# How do you find the roots, real and imaginary, of y=(x – 7 )^2-8x+4 using the quadratic formula?

Jan 25, 2016

$x = \frac{22 + 4 \sqrt{17}}{2} \approx 19.25$

or

$x = \frac{22 - 4 \sqrt{17}}{2} \approx 2.75$

#### Explanation:

If you would like to use the quadratic formula, you should expand your term first.

$y = {\left(x - 7\right)}^{2} - 8 x + 4$

use the formula ${\left(a - b\right)}^{2} = \textcolor{g r e e n}{{a}^{2}} \textcolor{b r o w n}{- 2 a b} + \textcolor{b l u e}{{b}^{2}}$:

$y = \textcolor{g r e e n}{{x}^{2}} \textcolor{b r o w n}{- 2 \cdot x \cdot 7} + \textcolor{b l u e}{{7}^{2}} - 8 x + 4$

$y = {x}^{2} - 14 x + 49 - 8 x + 4$

$y = {x}^{2} - 22 x + 53$

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

and in your case, $a = 1$, $b = - 22$ and $c = 53$.

Thus, you can apply the formula:

$x = \frac{22 \pm \sqrt{{\left(- 22\right)}^{2} - 4 \cdot 1 \cdot 53}}{2} = \frac{22 \pm \sqrt{484 - 212}}{2}$

$= \frac{22 \pm \sqrt{272}}{2} = \frac{22 \pm \sqrt{16 \cdot 17}}{2} = \frac{22 \pm 4 \sqrt{17}}{2}$

You have two solutions:

$x = \frac{22 + 4 \sqrt{17}}{2} \approx 19.25$

or

$x = \frac{22 - 4 \sqrt{17}}{2} \approx 2.75$