1. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-2 x-m=0$, and $2 x_{1}+x_{2}=0$. Then the value of $m$ is $\qquad$ .
By the relationship between roots and coefficients, we know $x_{1}+x_{2}=2$.
Also, $2 x_{1}+x_{2}=0$, then
$$
\begin{array}{l}
x_{1}+2=0 \\
\Rightarrow x_{1}=-2 \\
\Rightarrow(-2)^{2}-2 \times(-2)-m=0 \\
\Rightarrow m=8 .
\end{array}
$$