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

Mar 3, 2016

Multiply out and simplify to find a quadratic in standard form, then apply the formula to find zeros $x = 3 \pm i$

#### Explanation:

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

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

$= 2 {x}^{2} - 12 x + 20$

$= 2 \left({x}^{2} - 6 x + 10\right)$

${x}^{2} - 6 x + 10$ is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 6$ and $c = 10$.

This has zeros given by the quadratic formula:

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

$= \frac{6 \pm \sqrt{{\left(- 6\right)}^{2} - \left(4 \cdot 1 \cdot 10\right)}}{2 \cdot 1}$

$= \frac{6 \pm \sqrt{36 - 40}}{2}$

$= \frac{6 \pm \sqrt{- 4}}{2}$

$= \frac{6 \pm \sqrt{4} i}{2}$

$= \frac{6 \pm 2 i}{2}$

$= 3 \pm i$