Show that the equation $x^n=f(x)$ has one positive root

Question

Show that the equation $x^n=f(x)$ where $f(x)$ is a polynomial with positive coefficients of degree $n-1$, has only one positive root.

I found this problem but I'm having trouble solving it and I would really like some help.

I thought proof by contradiction by assuming that we have at least two positive roots that satisfy the equation but I don't really know where to go from there.

Sorry for any mistakes in my English. It's not my native language

Answer 1

This is (a particular case of) Descartes' rule of signs

Since $x^n-f(x)$ has exactly one sign change, the number of positive real roots is either 1 or an odd number less than 1. This means it has exactly 1 positive root.

You can find a proof for example here, or many other places

Answer 2

Let $g(x)=x^n -f(x)$. Let $M$ be the number of positive roots of $g(x)$. Observe that $g(0)=-k$ where $k>0$ is constant term in $f(x)$ . Also $lim_{x\rightarrow\infty}g(x)\rightarrow\infty$ so there is at least one root in $(0,\infty)$ which sets the lower bound for $M$ i.e. $M\ge 1$.

Descarte's rule of sign changes in $g(x)$ sets the upper bound for $M$ i.e. $M\le 1$. So $M=1$.

Furthermore, if $f(x)$ doesn't have a constant term then $g(x)=x^r.h(x)$ where $g(x)$ has a root at $x=0$ with multiplicity $r$. You can work along the same lines on $h(x)$ now.

Answer 3

HINT:

It's enough to show that $$1=\frac{f(x)}{x^n}$$ has one root in $(0,\infty)$. Note that the RHS is decreasing.