5. If $a>b>c, a+b+c=0$, and $x_{1}, x_{2}$ are the two real roots of $a x^{2}+b x+c=0$. Then the range of $\left|x_{1}^{2}-x_{2}^{2}\right|$ is $\qquad$
5. $[0,3)$.

From $a+b+c=0$, we know that the equation $a x^{2}+b x+c=0$ has a real root of 1. Without loss of generality, let $x_{1}=1$.
Then, by Vieta's formulas, we know $x_{2}=\frac{c}{a}$.
Given $a>b>c, a+b+c=0$, it follows that $a>0, c-a-c>c$.
Thus, $-2<\frac{c}{a}<-\frac{1}{2}$.
Therefore, $\frac{1}{4}<x_{2}^{2}=\left(\frac{c}{a}\right)^{2}<4$.
Hence, $\left|x_{1}^{2}-x_{2}^{2}\right| \in[0,3)$.