Liouville's Theorem states that for any algebraic $\alpha \in \mathbb{R}$ of degree $n$, there exists a positive constant $c:=c(\alpha)$ such that $$\left\lvert\alpha-\frac{p}{q}\right\rvert>\frac{c}{q^n}$$ for any $p \in \mathbb{Z}$ and $q \in \mathbb{N}.$


One can find an effective lower bound for $c(\alpha).$ In the special case that $\alpha$ is a quadratic irrational, Exercise 27 in the following set of notes of Jorn Steuding 


<http://www.math.uni-bremen.de/~bos/dioph.pdf>


yields $$c(\alpha) \gg \frac{1}{(1+|\alpha|)H(\alpha)} .$$
Here if $m\_{\alpha}(x):=x^2+bx+c \in \mathbf{Q}[x]$ is the minimal polynomial of $\alpha$, the height $H(\alpha)$ is defined as the maximum of $|b|$ and $|c|.$ My question is whether one can find a better lower bound for $c(\alpha)$ when $\alpha$ is a quadratic irrational or if this is best possible.



Liouville's Theorem in Diophantine Approximation
Let's assume $\alpha > 0$. If $\alpha$ is a quadratic irrational, its simple continued fraction $a\_0 + \dfrac{1}{a\_1 + \frac{1}{a\_2+\ldots}}$ is eventually periodic.
Every $p/q$ (in lowest terms) with $\left|\alpha - \dfrac{p}{q}\right| < \dfrac{1}{2q^2}$ is a convergent of $\alpha$,
and for the $n$'th convergent 
$$ \dfrac{1}{q\_n^2 (a\_{n+1}+2)} < \left| \alpha - \dfrac{p\_n}{q\_n} \right| \le \dfrac{1}{q\_n^2 a\_{n+1}}$$ 
Thus $\dfrac{1}{a\_M+2} \le c(\alpha) \le \dfrac{1}{a\_M}$ where $a\_M$ is the largest element in the continued fraction of $\alpha$. 


