9. A. Given real numbers $a, b, c, d$ satisfy: the quadratic equation $x^{2}+c x+d=0$ has roots $a, b$, and the quadratic equation $x^{2}+a x+b=0$ has roots $c, d$. Then all sets of $(a, b, c, d)$ that meet the conditions are $\qquad$ .
9. A. $(1,-2,1,-2),(t, 0,-t, 0)(t$ is any real number).
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a+b=-c, \\
a b=d, \\
c+d=-a, \\
c d=b .
\end{array}\right.
$$

It is easy to see that $b=-a-c=d$.
If $b=d \neq 0$, then
$$
\begin{array}{l}
a=\frac{d}{b}=1, c=\frac{b}{d}=1 \\
\Rightarrow b=d=-a-c=-2 ;
\end{array}
$$

If $b=d=0$, then $c=-a$
$\Rightarrow(a, b, c, d)=(t, 0,-t, 0)(t$ is any real number).

Upon verification, the arrays $(1,-2,1,-2)$ and $(t, 0,-t, 0)$ ( $t$ is any real number) satisfy the conditions.