1978 AHSME Problems/Problem 13

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).

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1978 AHSME Problems/Problem 13