Find all values of a for which the equation $x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$ possesses at least two distinct negative roots Find all values of a for which the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0 $$ possesses at least two distinct negative roots.
I am able to prove that all roots would be negative .How to proceed after this.
 A: I decided to transform my comment into an answer.
As $x=0$ is not a root, divide the equation by $x^2$ and solve for $x+\frac{1}{x}$ as
RGB has pointed out.
You will get two solutions $y_1$ and $y_2$ for $x+\frac{1}{x}$:
$$x+\frac{1}{x}=\frac{-(a-1)-\sqrt{(a-1)^2+4}}{2}=y_1$$
and
$$x+\frac{1}{x}=\frac{-(a-1)+\sqrt{(a-1)^2+4}}{2}=y_2$$
If we want two negative roots the line $y=y_1$ must intercept twice the function $f(x)=x+\frac{1}{x}$ in the third quadrant. That's only possible if $y_1<-2$. See the figure bellow:

So let's solve the inequality:
$$y_1<-2 \Rightarrow $$
$$\frac{-(a-1)-\sqrt{(a-1)^2+4}}{2}<-2 \Rightarrow $$
$$-(a-1)-\sqrt{(a-1)^2+4}<-4 \Rightarrow $$
$$(a-1)+\sqrt{(a-1)^2+4}>4 \Rightarrow $$
$$\sqrt{(a-1)^2+4}>5-a \Rightarrow $$
$$a^2-2a+5>25-10a+a^2 \Rightarrow $$
$$8a>20 \Rightarrow $$
$$a>\frac{5}{2}$$ 
Therefore for $a>\frac{5}{2}$ the original equation will have two negative roots.
A: This is a symmetric (the coefficients) polynomial. We may want to take advantage of that. We can divide by $x^2$ and get $$(x+\frac{1}{x})^2+(a-1)(x+\frac{1}{x})-1=0$$
From there we can solve for $x+\frac{1}{x}$ and get $$x+\frac{1}{x}=\frac{-(a-1)\pm\sqrt{(a-1)^2+4}}{2}.$$
This is a quadratic equation. Solve for $x$ and get the roots.
