$ax^2+bx+c=0$ 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.
 A: Another way to see it is to look at sum and product of roots. $x_1$ and $x_2$ are roots of $ax^2+bx+c=0$ so $x_1+x_2=-\frac{b}{a}$ and $x_1\cdot x_2=\frac{c}{a}$.
$$\frac{a}{c}=\frac{1}{x_1}\cdot \frac{1}{x_2}$$
$$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}=-\frac{b}{c}$$
and this shows that $\frac{1}{x_1}$ and $\frac{1}{x_2}$ are the roots of 
$$X^2+\frac{b}{c}X+\frac{a}{c}=0$$
Equivalently
$$cX^2+bX+a=0$$
A: 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 $\frac{1}{x_2}$
A: Given $ax^2 + bx + c = 0$,
$$\begin{align} ax^2 + bx + c& = 0\\ a + b\left(\frac{1}{x}\right) + c\left(\frac{1}{x^2}\right) & = 0 \tag{Divide both sides by $x^2$}\\ cy^2 + by + a & = 0\tag{Let $y = \frac{1}{x}$}\end{align}$$
We'd discover that this polynomial has roots $y_1$ and $y_2$. We know that $y = \frac{1}{x}$. So, without loss of generality, assume $y_1 = \frac{1}{x_1}$. Then $y_2 = \frac{1}{x_2}$. These are thus the roots.
