2. Given that $2 r$ is an integer, the equation
$$
\left(r^{2}+r\right) x^{2}+4-r^{2}=0
$$

has rational roots. Then $r=$ $\qquad$
2. $\pm 2,-4$.

Let $r=\frac{m}{2}(m \in \mathbf{Z})$. Then
$$
\begin{array}{l}
\left(r^{2}+r\right) x^{2}+4-r^{2}=0 \\
\Leftrightarrow\left(m^{2}+2 m\right) x^{2}+16-m^{2}=0 .
\end{array}
$$

If $m=0$ or $m=-2$, the equation clearly has no solution. Therefore, for the equation to have rational roots, the discriminant of equation (1) must be a perfect square, i.e., $\Delta=m^{4}+2 m^{3}-16 m^{2}-32 m$ must be a perfect square.
When $m \leqslant-5$,
$$
\begin{array}{l}
\left(m^{2}+m-9\right)^{2}<m^{4}+2 m^{3}-16 m^{2}-32 m \\
<\left(m^{2}+m-8\right)^{2},
\end{array}
$$

it cannot be a perfect square. Therefore, we enumerate $m$ between -4 and 18 (using the factorized form of the discriminant is more convenient), and solve to get $m= \pm 4,-8$, corresponding to $r= \pm$ $2,-4$.