AoPSWiki

Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

Personal tools

Search

Toolbox

1966 AHSME Problems/Problem 3

From AoPSWiki

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)
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1966_AHSME_Problems/Problem_3"

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us