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

Apr 6, 2018

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

$0 = {\left(x - 4\right)}^{2} + {\left(x - 4\right)}^{2}$

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 $a {x}^{2} + b x + c = 0$, so we need to expand our quadratic using the perfect square law.

$0 = {\left(x - 4\right)}^{2} + {\left(x - 4\right)}^{2}$
$0 = \left({x}^{2} - 8 x + 16\right) + \left({x}^{2} - 8 x + 16\right)$
$0 = 2 {x}^{2} - 16 x + 32$

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(- 16\right) \pm \sqrt{{\left(- 16\right)}^{2} - 4 \left(2\right) \left(32\right)}}{2 \left(2\right)}$

$x = \frac{16 \pm \sqrt{256 - 256}}{4}$

$x = \frac{16}{4}$

$x = 4$

The quadratic only has one root, 4.