# 1978 AHSME Problems/Problem 13

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## Problem 13

If $a,b,c$, and $d$ are non-zero numbers such that $c$ and $d$ are the solutions of $x^2+ax+b=0$ and $a$ and $b$ are the solutions of $x^2+cx+d=0$, then $a+b+c+d$ equals

$\textbf{(A) }0\qquad \textbf{(B) }-2\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }(-1+\sqrt{5})/2$

## Solution

By Vieta's formulas, $c + d = -a$, $cd = b$, $a + b = -c$, and $ab = d$. From the equation $c + d = -a$, $d = -a - c$, and from the equation $a + b = -c$, $b = -a - c$, so $b = d$.

Then from the equation $cd = b$, $cb = b$. Since $b$ is nonzero, we can divide both sides of the equation by $b$ to get $c = 1$. Similarly, from the equation $ab = d$, $ab = b$, so $a = 1$. Then $b = d = -a - c = -2$. Therefore, $a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{-2}$. The answer is (B).