3. Given the real-coefficient equation $x^{3}+a x^{2}+b x+c=0$ whose three roots can serve as the eccentricities of an ellipse, a hyperbola, and a parabola, the range of $\frac{b}{a}$ is $\qquad$
3. $\left(-2,-\frac{1}{2}\right)$.

Analysis: Since the eccentricity of the parabola is 1, we have $1+a+b+c=0$, so the equation becomes $(x-1)\left[x^{2}+(a+1) x+(1+a+b)\right]=0$, which means the equation $x^{2}+(a+1) x+(1+a+b)=0$ has roots in $(0,1),(1,+\infty)$. From the distribution of the roots, we get: $\left\{\begin{array}{l}f(0)=a+b+1>0 \\ f(1)=2 a+b+3<0\end{array}\right.$, and by linear programming, we know: $\frac{b}{a} \in\left(-2,-\frac{1}{2}\right)$