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

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

ReportAnswer 1

$x = 2 + \sqrt{2}$
$x = 2 - \sqrt{2}$

#### Explanation:

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

To write the quadratic formula, your equation first needs to be in standard form:

$y = a {x}^{2} + b x + c$

In order to get this, just simplify your equation.

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

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

$y = x - {x}^{2} + 3 x - 2$

$y = - {x}^{2} + 4 x - 2$

Now you can look turn this into the quadratic formula:

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

The variables in standard form are the same as the ones in the quadratic formula, which means you can easily transfer them over.

$a = - 1$
$b = 4$
$c = - 2$

All you have to do is plug in and solve!

$x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(- 1\right) \left(- 2\right)}}{\left(2\right) \left(- 1\right)}$

$x = \frac{- 4 \pm \sqrt{16 - 8}}{-} 2$

$x = \frac{- 4 \pm \sqrt{8}}{-} 2$

$x = \frac{- 4 \pm 2 \sqrt{2}}{-} 2$

$x = 2 \pm \sqrt{2}$

$x = 2 + \sqrt{2}$
$x = 2 - \sqrt{2}$

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