Question

The value of $m$ for which the equation $\frac{a}{x+a+m}+\frac{b}{x+b+m}=1$ has roots equal in magnitude but opposite in sign is

• A. $\frac{a+b}{a-b}$
• B. $0$
• C. $\frac{a-b}{a+b}$
• D. $\frac{2}{(}a-b)a+b$.

Answer 1

The correct option is B $0$

Equation has $\frac{a}{x+a+m}+\frac{b}{x+b+m}=1$ roots equal in magnitude but opposite in sign

$a(x+b+m)+b(x+a+m)=(x+a+m)(x+b+m)$

$x^2+x(a+b+2m)+(a+m)(b+m)=x(a+b)+a(b+m)+b(a+m)$

${\alpha }_1-\alpha \Rightarrow$ Sum of roots zero.

co-efficient of $x=0$

$a+b+2m-(a+b)=0$

$m=0$