If p and q are distinct primes and $x^2-px+q=0$ has distinct positive integral roots, then p+q equals

Let $\alpha$ and $\beta$ be two roots of $x^2-px+q=0$ and assume $\alpha$< $\beta$. We have $\alpha +\beta =p,\alpha \beta =q$ As q is prime, we must have $\alpha =1,\beta =q$.

Thus, $p=1+q\Rightarrow p-q=1$

As p and q are prime, this is possible if p = 3, q = 2.