The master formula for two generic weights (roots) is $$ 2 \frac{\vec{a} \cdot \vec{b} }{\vec{a} \cdot \vec{a} }=q-p $$ but if we require that the roots are simple then this reduces to $$ 2 \frac{\vec{a} \cdot \vec{b} }{\vec{a} \cdot \vec{a} }=q \neq 0 $$ i.e. $p=0$. Why is this and why $q\neq0$?
P.S. I understand why $q\neq0$ since it comes straight out from the requirement that the roots are simple.
The answer to your doubt can be found in chapter $8$ of Lie algebras and particle physics by Georgi.
In particular, a simple root is a root that cannot be written as a sum of positive roots.
Given two simple roots $\vec{a}, \vec{b}$ then $\vec{a}- \vec{b}$ is not a root, otherwise - say $\vec{b}$ is larger - $\vec{b} = \vec{a} + (\vec{b}-\vec{a})$ and $\vec{b}$ would be not simple.
So, since $\vec{a}- \vec{b}$ is not a root you have $$ 2 \frac{\vec{a}\cdot \vec{b}}{\vec{a}\cdot\vec{a}} = q-p$$
with $p = 0$, just because the combination of roots you have chosen is not a root. Recall that $p$ is the number of times you apply a (negative) root to your weight representation state by $E_{-a}$.