8. Given real numbers $a, b, c$ satisfy
$$
a \geqslant b \geqslant c, a+b+c=0 \text {, and } a \neq 0 \text {. }
$$

Let $x_{1}, x_{2}$ be the two real roots of the equation $a x^{2}+b x+c=0$. Then the maximum distance between two points $A\left(x_{1}, x_{2}\right)$ and $B\left(x_{2}, x_{1}\right)$ in the Cartesian coordinate system is $\qquad$
8. $3 \sqrt{2}$.

From the problem, we know $a>0, c<0$.
Thus, from $a \geqslant b \geqslant c$, we get $\frac{c}{a} \leqslant \frac{b}{c} \leqslant 1$.
Also, from $a+b+c=0$, we get $\frac{b}{a}=-1-\frac{c}{a}$.
Therefore, $\frac{c}{a} \leqslant-1-\frac{c}{a} \leqslant 1, -2 \leqslant \frac{c}{a} \leqslant-\frac{1}{2}$.
Since $a+b+c=0$, the roots of $a x^{2}+b x+c=0$ are $1, \frac{c}{a}$.
$$
\begin{array}{l}
\text { Hence }|A B|=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{1}\right)^{2}} \\
=\sqrt{2}\left|x_{1}-x_{2}\right|=\sqrt{2}\left(1-\frac{c}{a}\right) . \\
\text { When } \frac{c}{a}=-2 \text {, }|A B|_{\max }=3 \sqrt{2} .
\end{array}
$$