Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Answer is (B): -2)
 
 
Line 1: Line 1:
 +
== Problem 13 ==
 +
 +
If <math>a,b,c</math>, and <math>d</math> are non-zero numbers such that <math>c</math> and <math>d</math> are the solutions of <math>x^2+ax+b=0</math> and <math>a</math> and <math>b</math> are
 +
the solutions of <math>x^2+cx+d=0</math>, then <math>a+b+c+d</math> equals
 +
 +
<math>\textbf{(A) }0\qquad
 +
\textbf{(B) }-2\qquad
 +
\textbf{(C) }2\qquad
 +
\textbf{(D) }4\qquad
 +
\textbf{(E) }(-1+\sqrt{5})/2  </math>
 +
 +
== Solution ==
 
By Vieta's formulas, <math>c + d = -a</math>, <math>cd = b</math>, <math>a + b = -c</math>, and <math>ab = d</math>. From the equation <math>c + d = -a</math>, <math>d = -a - c</math>, and from the equation <math>a + b = -c</math>, <math>b = -a - c</math>, so <math>b = d</math>.
 
By Vieta's formulas, <math>c + d = -a</math>, <math>cd = b</math>, <math>a + b = -c</math>, and <math>ab = d</math>. From the equation <math>c + d = -a</math>, <math>d = -a - c</math>, and from the equation <math>a + b = -c</math>, <math>b = -a - c</math>, so <math>b = d</math>.
  
 
Then from the equation <math>cd = b</math>, <math>cb = b</math>. Since <math>b</math> is nonzero, we can divide both sides of the equation by <math>b</math> to get <math>c = 1</math>. Similarly, from the equation <math>ab = d</math>, <math>ab = b</math>, so <math>a = 1</math>. Then <math>b = d = -a - c = -2</math>. Therefore, <math>a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{-2}</math>. The answer is (B).
 
Then from the equation <math>cd = b</math>, <math>cb = b</math>. Since <math>b</math> is nonzero, we can divide both sides of the equation by <math>b</math> to get <math>c = 1</math>. Similarly, from the equation <math>ab = d</math>, <math>ab = b</math>, so <math>a = 1</math>. Then <math>b = d = -a - c = -2</math>. Therefore, <math>a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{-2}</math>. The answer is (B).
 +
 +
==See Also==
 +
{{AHSME box|year=1978|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 21:06, 13 February 2021

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).

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_13&oldid=146527"