1. Given two equations about $x$
$$
x^{2}-x+3 m=0, x^{2}+x+m=0(m \neq 0) \text {. }
$$

If one root of the first equation is three times a root of the second equation, then the value of the real number $m$ is $\qquad$
- 1. -2 .

Let one root of the latter equation be $\alpha$. And the former equation has a root $3 \alpha$, then
$$
\alpha^{2}+\alpha+m=0,
$$

and
$$
\begin{array}{l}
9 \alpha^{2}-3 \alpha+3 m=0 \\
\Rightarrow 3 \alpha^{2}-\alpha+m=0 .
\end{array}
$$
(2) - (1) gives
$$
2 \alpha^{2}-2 \alpha=0 \Rightarrow \alpha=0 \text { or } 1 \text {. }
$$

If $\alpha=0$, then $m=0$, which contradicts the assumption that $m \neq 0$. Therefore, $\alpha=1$. Consequently, $m=-2$.