Question 552366
If one root of a quadratic equation with real coefficients is 3-2i, the other root is 3+2i.
One equation with such roots would be
{{{(x-(3-2i))(x-(3+2i))=0}}} --> {{{(x-3+2i)(x-3-2i)=0}}} --> {{{((x-3)+2i)((x-3)+2i)=0}}} --> {{{(x-3)^2-(2i)^2=0}}} --> {{{x^2-6x+9-2^2*i^2=0}}} --> {{{x^2-6x+9-4(-1)=0}}} --> {{{x^2-6x+9+4=0}}} --> {{{x^2-6x+13=0}}}
If I am not told that the coefficients are real numbers, I cannot find the other root; it could be any complex number.