Problem of Complex Numbers in Geometry using Roots of Unity 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 am unable to find a good starting place for this problem. Should I start by working backwards using the roots?

$$w^7-1=(w-1)(w^6+w^5+w^4+\dots+w+1) $$
So $$w^6+w^5+w^4+\dots+w+1 =0 \tag{1}$$
Now use Vieta's Formula. 
First note that the sum of roots is $-p$, so we have that
$$-p=\alpha+\beta=w^6+w^5+w^4+w^3+w^2+w=-1$$
From $\text{(1)}$ Also, 
$$q=(\omega + \omega^2 + \omega^4) \times (\omega^3 + \omega^5 + \omega^6)$$
Expanding,
$$q=w^4(w^{6}+w^{5}+w^{4}+w^{3}+w^{2}+w+1)+2w^7=2$$
So $p=1, q=2$. 
