$\because 、\left(20\right.$ points) For a quadratic equation with real coefficients $a x^{2}+2 b x$ $+ c=0$ having two real roots $x_{1}, x_{2}$. Let $d=\left|x_{1}-x_{2}\right|$. Find the range of $d$ when $a>b>c$ and $a+b+c=0$.
By Vieta's formulas, $x_{1}+x_{2}=-\frac{2 b}{a}, x_{1} x_{2}=\frac{c}{a}$.
$$
\begin{aligned}
d^{2} & =\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2} \\
& =\frac{4 b^{2}-4 a c}{a^{2}}=\frac{4 b^{2}+4 a(a+b)}{a^{2}} \\
& =4\left[\left(\frac{b}{a}\right)^{2}+\frac{b}{a}+1\right] .
\end{aligned}
$$

From $a>b>c$ and $a+b+c=0$, we can get $a>0$ and $a>b>-a-b$. Therefore, $\frac{b}{a}>-1-\frac{b}{a}, \frac{b}{a}>-\frac{1}{2}$.
Thus, $-\frac{1}{2}<\frac{b}{a}<1$.
The quadratic function $\left(\frac{b}{a}\right)^{2}+\frac{b}{a}+1$ has a range of $\frac{3}{4}<\left(\frac{b}{a}\right)^{2}+\left(\frac{b}{a}\right)+1<\frac{12}{4}$ when $-\frac{1}{2}<\frac{b}{a}<1$.
$$
\text { Hence } \sqrt{3}<d<2 \sqrt{3} \text {. }
$$