System of equations from roots of polynomial I'm given the equation $3072x^4-2880x^3+840x^2-90x+3=0$ and told that its roots are $\alpha, \alpha r, \alpha r^2, \alpha r^3,$ for some $r\in \mathbb{R}$.
By considering the sum of the roots, the product, etc. I've found that \begin{gather}\alpha(1+r+r^2+r^3)=\frac{15}{16} \\ \alpha^2r(1+r+2r^2+r^3+r^4)=\frac{35}{128} \\ \alpha^3 r^3(1+r+r^2+r^3)=\frac{15}{512} \\ \alpha^4 r^6=\frac{1}{1024}\end{gather}
But this looks like a rather complex system and I can't see any obvious way solve this for $\alpha$ and $r$.
How can this system be solved?
EDIT
I can see that all of the denominators are powers of $2$, but I can't see how that will help me here.

Going with your equations... Dividing third equation by first, we get $\alpha^2 r^3=\frac{1}{32}$. This implies $r > 0$ because $\alpha$ must be real as well (follows from the first equation). Plugging this into the second equation we get
$$
\frac{35}{128}=\frac{1}{32}\left(\frac{1}{r^2}+\frac{1}{r}+2+r+r^2\right)=\frac{1}{32}\left(\left(r+\frac{1}{r}\right)^2+\left(r+\frac{1}{r}\right)\right).
$$
Letting $u=r+1/r$ gives quadratic equation
$$
u^2+u-\frac{35}{4}=0.
$$
This yields $u=\frac{5}{2}$ as we must have $u>0$. Then solving corresponding quadratic equation given by $\frac{5}{2}=r+1/r$ we see $r \in \{\frac{1}{2},2\}$. From the first equation we get $\alpha$ and so the two solutions are $r=\frac{1}{2}, \alpha=\frac{1}{2}$ and $r=2, \alpha=\frac{1}{16}$.
Clearly both solutions generate the same set of roots $\{\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16}\}$.
