# How do you find the roots, real and imaginary, of y=x(x-1)-(3x-1)^2  using the quadratic formula?

Jun 25, 2018

Expand and simplify to find the equation in standard form, then apply the quadratic formula to obtain the roots: $x = \frac{5 \pm i \sqrt{7}}{16}$.

#### Explanation:

The quadratic formula can be used with quadratic equations in standard form, $y = a {x}^{2} + b x + c$.

First expand and simplify the given equation to rearrange it into standard form.

$y = x \left(x - 1\right) - {\left(3 x - 1\right)}^{2}$
$y = {x}^{2} - x - \left(9 {x}^{2} - 6 x + 1\right)$
$y = {x}^{2} - x - 9 {x}^{2} + 6 x - 1$
$y = - 8 {x}^{2} + 5 x - 1$

This equation is now in standard form, where $a = - 8$, $b = 5$, and $c = - 1$.

To solve for the roots of the equation, set $y = 0$:
$0 = - 8 {x}^{2} + 5 x - 1$,

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(5\right) \pm \sqrt{{\left(5\right)}^{2} - 4 \left(- 8\right) \left(- 1\right)}}{2 \left(- 8\right)}$
$x = \frac{- 5 \pm \sqrt{- 7}}{-} 16$
$x = \frac{5 \pm \sqrt{- 7}}{16}$
$x = \frac{5 \pm \sqrt{- 1} \sqrt{7}}{16}$
$x = \frac{5 \pm i \sqrt{7}}{16}$