Question

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

Answer 1

First of all we change the quadratic to 0, or just change it to

`0 = (x-4)^2 + (x-4)^2`

Keep in mind this is not exactly equal to your initial problem, however whenever finding x-intercepts/roots, we substitute y with 0, and we are doing the same here. In order to use the quadratic equation, we need to get it in the form `ax^2+bx+c = 0`, so we need to expand our quadratic using the perfect square law.

`0 = (x-4)^2 + (x-4)^2`
`0= (x^2 -8x +16) + (x^2-8x + 16)`
`0= 2x^2 -16x +32`

We now know `a=2, b=-16, c=32`, so we substitute these values into our quadratic equation.

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

`x=(-(-16)+-sqrt((-16)^2-4(2)(32)))/(2(2))`

`x=(16+-sqrt(256-256))/4`

`x=16/4`

`x=4`

The quadratic only has one root, 4.