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>. | ||
| − | 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> | + | 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== | ||
Revision as of 10:29, 2 April 2008
Problem
If the arithmetic mean of two numbers is 
and their geometric mean is 
, then an equation with the given two numbers as roots is:
Solution
Let the numbers be 
and 
.
![\[\dfrac{\eta+\zeta}{2}=6\Rightarrow \eta+\zeta=12\]](http://latex.artofproblemsolving.com/a/f/e/afe6533bbea01adc1130bcd5c72896bb8a28d5a5.png)
.
![\[\sqrt{\eta\zeta}=10\Rightarrow \eta\zeta=100\]](http://latex.artofproblemsolving.com/d/e/7/de7a4ce3688ec5ffd545921a25adf78c83084bae.png)
.
The monic quadratic with roots and is 
. Therefore, an equation with and as roots is 
