$ω≠1$ and $ω^{13}=1$. If $\alpha=ω+ω^3+ω^4+ω^{−4}+ω^{−3}+ω^{−1}$ and $\beta=ω^2+ω^5+ω^6+ω^{−6}+ω^{−5}+ω^{−2}$, then quadratic equation, whose roots are $\alpha$ and $\beta$ is:
I tried finding the sum and product of roots and placing in the equation $x^2-(\alpha+\beta)x+(\alpha\cdot\beta)=0$ but I was not able to solve for it.
On
We have $$ 1+\alpha+\beta = 1+\omega+\omega^2+\cdots+\omega^{12}=0 $$ and so $\alpha+\beta=-1$. Therefore, $\alpha\beta=-(\alpha+\alpha^2)=-3$.
Unfortunately, I can't see shortcut to compute $\alpha+\alpha^2$ except by expanding it.
By the formula for a finite geometric series,
$$\alpha + \beta = \sum_{i=-6}^{-1} \omega^i + \sum_{i=1}^6 \omega^i = \sum_{i=7}^{12} \omega^i + \sum_{i=1}^{6} \omega^i =\frac{(1-\omega^{13})}{1 - \omega} -1 = -1 \tag{$\omega^{13} =1$}$$
For the product $\alpha \beta$, note that $\omega^n = \overline{\omega^{-n}}$. Taking $\omega$ as $e^{2 i \pi /13} = \cos(2\pi /13) + i \sin(2\pi /13)$ for example, $\alpha = 2 \left(\cos(2 \pi/13) + \cos(6 \pi/13)+\cos(8 \pi/13)\right)$ and you can do the same for $\beta$. Then you can use the product-to-sum formula $\cos A \cos B = \frac{1}{2} \left(\cos(A-B) + \cos(A+B) \right)$: by expanding the brackets, there are nine terms in this form.