Let $t$ be a real number and let $\frac{p_n}{q_n}$ be its continued fraction approximants. These have the property that $$ \left| t - \frac{p_n}{q_n} \right| < \frac{1}{q_n q_{n+1}} $$
In other words, $t$ is an approximate solution to the equation $q_n x - p_n = 0$, with error $< \frac{1}{q_{n+1}}$.
Suppose $1,t,t^2$ are rationally independent. Is there a sequences of triples of integers $(a_n,b_n,c_n)$ such that $a_n t^2 + b_n t + c_n$ tends to zero as $n$ tends to infinity, and each triple $(a_n,b_n,c_n)$ is the best such choice of coefficients, in some sense?
Presumably you're not allowing $a_n = b_n = c_n = 0$.
Clearly yes, in some sense: if you restrict the possible $(a, b, c)$ in some way (e.g. a bound on $\max(|a|,|b|,|c|)$) that leaves finitely many possible choices, one of them must have the smallest value of $|at^2 + b t + c |$.
If, for example, you impose $a = b$, you're essentially approximating $t^2 + t$ by rationals, so by Hurwitz's theorem you can achieve $|a t^2 + a t + c| \le 1/(\sqrt{5} a)$. Presumably you can do better by allowing $b \ne a$, but I don't know how much better.