System of equations from roots of polynomial

Since you see powers of $2$, you can use a factorization: $$3072x^4-2880x^3+840x^2-90x+3=3(1024x^4-960x^3+280x^2-30x+1)=$$ $$=3(1024x^4-64x^3-896x^3+56x^2+224x^2-14x-16x+1)=$$ $$=3(16x-1)(64x^3-56x^2+14x-1)=$$ $$=3(16x-1)(64x^3-8x^2-48x^2+6x+8x-1)=$$ $$=3(16x-1)(8x-1)(8x^2-6x+1)=2(16x-1)(8x-1)(4x-1)(2x-1).$$

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 $\{\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16}\}$.