=(6+-sqrt(36-36))/6`

`color(white)("XXXX")=1`