# How do you find the roots, real and imaginary, of y= (x-3)^2+(x-4)^2  using the quadratic formula?

Apr 24, 2016

$x = \frac{7}{2} \pm \frac{1}{2} i$

#### Explanation:

First multiply out the quadratic to get it into standard form:

$y = {\left(x - 3\right)}^{2} + {\left(x - 4\right)}^{2}$

$= {x}^{2} - 6 x + 9 + {x}^{2} - 8 x + 16$

$= 2 {x}^{2} - 14 x + 25$

This is now in the form $a {x}^{2} + b x + c$ with $a = 2$, $b = - 14$ and $c = 25$

This has zeros given by the quadratic formula:

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

$= \frac{14 \pm \sqrt{{\left(- 14\right)}^{2} - 4 \left(2\right) \left(25\right)}}{2 \cdot 2}$

$= \frac{14 \pm \sqrt{196 - 200}}{4}$

$= \frac{14 \pm \sqrt{- 4}}{4}$

$= \frac{14 \pm \sqrt{4} i}{4}$

$= \frac{7}{2} \pm \frac{1}{2} i$