2.55 If $a, b, c$ and $d$ are non-zero numbers, $c$ and $d$ are solutions of $x^{2}+a x+b=0$, and $a$ and $b$ are solutions of $x^{2}+c x+d=0$, then $a+b+c+d$ equals
(A) 0.
(B) -2.
(C) 2.
(D) 4.
(E) $=\frac{1+\sqrt{5}}{2}$.
(29th American High School Mathematics Examination, 1978)
［Solution］From the given conditions and Vieta's formulas, we have
$$
\left\{\begin{array}{l}
c+d=-a, \\
c d=b, \\
a+b=-c, \\
a b=d,
\end{array}\right.
$$

From (1), we get $a+c=-d$,
From (3), we get $a+c=-b, \therefore b=d$, substituting into (2) gives $c=1$. Substituting $b=d$ into (4) gives $a=1$.
Adding (1) and (3) gives $a+b+c+d=-(a+c)=-2$.

Therefore, the answer is $(B)$.