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

(New page: ==Problem== If the arithmetic mean of two numbers is <math>6</math> and their geometric mean is <math>10</math>, then an equation with the given two numbers as roots is...)
 
 
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<cmath>\sqrt{\eta\zeta}=10\Rightarrow \eta\zeta=100</cmath>.
 
<cmath>\sqrt{\eta\zeta}=10\Rightarrow \eta\zeta=100</cmath>.
  
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The [[monic]] [[quadratic]] with roots <math>\eta</math> and <math>\zeta</math> is <math>x^2-(\eta+\zeta)x+\eta\zeta</math>. Therefore, an equation with <math>\eta</math> and <math>\zeta</math> as roots is <math>x^2 - 12x + 100 = 0\Rightarrow \text{(D)}</math>
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The [[monic polynomial|monic]] [[quadratic]] with roots <math>\eta</math> and <math>\zeta</math> is <math>x^2-(\eta+\zeta)x+\eta\zeta</math>. Therefore, an equation with <math>\eta</math> and <math>\zeta</math> as roots is <math>x^2 - 12x + 100 = 0\Rightarrow \text{(D)}</math>
  
 
==See Also==
 
==See Also==
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{{AHSME box|year=1966|num-b=2|num-a=4}}
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{{MAA Notice}}

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

See Also

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

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