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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?
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ReportAnswer #1

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

Explanation:

#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)#

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