Quadratic using the roots of unity, where $\omega^7 = 1, \omega \neq 1$

Question

Say that $\omega$ is a complex number, where $\omega^7 = 1, \omega \neq 1$.

Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. $\alpha$ and $\beta$ are roots of the quadratic $x^2 + px + q = 0$, given for some integers $p$ and $q$. What is the ordered pair $(p,q)$?

The roots of unity is tricky for me, so hints to get me started would help.

Answer 1

Hints: $$1+\omega+\cdots+\omega^6 = 0, \qquad \alpha + \beta = -p, \quad \alpha\beta = q.$$

Answer 2

Remember that $p$ is the negative sum of the roots and $q$ is their product. The roots are equally spaced on the unit circle starting with $\omega^0=1$