Three, (10 points) A quadratic trinomial $x^{2}+p x+q$ with coefficients $p$ and $q$ as integers, and roots as irrational numbers $\alpha_{1}, \alpha_{2}$, is called an irrational quadratic trinomial. Find the minimum value of the sum of the absolute values of the roots for all irrational quadratic trinomials.
Three, let $x^{2}+p x+q$ be an irrational quadratic trinomial, and $D=p^{2}-4 q$ be its discriminant.
Notice,
$$
\begin{array}{l}
\left(\left|\alpha_{1}\right|+\left|\alpha_{2}\right|\right)^{2} \geqslant\left(\alpha_{1}-\alpha_{2}\right)^{2} \\
=\left(\frac{-p+\sqrt{D}}{2}-\frac{-p-\sqrt{D}}{2}\right)^{2}=D .
\end{array}
$$

By the definition of an irrational quadratic trinomial, we know that $D$ is a non-negative integer, but not a perfect square.
In particular, $D \neq 0,1,4$.
Also, $p^{2}-4 q \equiv p^{2}(\bmod 4)$, so $D \neq 2,3$.
Thus, $D \geqslant 5$.
From equation (1), we get $\left|\alpha_{1}\right|+\left|\alpha_{2}\right| \geqslant \sqrt{5}$.
In fact, the minimum value $\sqrt{5}$ can be achieved.
For example, when $p=1, q=-1$, $x^{2}+x-1$ is an irrational quadratic trinomial, with roots $\alpha_{1,2}=\frac{-1 \pm \sqrt{5}}{2}$, at this time,
$$
\left|\alpha_{1}\right|+\left|\alpha_{2}\right|=\left|\frac{-1+\sqrt{5}}{2}\right|+\left|\frac{-1-\sqrt{5}}{2}\right|=\sqrt{5} .
$$