Question

For what values of $m$ will the equation $\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}$ have roots equal in magnitude but opposite sign?

Answer 1

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

It can also be written as $(m+1)x^2-\left(b\right(m+1)+a(m-1\left)\right)x+c(m-1)=0$

Given that the equation have roots equal in magnitude but opposite in sign

Which means the sum of roots must be equal to zero

$\Rightarrow \frac{b(m+1)+a(m-1)}{m+1}=0$

$\Rightarrow (a+b)m+b-a=0$

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