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

Apr 18, 2016

$y = \left(x - 2\right) \left(x + 1\right) - 3 x$
$\textcolor{w h i t e}{\text{XXX}}$has two real roots at $x = 2 + \sqrt{6} \mathmr{and} x = 2 - \sqrt{6}$

#### Explanation:

In order to use the quadratic formula it is first necessary to convert the given equation into standard form.

$y = \left(x - 2\right) \left(x + 1\right) - 3 x$

$\rightarrow y = {x}^{2} - x - 2 - 3 x$

$\rightarrow y = {x}^{2} - 4 x - 2$ (standard form)

Roots for an equation of the form:
$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
are given by the quadratic formula
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case
$\textcolor{w h i t e}{\text{XXX}} a = 1 ,$
$\textcolor{w h i t e}{\text{XXX}} b = - 4$
$\textcolor{w h i t e}{\text{XXX}} c = - 2$

So the roots are
$\textcolor{w h i t e}{\text{XXX}} x = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(- 2\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{\text{XXX}} x = \frac{4 \pm \sqrt{24}}{2}$

$\textcolor{w h i t e}{\text{XXX}} x = 2 \pm \sqrt{6}$

Apr 18, 2016

We have only real roots, which are $2 - \sqrt{6}$ and $2 + \sqrt{6}$

#### Explanation:

$y = \left(x - 2\right) \left(x + 1\right) - 3 x = {x}^{2} - 2 x + x - 2 - 3 x = {x}^{2} - 4 x - 2$

The roots of ${x}^{2} - 4 x - 2$

given by quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ are

$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \times 1 \times \left(- 2\right)}}{2 \times 1} = \frac{4 \pm \sqrt{16 + 8}}{2}$ or

$x = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2 \sqrt{6}}{2} = 2 \pm \sqrt{6}$

Hence, roots are $2 - \sqrt{6}$ and $2 + \sqrt{6}$