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 
 A: Your strategy is good. The equation $t^2+at-8=0$ has two real roots, namely
$$
r_1=\frac{-a+\sqrt{a^2+32}}{2}
\qquad
r_2=\frac{-a-\sqrt{a^2+32}}{2}
$$
Note that $r_1>0$ and $r_2<0$. Consider now
$$
x+\frac{1}{x}=r_i
$$
which becomes
$$
x^2-r_ix+1=0
$$
If we take $i=2$, we see that the equation either has no real root or it has two negative real roots. So we need to take $i=1$. The equation has (distinct) positive roots if and only if its discriminant is positive:
$$
r_1^2-4>0
$$

This becomes$$(\sqrt{a^2+32}-a)^2-16>0$$and, doing some simplifications,$$a^2+8>a\sqrt{a^2+32}$$This is true whenever $a<0$. If $a\ge0$ we can square and get$$a^4+16a^2+64>a^4+32a^2$$that is, $a^2<4$, which gives $0\le a<2$. Thus the final solution is $a<2$.

A: 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 $a<2$.
