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......
 A: 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,
$$\fbox{$-\frac{13}{2}<k<-6$}$$
A: Easiest way is to draw a graph $$f(x) = {x^4+11x^2+1\over -x(x^2+1)}$$
and we want to find for which $k$ line $y=k$ cuts graph of $f$ for positive $x$ four times. If you do some calculus you see that $f$ has local minimum $-6,5$ for positive $x$ and local maximum $-6$ so the finally answer is for $$-6,5<k<-6$$ 

If you don't know calculus then you can try like this. Write 
$$f(x) =  {x^4+2x^2+1\over -x(x^2+1)} +{9x^2\over -x(x^2+1)}$$ 
$$={x^2+1\over -x} +{9x\over -(x^2+1)}$$
$$=t +{9\over t}$$
where $t={x^2+1\over -x}$. Since $f$ is odd it is the same question if we ask our self for which negative $x$ the line $y= k$ cuts graph of $f$ four times. So say $x<0$ and thus $t>0$. Now it is easy to see that $f(x)\geq 6$ (for all negative $x$) and now we have to prove that the line $y=6,5$ touch a graph of $f$ (in some $x_0$ and thus it will have a local extremum at that $x_0$). To find this $x_0$ we solve this $$ f(x) = 13/2 \implies (x+1)^2(2x^2+9x+2) =0$$
since we have double root at $x=-1$ we see that $x_0=-1$.
A: By your work let $f(u)=u^2+ku+13,$ where $u=x+\frac{1}{x}.$
Hence, $|u|>2$ and we need to solve the following system:
$$f(2)>0,$$
$$\frac{-k}{2}>2$$ and
$$k^2-52>0$$ or
$$f(-2)>0,$$
$$\frac{-k}{2}<-2$$ and
$$k^2-52>0,$$ which gives
$$-8.5<k<-\sqrt{52}$$ or
$$\sqrt{52}<k<8.5.$$
