9. Given $f(x)=2^{x} m+x^{2}+n x$. If
$$
\{x \mid f(x)=0\}=\{x \mid f(f(x))=0\} \neq \varnothing,
$$

then the range of values for $m+n$ is . $\qquad$
9. $[0,4)$.

Let $x_{1} \in\{x \mid f(x)=0\}$. Then
$$
\begin{array}{l}
f\left(x_{1}\right)=2^{x_{1}} m+x_{1}^{2}+n x_{1}=0 \\
\Rightarrow f\left(f\left(x_{1}\right)\right)=f(0)=m=0 \\
\Rightarrow f(x)=x^{2}+n x \\
\Rightarrow f(f(x))=f\left(x^{2}+n x\right) \\
\quad=\left(x^{2}+n x\right)\left(x^{2}+n x+n\right) .
\end{array}
$$

Thus, the solution set $A$ of the equation $x^{2}+n x+n=0$ is a subset of the solution set $B$ of the equation $x^{2}+n x=0$.
If $A=\varnothing$, then $\Delta=n^{2}-4 n<0 \Rightarrow 0<n<4$.
If $A \neq \varnothing$, then $A=B \Rightarrow n=0$.
Therefore, $0 \leqslant n<4$.

Hence, the range of $m+n$ is $[0,4)$.