Complex Numbers And Quadratic Equations question 599
Question: If the roots of the quadratic equation $ \frac{x-m}{mx+1}=\frac{x+n}{nx+1} $ are reciprocal to each other, then [MP PET 2001]
Options:
A) $ n=0 $
B) $ m=n $
C) $ m+n=1 $
D) $ m^{2}+n^{2}=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
Given, $ \frac{x-m}{mx+1}=\frac{x+n}{nx-1} $
Þ $ x^{2}(m-n)+2mnx+(m+n)=0 $ Roots are $ \alpha ,\frac{1}{\alpha } $ respectively, then $ \alpha .\frac{1}{\alpha }=\frac{m+n}{m-n} $
Þ $ m-n=m+n $
Þ $ n=0 $ .
BETA
