2006 Cyprus MO/Lyceum/Problem 16

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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
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