4. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. Given that $\angle C=2 \angle B, a b=24$. Then the range of values for $c$ is $\qquad$ .
4. $(4 \sqrt{2},+\infty)$.

As shown in Figure 4, draw the angle bisector $CD$. Then
$$
\begin{array}{l}
b^{2}a \Rightarrow c>\frac{a+b}{2} \text{. }
\end{array}
$$

Therefore, $c^{2}=a b+b^{2}>\frac{1}{4}(a+b)^{2}$, which means
$$
3 b^{2}+2 a b-a^{2}>0 \Rightarrow a8$.
Thus, $c^{2}=24+b^{2}>32 \Rightarrow c>4 \sqrt{2}$.
When $b \rightarrow 2 \sqrt{2}, a \rightarrow 6 \sqrt{2}, \alpha \rightarrow 0$, then $c \rightarrow 4 \sqrt{2}$. Hence, $c \in(4 \sqrt{2},+\infty)$.