Example 2 If the equation $\left(x^{2}-1\right)\left(x^{2}-4\right)=k$ has four non-zero real roots, and the four points corresponding to them on the number line are equally spaced, then $k=$ $\qquad$.
Solution: Let $x^{2}=y$, the original equation becomes
$$
y^{2}-5 y+(4-k)=0 \text {. }
$$

Assume this equation has real roots $\alpha, \beta(0<\alpha<\beta)$, then the four real roots of the original equation are $\pm \sqrt{\alpha} 、 \pm \sqrt{\beta}$. Since the four points corresponding to them on the number line are equally spaced, we have
$$
\sqrt{\beta}-\sqrt{\alpha}=\sqrt{\alpha}-(-\sqrt{\alpha}) \text {. }
$$

This gives $\beta=9 \alpha$.
By Vieta's formulas, we get
$$
\left\{\begin{array}{l}
\alpha+\beta=5, \\
\alpha \beta=4-k .
\end{array}\right.
$$

Solving (1) and (2), we get $\alpha=\frac{1}{2}, \beta=\frac{9}{2}$. Substituting into (3), we get
$$
\begin{array}{l}
4-k=\frac{9}{4} . \\
\therefore k=\frac{7}{4} .
\end{array}
$$