How do you find the roots, real and imaginary, of y= x^2 - 5x - (2x-2)^2  using the quadratic formula?

Feb 17, 2016

There are two imaginary roots:
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} + \frac{\sqrt{39}}{6} i$ and $\frac{1}{2} - \frac{\sqrt{39}}{6} i$

Explanation:

First expand the expression given on the left into standard form:
$y = {x}^{2} - 5 x - {\left(2 x - 2\right)}^{2}$
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 5 x - \left(4 {x}^{2} - 8 x + 4\right)$

$\textcolor{w h i t e}{\text{XXX}} y = - 3 {x}^{2} + 3 x - 4$

This is in standard form: $y = a {x}^{2} + b x + c$ for which the quadratic formula tells us the roots are:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

For this specific example, the roots are
color(white)("XXX")x=(-3+-sqrt(3^2-4(-3)(-4)))/(2(-3)

$\textcolor{w h i t e}{\text{XXXX}} = \frac{- 3 \pm \sqrt{- 39}}{- 6}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{1}{2} \pm \frac{\sqrt{39}}{6} i$