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 "1966 AHSME Problems/Problem 3"

(→‎See Also)
Line 16: Line 16:
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1966|num-b=2|num-a=3}}
 
{{AHSME box|year=1966|num-b=2|num-a=3}}
 +
{{MAA Notice}}

Revision as of 11:39, 5 July 2013

Problem

If the arithmetic mean of two numbers is $6$ and their geometric mean is $10$, then an equation with the given two numbers as roots is:

$\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0$ $\text{(D)} \ x^2 - 12x + 100 = 0 \qquad \text{(E)} \ x^2 - 6x + 100 = 0$

Solution

Let the numbers be $\eta$ and $\zeta$.

\[\dfrac{\eta+\zeta}{2}=6\Rightarrow \eta+\zeta=12\].

\[\sqrt{\eta\zeta}=10\Rightarrow \eta\zeta=100\].

The monic quadratic with roots $\eta$ and $\zeta$ is $x^2-(\eta+\zeta)x+\eta\zeta$. Therefore, an equation with $\eta$ and $\zeta$ as roots is $x^2 - 12x + 100 = 0\Rightarrow \text{(D)}$

See Also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 3
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=1966_AHSME_Problems/Problem_3&oldid=56238"