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

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Answer 1 · 2017-06-23T17:36:59

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

$y = x - (x - 2) (x - 1)$ $y=x-(x-2)(x-1)$

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

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

In order to get this, just simplify your equation.

$y = x - (x - 2) (x - 1)$ $y=x-(x-2)(x-1)$

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

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

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

Now you can look turn this into the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

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

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

All you have to do is plug in and solve!

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

$x = \frac{- 4 \pm \sqrt{16 - 8}}{- 2}$ $x=(-4+-sqrt(16-8))/-2$

$x = \frac{- 4 \pm \sqrt{8}}{- 2}$ $x=(-4+-sqrt(8))/-2$

$x = \frac{- 4 \pm 2 \sqrt{2}}{- 2}$ $x=(-4+-2sqrt(2))/-2$

$x = 2 \pm \sqrt{2}$ $x=2+-sqrt(2)$

You should get two real roots as your answer:

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

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