Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 16"

(solution)
m (Style)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
If <math>x_1,x_2</math> are the roots 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
+
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
  
−
A. <math>x^2-2k^2x+2m^2=0</math>
+
<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>
−
 
 
−
B. <math>x^2-\frac{k}{m}x+\frac{1}{2m}=0</math>
 
−
 
 
−
C. <math>x^2-\frac{m}{k}x+\frac{1}{2m}=0</math>
 
−
 
 
−
D. <math>2mx^2-kx+1=0</math>
 
−
 
 
−
E. <math>2kx^2-2mx+1=0</math>
 
  
 
==Solution==
 
==Solution==
−
By [[Vieta’s Formulas|Vieta’s]], <math>x_1 + x_2 = 2k,\ x_1 \cdot x_2 = 2m</math>.
+
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>.

Latest revision as of 09:28, 27 April 2008

Problem

If $x_1,x_2$ are the roots of the equation $x^2-2kx+2m=0$, then $\frac{1}{x_1},\frac{1}{x_2}$ are the roots of the equation

$\mathrm{(A)}\ x^2-2k^2x+2m^2=0\qquad\mathrm{(B)}\ x^2-\frac{k}{m}x+\frac{1}{2m}=0\qquad\mathrm{(C)}\ x^2-\frac{m}{k}x+\frac{1}{2m}=0\\ \mathrm{(D)}\ 2mx^2-kx+1=0\qquad\mathrm{(E)}\ 2kx^2-2mx+1=0$

Solution

By Vieta’s, $x_1 + x_2 = 2k,\ x_1 \cdot x_2 = 2m$.

The equation with roots $x = \frac{1}{x_1}, \frac{1}{x_2}$ is $0 = \left(x - \frac{1}{x_1}\right)\left(x - \frac{1}{x_2}\right)$ $= x^2 - \left(\frac{1}{x_1} + \frac{1}{x_2}\right) + \frac{1}{x_1x_2} = x^2 - \left(\frac{x_1 + x_2}{x_1x_2}\right) + \frac{1}{x_1x_2}$. Substituting from above, we get $x^2 - \frac{k}{m}x + \frac{1}{2m}= 0 \Longrightarrow \mathrm{(B)}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_Cyprus_MO/Lyceum/Problem_16&oldid=25592"