If #alpha# is a root of #x(2-x) = 3# then can you define a cubic equation with integer coefficients and roots including #-2# and #alpha# ?

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Answer 1 · 2017-02-18T07:48:39

Yes

Given:

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

Add #x^2-2x# to both ends to get

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

Multiply both sides by #(x+2)# to get:

#0 = (x^2-2x+3)(x+2) = x^3-x+6#

Add #x-6# to both ends and transpose to get:

#x^3 = x-6#

So the roots of this cubic equation are #x = -2# and the roots of #x(2-x) = 3#, including #alpha#.