• question_answer 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] A) $\frac{a-b}{a+b}$ B) $\frac{b-a}{a+b}$ C) $\frac{a+b}{a-b}$ D) $\frac{b+a}{b-a}$
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 $m=\frac{a-b}{a+b}$.