First slide

Theory of equations

Question

$If the roots of the quadratic equation \frac{x - m}{m x + 1} = \frac{x + n}{n x + 1} are reciprocal to each other, then$

Moderate

A

$m = n$
B

$m + n = 1$
C

$m^{2} + n^{2} = 1$
D

$n = 0$

Solution

$\begin{array}{l} Given \frac{x - m}{m x + 1} = \frac{x + n}{n x + 1} \\ (x - m) (n x + 1) = (x + n) (m x + 1) \\ \Rightarrow x^{2} (m - n) + 2 m n x + (m + n) = 0 \\ Roots are α, \frac{1}{α} respectively, then α \cdot \frac{1}{α} = \frac{m + n}{m - n} \\ \Rightarrow m - n = m + n \Rightarrow n = 0 . \end{array}$

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