1966 AHSME Problems/Problem 3

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 $x^2 - 12x + 100 = 0\Rightarrow \text{(D)}$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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.

Page

Toolbox

Search

1966 AHSME Problems/Problem 3

Problem

Solution

See Also