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

1 Answer
E P.
Apr 6, 2018

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.

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