<s>[INST] $ax^2+bx+c=0$ 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.
 [/INST] 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.
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