equation $x^4+ax^3-6x^2+ax+1 = 0$ has two distinct positive roots Finding  parameter of $a$ for which equation $x^4+ax^3-6x^2+ax+1 = 0$ 
has two distinct positive roots
Attempt: writing equation $\displaystyle \bigg(x^2+\frac{1}{x^2}\bigg)+a\bigg(x+\frac{1}{x}\bigg)-6=0\;,$ where $x\neq 0$
So $\displaystyle \bigg(x+\frac{1}{x}\bigg)^2+a\bigg(x+\frac{1}{x}\bigg)-8=0$
put $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t\;,$ where $|t|\geq 2$
so $t^2+at-8=0.$ So for real roots $D\geq 0.$ So $a^2+32\geq 0$
So $\displaystyle t = \frac{-a\pm \sqrt{a^2+32}}{2}.$
could some help me how to solve it, thanks 

With $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t$ the roots occur when $t^2+at-8=0$. 
$f(x)=x^4+ax^3-6x^2+ax+1 = 0$ shows $f(0)=1>0$ and $\displaystyle \lim_{x\to+\infty} f=+\infty$, which show $f$ has two distinct positive roots if there is $x_0>0$ for which $f(x_0)<0$. Let $\displaystyle \bigg(x_0+\frac{1}{x_0}\bigg)=t_0\geq2$ and $f(x_0)=t_0^2+at_0-8<0$ or
$$4+2a-8\leq t_0^2+at_0-8<0$$
Shows $a<2$.
