6. If the two quadratic equations $x^{2}+x+m=$ 0 and $m x^{2}+x+1=0$ each have two distinct real roots, but one of them is a common real root $\alpha$, then the range of the real root $\alpha$ is $\qquad$ .
6. $\alpha=1$.

From the problem, we know $m \neq 0$, and $\Delta=1-4 m>0$, i.e., $m<\frac{1}{4}$ and $m \neq 0$; simultaneously, $\alpha^{2}+\alpha+m=0$, and $m \alpha^{2}+\alpha+1=0$.
Subtracting these equations gives $(1-m) \alpha^{2}+(m-1)=0$.

Since $m \neq 1$, we have $\alpha^{2}=1$. Solving this, we get $\alpha= \pm 1$.
If $\alpha=-1$, then $m=0$, which contradicts the problem statement.
If $\alpha=1$, then $m=-2$.