Solution of polynomial Find the set of values of $k$ for which the equation $x^4+kx^3+11x^2+kx+1=0$ has four distinct positive root.
Attempt: 
$x^4+kx^3+11x^2+kx+1=0$
$x^2+kx+11+{k\over x}+{1\over x^2}=0$
$x^2 + {1\over x^2} +k(x+{1\over x})+11=0$
$(x + {1\over x})^2 +k(x+{1\over x})+13=0$
I don't know how to proceed after this......

The reciprocal quartic should be transformed as:
$$\left( x+\frac{1}{x} \right)^2+k\left( x+\frac{1}{x} \right)+\color{red}{9}=0$$
Now,
\begin{align}
  x+\frac{1}{x} &= \frac{-k \color{red}{\pm} \sqrt{k^2-36}}{2} \\
  x &=
  \frac{-k\color{red}{\pm} \sqrt{k^2-36}}{2(2)} \color{blue}{\pm}
  \frac{1}{2} \sqrt{
  \left( \frac{-k\color{red}{\pm} \sqrt{k^2-36}}{2} \right)^2-4} \\
  &=
  \frac{-k \color{red}{\pm} \sqrt{k^2-36}}{4} \color{blue}{\pm}
  \frac{\sqrt{k^2 \color{red}{\mp} k\sqrt{k^2-36}-26}}{2\sqrt{2}}
\end{align}
For four distinct real roots,


*

*$k^2-36>0$ and

*$k^2 \pm k\sqrt{k^2-36}-26>0$
$\implies k^2-26>|k|\sqrt{k^2-36}>0$
$\implies (k^2-26)^2 > k^2(k^2-36)$
$\implies k^4-4\times 13k^2+4\times 13^2 > k^4-4\times 9k^2$
$\implies 13^2 > 4k^2$
And the roots are positive, $\text{sum of roots}=-k>0$
Hence,
$$\fbox{$-\frac{13}{2}<k<-6$}$$
