How do you find the roots, real and imaginary, of #y=-(x - 6)^2-x^2+x + 3# using the quadratic formula?

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Answer 1 · 2017-11-09T23:20:57

Expand the brackets and use the Formula

#x=1/4(13+sqrt95i)# or #x=1/4(13-sqrt95i)#

Firstly, expand the brackets and collect like terms.

#y=-(x^2-12x+36)-x^2+x+3#
#y=-x^2+12x-36-x^2+x+3#
#y=-2x^2+13x-33#

Now we sub into the quadratic formula #a=-2, b=13, c=-33#

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

#x=(-13+-sqrt(13^2-4(-2)(-33)))/(2(-2)#

#x=(-13+-sqrt(169-264))/(-4)#

#x=(13+-sqrt(-95))/4#

Ok, we can see here our roots will be imaginary, but that's ok.

#x=(13+-sqrt(95)i)/(4)#

#x=1/4(13+sqrt95i)# or #x=1/4(13-sqrt95i)#