1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Certainly $f(x)=0$. Does $f$ have a tangent at $x=0$? $\endgroup$ Jun 26 '17 at 19:39
  • 1
    $\begingroup$ Descartes Rule of Signs looks easier to apply. $\endgroup$
    – hardmath
    Jun 26 '17 at 19:45
6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

Because $(x^3+mx+1)'=3x^2+m>0$.

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.