# 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......

• Besides the answers below, see DeCartes' Rule of Signs. Commented Jul 2, 2018 at 13:59

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,

$$\fbox{-\frac{13}{2}<k<-6}$$

• The question says 4 distinct positive roots so obviously positive values of k won't work here. Commented Jul 24, 2018 at 22:38
• Amended, thanks for your response. Commented Jul 25, 2018 at 11:35

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 $x_0=-1$.

• How to check if roots are positive.? Commented Jul 2, 2018 at 13:48
• I don't know calculus so that will be bit difficult for me......I am thinking since t > or equal to 2 for positive x and less than or equal to -2 for negative x then surely this can help in finding the roots Commented Jul 2, 2018 at 14:19

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 $$\sqrt{52}<k<8.5.$$