5. If three non-zero real numbers $x(y-z), y(z-$ $x), z(y-x)$ form a geometric sequence, then its common ratio is $\qquad$
5. $\frac{1 \pm \sqrt{5}}{2}$.

Notice that $x(y-z)+y(z-x)=z(y-x)$, so, $1+\frac{y(z-x)}{x(y-z)}=\frac{z(y-x)}{x(y-z)}=\frac{y(z-x)}{x(y-z)} \cdot \frac{z(y-x)}{y(z-x)}$,
i.e., $1+q=q^{2} (q$ is the common ratio $)$.
Solving for $q$ yields $q=\frac{1 \pm \sqrt{5}}{2}$.