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

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Answer 1 · 2017-11-05T03:31:55

#y# has two complex roots: #x=1/4(-1+- sqrt31 i)#

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

The roots of #y# occur where #y=0#

#:. x^2-3x+(-x-2)^2 =0#

Expanding:

#x^2-3x+x^2+4x+4=0#

Combining like terms:

#2x^2+x+4=0#

#x= (-1+-sqrt((1)^2 - 4xx2xx4))/(2xx2)#

#= (-1+-sqrt(1-32))/4#

#= (-1+-sqrt(-31))/4#

#= (-1+-sqrt(31xx(-1)))/4#

#=1/4(-1+- sqrt31 i)#

#y# has two complex roots: #x=1/4(-1+- sqrt31 i)#