Solving the quartic polynomials

Question

I have three quartic polynomials: $$1)\ x^4-(1+r+s+rs)x^2-2rsx+rs=0, s\geq r\geq 1,$$ $$2)\ x^4-(2q+qs+s)x^2-2qsx+qs=0,q\geq 2, s\geq 1,$$ $$3)\ x^4-(6+4s)x^2-8sx+4s=0,s\geq 2.$$ $q,r,$ and $s$ are positive integers. Also, I have for any $\varepsilon>0$ and $n\geq 4$, $x_1,x_2,x_3,x_4$: $$2n-\frac{\sqrt{3}}{3}-\frac{\varepsilon}{2}< x_1 < 2n+\frac{\sqrt{3}}{3}+\frac{\varepsilon}{2},$$ $$0<x_2\leq \sqrt{2}-1,$$ $$-n-\frac{\sqrt{3}}{3}-\frac{\varepsilon}{2}< x_3< -n+\frac{\sqrt{3}}{3}+\frac{\varepsilon}{2}$$ $$-n-\frac{\sqrt{3}}{3}-\frac{\varepsilon}{2}<x_4< -n+\frac{\sqrt{3}}{6}+\frac{\varepsilon}{4}$$ Suppose $x_2$ is a root of these equations. I want to prove other $x_i$'s can't be the roots of the above each quartic equations. I know $x_i$'s are real. I start with the easiest equation, third one. I studied the methods and I tried to obtain the roots but I was confused. Then I try to get some contradictions by the conditions of $x_i$'s and Vieta's formula, but I couldn't. Also I used the rational root theorem and I found the third equation dose not have the rational roots. If $x_i$'s make you confused, you can ignore them and help me how to solve the equations. Thanks for the help.