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?


2 Answers 2


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.


Your method is hard to understrand and ts unclear what you are trying to do

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


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