I have been trying this sum taking the zeros as $\alpha$ and $\alpha^{-1}$, using the formula $x^2-(\alpha+\beta)x+\alpha \beta$. However I am stuck in this step: $$ \alpha x^2-(\alpha^2+1)x+\alpha=0. $$ This could have been the answer, but $\alpha$ is not defined in the question and hence cannot be a part of the answer. I have searched the internet for this question but nowhere the answer seems to be available.
The source of the question is unknown, I found this question in a really old school question paper and that school is not in existence anymore.
Your $\alpha x^2-(\alpha^2+1)x+\alpha,$ or dividing by $\alpha$ if you want a monic polynomial: $$x^2-\left(\alpha+\frac1\alpha\right)x+1$$ is the answer, with $\alpha\ne0$ being an arbitrary constant. Indeed, for an arbitrary quadratic polynomial $x^2-sx+p,$ the product of the two roots is $p$ (their sum being $s$).
If you want $\alpha\in\Bbb R,$ this (family of) polynomial(s) coincides with $$x^2-sx+1,$$ with $|s|\ge2.$