101. Given two quadratic equations $x^{2}-x+m=0, x^{2}-x+$ $+3 m=0, m \neq 0$. Find the value of $m$ for which one of the roots of the second equation is twice the root of the first equation.
Solution. Let $x_{0}$ be the root of the first equation, and $2 x_{0}$ be the root of the second. Then the following numerical equalities hold:

$$
\begin{gathered}
x_{0}^{2}-x_{0}+m=0 \\
4 x_{0}^{2}-2 x_{0}+3 m=0
\end{gathered}
$$

Subtracting the second equality from three times the first, we get $-x_{0}^{2}-x_{0}=0$, from which $x_{0}=0$ or $x_{0}=-1$. If $x_{0}=0$, then from the first equality it follows that $m=0$, which contradicts the condition. If $x_{0}=-1$, then we get $m=-2$. (Verification shows that the value $m=-2$ indeed satisfies the condition of the problem.)

Answer: $m=-2$.