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Problem 551

AMC 12 late, AIME early

Algebra Difficulty 5.0 Find the answer

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)

Your answer

Official solution

［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 $(B)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.