Real values of $a$ for $x^4+(a-1)x^3+x^2+(a-1)x+1=0$ to have at least two negative roots 
Problem Statement:-
Find all the values of $a$ for which the equation 
  $$x^4+(a-1)x^3+x^2+(a-1)x+1=0$$
  possess at least two negative roots.


I know that there has been a post regarding this same problem here, but I have a different issue with the problem than that posted in the OP in the above post. 
So, I don't have an issue of finding a solution and coincidentally I had the same approach as the one posted by RicardoCruz. 
But still if you have a solution than you are more than welcome to post one, well come on who doesn't want some good reputation change ;P.

My Attempt at a solution:-

For those who wanna skip to my main issue may start to read from the next quote.

We can see very clearly that $x=0$ isn't a root of the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$$
hence it is safe to divide the equation throughout by $x^2$ for arriving at the condition which the problem wants.
So, we get
$$\left(x^2+\dfrac{1}{x^2}+2\right)+(a-1)(x+\dfrac{1}{x})-1=0\\
\implies \left(x+\dfrac{1}{x}\right)^2+(a-1)(x+\dfrac{1}{x})-1=0$$.
Now, let $z=x+\dfrac{1}{x}$ then the equation becomes
$$z^2+(a-1)z-1=0\\
\implies z=\dfrac{-(a-1)\pm\sqrt{\left(a-1\right)^2+4}}{2}$$
Now, its necessary to note that $\sqrt{\left(a-1\right)^2+4}\gt(a-1)$, hence for the roots to be negative $$\left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}$$
As, we know that $\text{A.M. $\ge$ G.M.}$ for non negative real numbers, and as we are dealing with negative roots of the equation, so we will be applying $\text{A.M. $\ge$ G.M.}$ to $-x,-\dfrac{1}{x}$, we get 
$$\dfrac{-\left(x+\dfrac{1}{x}\right)}{2}\ge1\implies x+\dfrac{1}{x}\le-2\\
\implies \left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\le -2$$
But as the roots are supposed to be distinct hence, we have to exclude the case of equality, because it occurs when $x=\dfrac{1}{x}=-1$, which results in $x+\dfrac{1}{x}=-2\implies (x-1)^2=0$, hence $x$ has a repeated root. So, we get
$$\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\lt -2\implies \sqrt{\left(a-1\right)^2+4}\gt 5-a$$

Now, this is where I am having problem, that's right solving the "inequality". I always had thought of this before but never tried to test it before, so bear with it and follow me on to a trip to solve inequalities in a weird way.

I did a case study for the inequality, which is as follows:-
Case-1:-
If $(5-a)\gt 0\implies a\lt 5$, then there is no problem in squaring the inequality as both sides are positive.
$$\left(a-1\right)^2+4\gt a^2+25-10a\implies a\gt \dfrac{5}{2}$$
$$\therefore a\in\left(\dfrac{5}{2},5\right)$$
Case-2:-
If $5-a\lt0$, then we might be faced with a problem cause, there might be a case where $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$, which would make the inequality $\sqrt{\left(a-1\right)^2+4}\gt 5-a$ on squaring change the sign of inequality to be in accordance with the condition $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$.
So, on solving the inequality we get,
$$\sqrt{\left(a-1\right)^2+4}\gt 5-a\implies (a-1)^2+4\lt(5-a)^2$$
Now, why did I change the sign of inequality was due to the fact that I was finding the interval where(as per the condition I stated before) $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$. So, just like $5\gt-6$ but on squaring the inequality we get $25\lt36$. So that's what I did.
This results in $a\lt\dfrac{5}{2}\cup a\gt5\implies a\in\emptyset$
So, on combining both the cases we get $a\in\left(\dfrac{5}{2},5\right)$
But from RobertCruz's solution and the books answer it seems that $ a\in\left(\dfrac{5}{2},\infty\right)$ is the correct solution, so what am I doing wrong here.

From the place where you have $x+1/x\; (=z)$ in terms of $a,$ write $x+1/x=r$ where $$r=(-(a-1)-\sqrt {(a-1)^2+4})/2 <0.$$ So $x^2-rx+1=0,$ giving $$x=(r\pm \sqrt {r^2-4}\;)/2<0.$$ This yields two negative roots iff $r^2-4>0,$ that is, iff $|r|>2$ because $r^2-4<0$ yields no real negative $x$, and $r^2-4=0$ yields exactly one negative root. In terms of $a,$ this is $$2<|(-(a-1)-\sqrt {(a-1)^2+4}\;)/2|=|(\;a-1+\sqrt {(a-1)^2+4}\;)/2| .$$ 
(Because we must have $a>1,$ else we have $ x^4+(a-1)x^3+x^2+(a-1)x+1>0$ whenever $x<0$.) 
For $a>1$ this is equivalent to $a-1>3/2,$ that is, $a>5/2.$
Another method would be: Let $f(x)=x^4+(a-1)x^3+x^2+(a-1)x+1$ with $a>1.$ Then $$(f(x)<0\land x<0)\iff$$ $$  a-1=-\frac {x^4+x^2+1}{-x^3-x}=$$ $$=\frac {(x+1/x)^2x^2-x^2}{-x^2(x+1/x)}=$$ $$=\frac {u^2-1}{u}$$ where $u=|x|+|1/x|=-x-1/x.$
The AGM inequality implies that $|x|+|1/x|\geq 2.$ Now show that $\frac {u^2-1}{u}\geq \frac {3}{2}$ for $u\geq 2.$ And I leave the rest to you. 
Remark. For $u\geq 2$ let $u=2+b.$ Then  $\frac {u^2-1}{u}-\frac {3}{2}=\frac {8b^2+2b}{2b+4}.$  
