2. Given that $p$ and $q$ are integers that satisfy the equation $36 x^{2}-4\left(p^{2}+11\right) x+$ $135(p+q)+576=0$, find the value of $p+q$.
2. Ans: 20
$$
p+q=\frac{p^{2}+11}{9} \text { and } p q=\frac{135(p+q)+576}{36}=\frac{15(p+q)}{4}+16 .
$$

So $p+q>0$ and is a multiple of 4. Also $p+q=1+\frac{p^{2}+2}{9}$. So $p^{2}+2$ is a multiple of 9. So $p=5,13,14, \ldots$ If $p=5, p+q=4$ and $p q=31$. If $p=13, p+q=20, p q=91$, $q=7$. Thus $p+q=20$.