Prime Coefficients

Eh, I attacked it a little differently. Letting the roots be $x_1$ and $x_2$ , I used Vieta's to construct $x_1+x_2=p$ and $x_1x_2=q$ . Obviously, for $q$ to be divisible by only $1$ and itself, one of the roots is $q$ and the other is $1$ . Substituting, $q+1=p$ . Thus, we're looking for a pair of consecutive primes. The only pair is $2, 3$ . Plugging that into the desired expression, we obtain $6+6=\boxed{12}$ .

Since $p,q > 0$ , therefore $x^2 -px + q = \left(x-1\right)\left(x-q\right) \Rightarrow p = q+1$ This means that $p,q$ are two consecutive numbers that are prime.

$\therefore q = 2, p=3 \Rightarrow 2p+3q = 6+ 6=12$