Example 2 If the two quadratic equations
$$
x^{2}+x+m=0 \text { and } m x^{2}+x+1=0
$$

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 $\qquad$.
Given that the common real root of the two equations is $\alpha$, and $m \neq 1$. Then
$$
\begin{array}{l}
\alpha^{2}+\alpha+m=0, \\
m \alpha^{2}+\alpha+1=0 .
\end{array}
$$

From equation (1), we get $m=-\alpha^{2}-\alpha$.
Substituting into equation (2), we get
$$
\begin{array}{l}
\alpha^{4}+\alpha^{3}-\alpha-1=0 \\
\Rightarrow\left(\alpha^{2}-1\right)\left(\alpha^{2}+\alpha+1\right)=0 .
\end{array}
$$

Since $\alpha^{2}+\alpha+1=\left(\alpha+\frac{1}{2}\right)^{2}+\frac{3}{4}>0$, then
$$
\alpha^{2}-1=0 \Rightarrow \alpha= \pm 1 \text {. }
$$

When $\alpha=1$, from equation (1) we get $m=-2$, at this time, the given two equations become
$$
x^{2}+x-2=0 \text { and } 2 x^{2}-x-1=0,
$$

with distinct roots -2 and $\frac{1}{2}$.
When $\alpha=-1$, from equation (1) we get $m=0$, at this time, the equation $m x^{2}+x+1=0$ becomes a linear equation $x+1=0$, which contradicts the problem statement.
In summary, the value of the real root $\alpha$ is $\alpha=1$.