Difference between revisions of "1978 AHSME Problems/Problem 13"

Revision as of 12:06, 5 March 2025

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).

1978 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 \textbf{(C) }2\qquad
 \textbf{(D) }4\qquad
 \textbf{(E) }(-1+\sqrt{5})/2  </math>
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Difference between revisions of "1978 AHSME Problems/Problem 13"

Revision as of 12:06, 5 March 2025

Problem 13

Solution

See Also