Question

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

Answer 1

$\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}$

$\left(m+1\right)\left(x^2\right)-\left(m+1\right)bx=\left(m-1\right)ax-c\left(m-1\right)$

$\left(m+1\right)x^2-x\left[\left(m+1\right)b+\left(m-1\right)a\right]$

$\left(m+1\right)b+\left(m-1\right)a=0$

$mb+b+ma-a=0$

$m\left(a+b\right)=a-b$

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