Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $m, n, p\in \mathbb R$, $n>0, p>0$. Prove that the following equation has exactly one positive solution:


Here is my attempt: Let $f(x)=x^5-mx^3-nx-p$, $f$ is continuous on $\mathbb R$ and $f(0)=-p<0, \lim_{x\to +\infty} f(x)=+\infty$. This implies, there exists $\xi>0$ such that $f(\xi)=0$ as a consequence of Bolzano-Cauchy theorem. Moreover, $$f'(x)=5x^4-3mx^2-n.$$ Since $(3m)^2+20n>0$ and $-5n<0$, we can easily see that $f'(x)=0$ has two roots $$x=\pm \frac{3m+\sqrt{9m^2+20n}}{10}.$$

I cannot continue to verify that $f(x)=0$ has only one positive solution?

share|cite|improve this question
Actually "$x^2= \frac {3m \pm \sqrt {9m^2 +20n}}{10}$" in your post. – Vikrant Desai Jan 11 at 6:52

I think your result follows from the Descartes' rule of signs. The number of sign differences in your polynomial is 1, irrespective of whether $m$ is strictly positive, strictly negative or zero.

share|cite|improve this answer
This looks like a good argument to me. Perhaps the downvoter can explain where the problem is? – Martin R Jan 11 at 6:35

Here is how you can make your method work:

$f'(x) = 0$ has exactly one positive solution $$ x=\sqrt{ \frac{3m+\sqrt{9m^2+20n}}{10} } \, . $$

$f(0) = -p < 0$ and $f'(0) = -n < 0$, therefore $f$ has a local minimum in the interval $(0, r)$ where $r$ is the smallest positive root of $f$. So $f'(x) = 0$ for some $x \in (0, r)$.

If $f$ has another positive root $s > r$ then $f'(x) = 0$ for some $x \in (r, s)$, in contradiction to the fact that $f'(x) = 0$ has only one positive solution.

share|cite|improve this answer
Can you explain why "f(0)=−p<0 and f′(0)=−n<0, therefore ff has a local minimum in the interval (0,r)" – Richkent Jan 11 at 6:43
@Richkent: $f$ has a minimum on the interval $[0, r]$. Since $f$ is decreasing near zero, the minimum can not be at $x = 0$. It can also not be at $x = r$, therefore it is in the interior of the interval, and then the derivative vanishes at the minimum. – Martin R Jan 11 at 6:47
@Richkent Try to visualize this explanation. Draw a rough graph. – Vikrant Desai Jan 11 at 7:09

Let $f(x)=x^5-mx^3-nx-p$.

$f(0)<0, \lim_{x\to\infty}f(x)=\infty$, so the IVT gives that $f$ has a positive root.

Let's call the least such root $a$, noting thus that $a^5=ma^3+na+p$

Then, for $x\ge 0, f'(x)=5x^4-3mx^2-n$.

Note that $f'(a)=5a^4-(3ma^2+n)>5a^4-3(ma^2+n)=5a^4-3(a^4-\frac{p}{a})=2a^4+\frac{3p}{a}>0$

Furthermore, for $x\ge a$, $f'(x)=5x^4-3mx^2-n\ge x^2(5a^2-3m)-n\ge a^2(5a^2-3m)-n$.


  • $a^5=ma^3+na+p\implies a^2=m+\frac{n}{a^2}+\frac{p}{a^3}$

  • $\implies a^2(5a^2-3m)=a^2(2m+\frac{5n}{a^2}+\frac{5p}{a^3})\ge 5n$

  • $\implies f'(x)\ge5n-n=4n>0$ for $x\ge a$

So the function has a root at $a$, and increases to the right of $a$, giving uniqueness.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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