5. Let $a, b \in \mathbf{R}$, the equation $\left(x^{2}-a x+1\right)\left(x^{2}-b x+1\right)=0$ has 4 real roots that form a geometric sequence with common ratio $q$. If $q \in\left[\frac{1}{3}, 2\right]$, then the range of $a b$ is $\qquad$.
5. $\left[4, \frac{112}{9}\right]$ Detailed Explanation: From the conditions, it is easy to see that the two quadratic equations $x^{2}-a x+1=0$ and $x^{2}-b x+1=0$ have roots that correspond to the first and fourth terms of a geometric sequence for one equation, and the second and third terms for the other. Thus, $x_{1}^{2} \cdot q^{3}=1$ and $x_{1} q \cdot x_{1} q^{2}=1$, and $a b = x_{1}\left(1+q^{3}\right) \cdot x_{1}\left(q+q^{2}\right)$, where $x_{1}$ is the first term of the geometric sequence. Since $q \in\left[\frac{1}{3}, 2\right]$, it follows that $q+\frac{1}{q} \in\left[2, \frac{10}{3}\right]$, and $a b = x_{1}^{2} \cdot q(1+q)\left(1+q^{3}\right) = \frac{1+q+q^{3}+q^{4}}{q^{2}} = \left(q^{2}+\frac{1}{q^{2}}\right) + \left(q+\frac{1}{q}\right) = \left(q+\frac{1}{q}\right)^{2} + \left(q+\frac{1}{q}\right) - 2$. Using the fact that $t^{2}+t-2$ is an increasing function of $t$ for $2 \leqslant t \leqslant \frac{10}{3}$, we get $a b \in\left[4, \frac{112}{9}\right]$.