I have a quadratic polynomial $f(x)\in\mathbb Z[x]$. I want to compute its roots modulo $q^2$.
The polynomial $f(x)\bmod q$ has a double root at $r_1$ because $disc\equiv0\bmod q$.
In my case $q|f'(r_1)$ (because of double root) and so there is no inverse of $f'(r_1)$.
It satisfies the condition $|f(r_1)|_q\leq |f'(r_1)|^2_q$ and so there is an unique root satisfying $|r_2-r_1|_q=|f(r_1)/f'(r_1)|_p<|f'(r_1)|_p$ and $|f'(r_2)|_p=|f'(r_1)|_p$ by https://kconrad.math.uconn.edu/blurbs/gradnumthy/hensel.pdf. But I do not understand how the unique root satisfying these conditions is derived. Can someone explain?
How should I do the Hensel lifting process to compute this unique root $r_2$ so that $f(r_2)\equiv0\bmod q^2$?
Either there are no solutions or every lift modulo $q^2$ is a solution, depending on whether $f(r_1)$ has valuation $1$ or more than $1$ (for any lift of $r_1$).
You have
$$f(r_1 + x) = f(r_1) + f'(r_1) x + O(x^2),$$
If you want to lift $r_1$ to a solution modulo $q^2$ you need $q|x$, but then $f'(r_1) x$ is divisible by $q^2$ because $q$ divides $f'(r_1)$. Hence all lifts have the same value modulo $q^2$, and thus either very lift is a solution or there are no solutions.
Of course both possibilities can happen (take $f(x) = x^2$ or $f(x) = x^2 - q$).