Proving a polynomial has only one positive root. Let $m, n, p\in \mathbb R$, $n>0, p>0$. Prove that the following equation has exactly one positive solution:
$$x^5-mx^3-nx-p=0.$$  
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?  
 A: 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.
A: 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.
A: 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 $x\ge a$, $f'(x)=5x^4-3mx^2-n\ge x^2(5a^2-3m)-n\ge a^2(5a^2-3m)-n$.
Now:


*

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

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

*$\implies f'(x)\ge5n-n=4n>0$ for $x\ge a$
So the function has a root at $a$, and increases to the right of $a$, giving uniqueness.
