Difference between revisions of "1966 AHSME Problems/Problem 3"

Revision as of 00:27, 26 June 2016

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)
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All AHSME Problems and Solutions

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

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 ==See Also==
-{{AHSME box|year=1966|num-b=2|num-a=3}}
+{{AHSME box|year=1966|num-b=3|num-a=4}}
 {{MAA Notice}}

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Difference between revisions of "1966 AHSME Problems/Problem 3"

Revision as of 00:27, 26 June 2016

Problem

Solution

See Also