this locus is not explicitly computed in the Voisin's paper

Dear Remke, According to the above paper by Voisin every general hypersurface in $P^n$ of degree $d\ge 2n-1$ contains no rational curve. In my case the polynomials P and Q can be taken general (e.g. their coefficients are general) can I deduce that by Voisin theorem the above hypersurface for $d\ge 5$ contains no rational curve?

Dear Will, in my case $P$ and $Q$ are as follows: $P(x)=(x-a_1)\cdots (x-a_n)$ and $Q(x)=(x-\overline{a}_1)\cdots (x-\overline{a}_n)$ where $a_i$ are general complex numbers. I think, I could derive contradiction if we could prove a certain proportion of the zeros of P(f) and Q(g) are single roots. I mainly apply complex analysis specially nevanlinna theory.

May be it works, but note that these surfaces are not K3 surfaces or elliptic surfaces (like in Ulmer's paper . In fact they are general type surfaces.

I mean that if $P(x)=(x-a_1)\cdots(x-a_n)$ then want to see that all roots of the equation, for example,$f(z)=a_1$ are of multiplicity 1.

thanks Will, but by applying Riemann-Hurwitz to the map $P(f):\mathbb{P}^1 \rightarrow \mathbb{P}^1$ we have: $2g_{\mathbb{P}^1} -2=(2g_{\mathbb{P}^1}-2)deg(P(f))+\sum_{p\in\mathbb{P}^1}(e_p -1)$ so we have $-2=-2d +\sum (e_p-1)$ where $d=deg(P(f))$. How we can deduce from the above that there are simple roots?

Yes, you are right, this is the trivial part. But do you have any idea about this problem? As I mentioned in above, how to prove that the above functional equation has no non-constant solution. Or maybe you want to think by using the algebrogeometric tools to solve the problem.