How do you find the zeros, real and imaginary, of #y=x^2 +36x +81# using the quadratic formula?

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Answer 1 · 2016-04-20T06:15:35

Zeros are #-18-9sqrt3# and #-18+9sqrt3#

Zeros of #y=x^2+36x+81# are given by solution to the equation

#x^2+36x+81=0#

Hence using quadratic formula #x# is given by #(-b+-sqrt(b^2-4ac))/(2a)#

Hence #x=(-36+-sqrt(36^2-4xx1xx81))/2=(-36+-sqrt(1296-324))/2#

= #(-36+-sqrt(972))/2=(-36+-sqrt(3xx3xx3xx3xx3xx2xx2))/2=(-36+-18sqrt3)/2#

= #-18+-9sqrt3#

Hence zeros are #-18-9sqrt3# and #-18+9sqrt3#