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

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Answer 1 · 2017-06-23T17:36:59

$x=2+sqrt(2)$
$x=2-sqrt(2)$

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

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

$y=ax^2+bx+c$

In order to get this, just simplify your equation.

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

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

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

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

Now you can look turn this into the quadratic formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

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

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

All you have to do is plug in and solve!

$x=(-4+-sqrt(4^2-4(-1)(-2)))/((2)(-1))$

$x=(-4+-sqrt(16-8))/-2$

$x=(-4+-sqrt(8))/-2$

$x=(-4+-2sqrt(2))/-2$

$x=2+-sqrt(2)$

You should get two real roots as your answer:

$x=2+sqrt(2)$
$x=2-sqrt(2)$

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?