Prove: $x^4+mx^2+x$ have only two roots when $m>0$

I got the question:

Prove that $x^4+mx^2+x$ have only two roots when $m>0$.

I know that it is a continuous function.

I tried to use solve this question with two steps:

1. Use intermediate value theorem to prove that there are at least two roots.
2. Use Rolle's theorem to prove that there are not more then two roots.

I am stuck on first step. I can find a positive value of the function, but I can't find $x$ that give me a negative value. I assume that the $x$ that gives the negative value depends on $m$ but we know only that $m>0$, and there are many cases to check.

Any idea how to solve it?

• Certainly $f(x)=0$. Does $f$ have a tangent at $x=0$? Jun 26 '17 at 19:39
• Descartes Rule of Signs looks easier to apply. Jun 26 '17 at 19:45

Since the polynomial factors as $x(x^3+mx+1)$, you have one root at $x=0$. So now you just have to show that $x^3+mx+1$ has exactly one root, and your plan of action above should do the trick.
Because $(x^3+mx+1)'=3x^2+m>0$.
Factor out an $x$ to get $x^3+mx+1$. Need to show that this has only 1 real root if $m>0$. Take the derivative $3x^2+m$. Assume $x^3+mx+1$ has more than 1 real root, say $a$ and $b$. Then by Rolle's Theorem there is a $c$ in $(a,b)$ such that $3c^2+m=0$, but this is clearly impossible if $m>0$. Hence $x^3+mx+1$ has only 1 real root. (Since it has odd degree, it must have 1.)