How do you find the zeros of # y = -2x^2 + 8x -2 # using the quadratic formula?

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Answer 1 · 2018-03-15T22:32:00

#x=2+-sqrt3#

The Quadratic Formula states that if a quadratic is of the form #ax^2+bx+c#, it's zeroes are at

#(-b+-sqrt(b^2-4ac))/(2a)#

Our quadratic is in standard form, therefore we know

#a=-2#
#b=8#
#c=-2#

Now, we can just plug in. We get:

#(-8+-sqrt(8^2-4(-2)(-2)))/(2(-2)#

Which simplifies to

#(-8+-sqrt(64-16))/-4#

Which can be further simplified to

#(-8+-sqrt48)/-4#

At this point, we can factor #sqrt48# into #color(blue)(sqrt16*sqrt3)#, because #sqrt(ab)=sqrta*sqrtb#. This simplifies to

#(-8+-color(blue)(4sqrt3))/-4#

We can factor out a #4# because all of the terms have a #4# in common. We get:

#(-2+-sqrt3)/-1#

Which can be simplified as

#2+-sqrt3#