3. Given the quadratic function $y=a x^{2}+b x+c(a, b, c$ are real numbers) whose graph intersects the $x$-axis at points $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $a>0, a>2 c>b$, when $x=1$, $y=-\frac{a}{2}$. Then the range of $| x_{1}-x_{2} |$ is ( ).
(A) $(0,1)$
(B) $(1,2)$
(C) $\left(\frac{1}{2}, \frac{3}{2}\right)$
(D) $(\sqrt{2}, \sqrt{3})$
3. D.

Since when $x=1$, $y=-\frac{a}{2}$, we have
$$
3 a+2 b+2 c=0,
$$

which means $2 c=-3 a-2 b$.
From $a>2 c>b$, we know $a>-3 a-2 b>b$.
Since $a>0$, then $1>-3-2 \times \frac{b}{a}>\frac{b}{a}$, which means
$$
-2<\frac{b}{a}<-1 \text {. }
$$

Given that the graph of the quadratic function intersects the $x$-axis at points $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, we have
$$
x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=-\frac{3}{2}-\frac{b}{a} \text {. }
$$

Thus, $\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}$
$$
=\sqrt{\left(\frac{b}{a}+2\right)^{2}+2} \text {. }
$$

Therefore, $\sqrt{2}<\left|x_{1}-x_{2}\right|<\sqrt{3}$.