Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 16"
(New page: ==Problem== {{empty}} ==Solution== {{solution}} ==See also== {{CYMO box|year=2006|l=Lyceum|num-b=15|num-a=17}}) |
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==Problem== | ==Problem== | ||
| − | {{ | + | If <math>x_1,x_2</math> are the [[root]]s of the [[equation]] <math>x^2-2kx+2m=0</math>, then <math>\frac{1}{x_1},\frac{1}{x_2}</math> are the roots of the equation |
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| + | <math>\mathrm{(A)}\ x^2-2k^2x+2m^2=0\qquad\mathrm{(B)}\ x^2-\frac{k}{m}x+\frac{1}{2m}=0\qquad\mathrm{(C)}\ x^2-\frac{m}{k}x+\frac{1}{2m}=0\\ \mathrm{(D)}\ 2mx^2-kx+1=0\qquad\mathrm{(E)}\ 2kx^2-2mx+1=0</math> | ||
==Solution== | ==Solution== | ||
| − | {{ | + | By [[Vieta’s Formulae|Vieta’s]], <math>x_1 + x_2 = 2k,\ x_1 \cdot x_2 = 2m</math>. |
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| + | 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>. | ||
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=15|num-a=17}} | {{CYMO box|year=2006|l=Lyceum|num-b=15|num-a=17}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 09:28, 27 April 2008
Problem
If 
are the roots of the equation 
, then 
are the roots of the equation
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 | ||
