if the given equation has one real solution then the other two complex roots are conjugates of each other, hence the sum of the roots will always be real, but in this equation the sum of the roots is (5 + 6i)/2, how is that possible?

What makes you think that the nonreal roots must be conjugates of each other?

$2z^3 -5z^2 +1 = 0$ for real part to be 0 and $ -6z^2 +9z-3=0 $for imaginary part to be 0 Both equations have common root 1/2 So first real root =1/2 Using synthetic division, $2z^3-(5+6i)z^2+9iz+1-3i= (z-1/2)(2z^2 -(4+6i) -2+6i)$ Solving, other roots become $1+2i$ and $1+i$

agirlhasnoname wrote: if the given equation has one real solution then the other two complex roots are conjugates of each other, hence the sum of the roots will always be real, but in this equation the sum of the roots is (5 + 6i)/2, how is that possible? Complex roots have conjugates only if coefficients of polynomial are real.