How do you find the roots, real and imaginary, of #y=-3(x -1 )^2-3x-1# using the quadratic formula?

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Answer 1 · 2018-04-07T17:08:33

The roots are #x=1/2+isqrt39/6# and #x=1/2-isqrt39/6#.

First, express #y# in standard form.

#y=color(red)(-3(x-1)^2)-3x-1=color(red)(-3x^2+6x-3)-3x-1=-3x^2+3x-4#

The quadratic formula finds when #y=0#. #y=0# when

#x=(-3+-sqrt(3^2-4(-4)(-3)))/(2*(-3))#

#=(-3+-sqrt(9-48))/(2*(-3))=(-3+-sqrt(-39))/(2*(-3))=(-3+-isqrt(39))/(2*(-3))#

#=1/2+-isqrt(39)/6#

Therefore the roots are

#x=1/2+isqrt39/6# and #x=1/2-isqrt39/6#.