Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
[x^2 + px + q = 0]
for some integers $p$ and $q$. Find the ordered pair $(p,q)$.

I got that $p=-1$ but does not know how to go on to finding $q$.  All help is appreciated!

$$q=\alpha\beta=3+\sum_{j=1}^6w^j=3+w\frac{w^6-1}{w-1}=3+\frac{1-w}{w-1}=2$$
