Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 16"
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==Solution== | ==Solution== | ||
| − | By [[Vieta’s | + | By [[Vieta’s Formulae|Vieta’s]], <math>x_1 + x_2 = 2k,\ x_1 \cdot x_2 = 2m</math>. |
The equation with roots <math>x = \frac{1}{x_1}, \frac{1}{x_2}</math> is <math>0 = \left(x - \frac{1}{x_1}\right)\left(x - \frac{1}{x_2}\right)</math> <math> = x^2 - \left(\frac{1}{x_1} + \frac{1}{x_2}\right) + \frac{1}{x_1x_2} = x^2 - \left(\frac{x_1 + x_2}{x_1x_2}\right) + \frac{1}{x_1x_2}</math>. Substituting from above, we get <math>x^2 - \frac{k}{m}x + \frac{1}{2m}= 0 \Longrightarrow \mathrm{(B)}</math>. | The equation with roots <math>x = \frac{1}{x_1}, \frac{1}{x_2}</math> is <math>0 = \left(x - \frac{1}{x_1}\right)\left(x - \frac{1}{x_2}\right)</math> <math> = x^2 - \left(\frac{1}{x_1} + \frac{1}{x_2}\right) + \frac{1}{x_1x_2} = x^2 - \left(\frac{x_1 + x_2}{x_1x_2}\right) + \frac{1}{x_1x_2}</math>. Substituting from above, we get <math>x^2 - \frac{k}{m}x + \frac{1}{2m}= 0 \Longrightarrow \mathrm{(B)}</math>. | ||
Revision as of 20:58, 17 October 2007
Problem
If 
are the roots of the equation 
, then 
are the roots of the equation
A. 
B. 
C. 
D. 
E. 
Solution
By Vieta’s, 
.
The equation with roots 
is 
. Substituting from above, we get 
.
See also
| 2006 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
