Question

If the equation $\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}$ has roots equal in magnitude but opposite in sign, then $m=$

• A. $\frac{a+b}{a-b}$
• B. $\frac{a-b}{a+b}$
• C. $\frac{b-a}{b+a}$
• D. None of these

Answer 1

Answer: (B)

$\frac{a-b}{a+b}$

$\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}$
Taking componendo and dividendo
$\frac{x^2-bx+ax-c}{x^2-bx-ax+c}=\frac{m-1+m+1}{m-1-m-1}$
$\Rightarrow \frac{x^2-bx+ax-c}{x^2-bx-ax+c}=-m$
$\Rightarrow (1+m)x^2+(a-b-mb-ma)x-c+cm=0$
For roots to be equal in magnitude and opposite in sign, the sum of roots will be zero.
$\Rightarrow a-b-mb-ma=0\Rightarrow a-b=ma+mb$
$\Rightarrow m=\frac{(a-b)}{(a+b)}$.