find roots

make a substitution $\quad t=2x\quad$ the polynomial becomes $G(x)= t^4 +t^3+t^2+t+1$ which is the same thing as $G(x)=\dfrac{t^5-1}{t-1}\longrightarrow t=\sqrt[5]{1}\neq 1$ now, we know that $1=e^{2\pi i}=e^{4\pi i}=e^{6\pi i}=e^{8\pi i}$ so, $t=\sqrt[5]{e^{2\pi i}},\sqrt[5]{e^{4\pi i}},\sqrt[5]{e^{6\pi i}},\sqrt[5]{e^{8\pi i}}$ $t= \cos(\dfrac{2}{5}\pi)+i\sin(\dfrac{2}{5}\pi)\dots\dots$ since $\quad x=\dfrac{t}{2}$ $x=\dfrac{1}{2}(\cos(\dfrac{2}{5}\pi)+i\sin(\dfrac{2}{5}\pi))\dots\dots$ $\sum_{n=1}^4 \dfrac{\dfrac{1}{2}}{\dfrac{2x}{5}} =\dfrac{125}{48}$ $125+48=\boxed{173}$

I think we must have $x_n > 0$ . Notice that
$x_n (\cos (z_n \pi) + i \sin (z_n \pi) ) = -x_n (\cos(z_n\pi + \pi) + i \sin (z_n \pi + \pi))$