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Question :

If $x_1, x_2, \cdots, x_n$ are the roots of the equation: $$\mathcal{f}(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_0 ~,$$ find the roots of the following equation : $$\mathcal{g}(x)=a_0x^n-a_1x^{n-1}+\cdots + (-1)^na_n$$

I've stated the question above.

I noticed that $\mathcal{g}(x)= x^n \mathcal{f}\left(-\frac{1}{x}\right)$. It is essentially same as writing $\mathcal{f}\left(-\frac{1}{x}\right)$ to avoid ambiguity of $g(0)=0$ which isn't true unless $a_n=0$.

Basically, we have to find the roots of $\mathcal{f}\left(-\frac{1}{x}\right)$. How to find that?

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Mathejunior
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  • Sorry I hadn't seen... I erase my silly comment. But if $f(x)=0$ has root $2$ don't you think that $-\tfrac12$ is a root of $f(-\tfrac1x)=0$? – Jean Marie Sep 20 '17 at 14:45
  • @JeanMarie Yeah I think something like that but couldn't prove it. I can prove that the roots of $f(-\tfrac 1x)$ are $\tfrac{1}{x_i}$ (using Vieta's formula) and then since roots are unique, it holds. But I am trying to find a way of finding the roots in a straight forward way by substituting​ the equations or whatsoever. – Mathejunior Sep 20 '17 at 15:02

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For $x_i\not=0$, we have $$\begin{align}g\left(-\frac{1}{x_i}\right)&=a_0\left(-\frac{1}{x_i}\right)^n-a_1\left(-\frac{1}{x_i}\right)^{n-1}+\cdots +(-1)^na_n\\\\&=\left(-\frac{1}{x_i}\right)^n\left(a_0+a_1x_i+\cdots +a_nx_i^{n}\right)\\\\&=\left(-\frac{1}{x_i}\right)^nf(x_i)\\\\&=0\end{align}$$

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