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 ax2+bx+c=0, so we need to expand our quadratic using the perfect square law.

0=(x−4)2+(x−4)2
0=(x2−8x+16)+(x2−8x+16)
0=2x2−16x+32

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

x=−b±√b2−4ac2a

x=−(−16)±√(−16)2−4(2)(32)2(2)

x=16±√256−2564

x=164

x=4

The quadratic only has one root, 4.

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