8. Let real numbers $a, b, c, d$ satisfy $ad + bc = 1$. Then
$$
u=a^{2}+b^{2}+c^{2}+d^{2}+(a+c)^{2}+(b-d)^{2}
$$

the minimum value of $u$ is $\qquad$.
8. $2 \sqrt{3}$.
$$
\begin{array}{l}
\text { Given } \frac{u}{2}-t \\
=a^{2}+b^{2}+c^{2}+d^{2}+a c-b d-t(a d+b c) \\
=\left(a+\frac{c-t d}{2}\right)^{2}+\left(b-\frac{d+t c}{2}\right)^{2}+\left(1-\frac{t^{2}+1}{4}\right)\left(c^{2}+d^{2}\right),
\end{array}
$$

Let $t=\sqrt{3}$, then $u \geqslant 2 \sqrt{3}$.
When $a+\frac{c-t d}{2}=0, b-\frac{d+t c}{2}=0$, and $a d+b c=1$, we can take $a=d, b=-c$, i.e., $a^{2}=d^{2}=\frac{2 \sqrt{3}+3}{6}$, $b^{2}=c^{2}=\frac{2 \sqrt{3}-3}{6}$, then $u_{\text {min }}=2 \sqrt{3}$.