# How to prove there is only one positive root?

If I want to prove $$x^n+x^{n-1}-a=0$$ has only one positive root for $$a> 0$$ and $$n \ge 2$$ Can I say for $$n=2$$ : $$x^2+x-a=0$$ and we know only one of the roots is positive. Now if we we know there is only one root for $$x^n+x^{n-1}-a=0$$ now I prove for $$x^{n+1}+x^{n}-a=0$$ In the end I get to $$x=1$$ , but I don't think I’m doing it correctly. Should I use different $$a$$ s?

Is my way wrong? If it is, could you tell me what I should do?

Let $$p(x) = x^n + x^{n-1} - a.$$
That it has at least one root is clear by Intermediate Value Property. (The value at $$0$$ is negative and $$p(x) \to \infty$$ as $$x \to \infty$$.)
Now, note that $$p'(x) = nx^{n-1} + (n - 1)x^{n-2} > 0$$ for all $$x > 0.$$
Thus, if $$p$$ had two positive roots $$a < b$$, then $$p'(c) = 0$$ for some $$a < c < b$$, by Rolle's theorem. But $$c > 0$$ and hence, $$p'(c) = 0$$ is not possible.
Suppose by contradiction ther is more than one possible positive root say $$c,b$$ are two of them.WLOG $$c>b$$ now $$c^n+c^{n-1}=a$$ $$b^n+b^{n-1}=a$$ subtract $$\color{red}{c^n-b^n}+\color{blue}{c^{n-1}-b^{n-1}}=0$$ which is not possible as both colored terms are positive