8. If the three sides of a triangle form a geometric sequence, then the range of the common ratio $q$ is $\qquad$ .
8. $\frac{\sqrt{5}-1}{2}$ $0, q \geqslant 1$. Then $a + aq > aq^2$, which means $q^2 - q - 1 < 0, 0 < q < 1$, and $q^2 + q - 1 > 0$. Solving this, we get
$$
\frac{\sqrt{5}-1}{2} < q < 1 \text{. }
$$

In summary, $\frac{\sqrt{5}-1}{2} < q < \frac{\sqrt{5}+1}{2}$.