Complex Numbers And Quadratic Equations question 174
Question: If the roots of equation $ \frac{x^{2}-bx}{ax-c}=\frac{m-1}{m+1} $ are equal but opposite in sign, then the value of $ m $ will be [RPET 1988, 2001; MP PET 1996, 2002; Pb. CET 2000]
Options:
A) $ \frac{a-b}{a+b} $
B) $ \frac{b-a}{a+b} $
C) $ \frac{a+b}{a-b} $
D) $ \frac{b+a}{b-a} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation can be written as $ (m+1)x^{2}-{m(a+b)-(a-b)}x+c(m-1)=0 $ . Roots are equal and of opposite sign. So sum of roots is equal to zero. Þ $ 0=m(a+b)-(a-b) $ Þ $ m=\frac{a-b}{a+b} $ .
BETA
