$ax^2+bx+c=0$ has roots $x_1,x_2$. what are the roots of $cx^2+bx+a=0$. Given solution: Dividing the first equation by $x^2$ we get
$c(\frac{1}{x^2})+b(\frac{1}{x})+a=0$
so $(\frac{1}{x_1}),(\frac{1}{x_2})$ are the roots of $cx^2+bx+a=0$.{How?It is not obvious to me.}
The answers so far are proving retrospectively that the roots are indeed those given above.I would like to know how the relation of the roots is derived.

The definition says: $t$ is a root of a polynomial $P$ if and only if $P(t)=0$.
so let $P(x)=cx^2+bx+a$, we have :
$$P(\frac{1}{x_1})=c(\frac{1}{x_1^2})+b(\frac{1}{x_1})+a=\frac{c+bx_1+ax_1^2}{(x_1)^2}$$
and as we know $x_1$ is a root of $ax^2+bx+c=0$ hence $ax_1^2+bx_1+c=0$ and finally:
$P(\frac{1}{x_1})=0$
as a conclusion $\frac{1}{x_1}$ is a root of $cx^2+bx+a$.
You can do the same for $\frac{1}{x_2}$
