7. Given a quartic polynomial $f(x)$ whose four real roots form an arithmetic sequence with a common difference of 2. Then the difference between the largest and smallest roots of $f^{\prime}(x)$ is $\qquad$
$7.2 \sqrt{5}$.
Let the four real roots of the quartic polynomial $f(x)$ be $a, a+2, a+4, a+6$. Then,
$$
f(x)=k(x-a)(x-a-2)(x-a-4)(x-a-6)
$$
$(k \neq 0)$.
Let $t=x-a-3$. Thus,
$$
\begin{array}{l}
f(x)=k(t+3)(t+1)(t-1)(t-3) \\
=k\left(t^{4}-10 t^{2}+9\right)=g(t) .
\end{array}
$$

Then, $g^{\prime}(t)=k\left(4 t^{3}-20 t\right)=4 k t\left(t^{2}-5\right)$.
Therefore, the three roots of $g^{\prime}(t)$ are $-\sqrt{5}, 0, \sqrt{5}$.
Thus, the three roots of $f^{\prime}(x)$ are
$$
\begin{array}{l}
a+3-\sqrt{5}, a+3, a+3+\sqrt{5}. \\
\text { Hence }(a+3+\sqrt{5})-(a+3-\sqrt{5})=2 \sqrt{5} \text {. }
\end{array}
$$