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

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

• @NoChance $f(x)$ is supposed to have +ve coeffs – Anvit Jun 18 '19 at 11:08
• @OK, thanks for the remark. – NoChance Jun 18 '19 at 11:09

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

• I think that the sign of the constant term must be considered. For example $x^3-2x^2+1=0$ has 2 positive roots, (1 and 1.618) whereas $x^3-2x^2-1=0$ has only 1 positive root of value 2.206 (assuming no complex roots). – NoChance Jun 18 '19 at 11:21
• @NoChance What do you mean? In $$x^3=2x^2-1=:f(x)$$ $f(x)$ does not have positive coefficients. The constant term is the coefficient of $x^0$, it is still called a coefficient. – N. S. Jun 18 '19 at 11:24
• You are correct, thank you for your reply. – NoChance Jun 18 '19 at 11:24
• @NoChance No problem... – N. S. Jun 18 '19 at 11:26

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.

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.