If the roots of the equation frac{x^2 - bx}{a | vedclass

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(B) Given the equation: $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$.Cross-multiplying,we get: $(m + 1)(x^2 - bx) = (m - 1)(ax - c)$.Expanding both

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

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(B) Given the equation: $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$.Cross-multiplying,we get: $(m + 1)(x^2 - bx) = (m - 1)(ax - c)$.Expanding both sides: $(m + 1)x^2 - b(m + 1)x = a(m - 1)x - c(m - 1)$.Rearranging into standard quadratic form $Ax^2 + Bx + C = 0$:$(m + 1)x^2 - [b(m + 1) + a(m - 1)]x + c(m - 1) = 0$.Let the roots be $\alpha$ and $-\alpha$. For roots to be equal in magnitude but opposite in sign,the sum of roots must be zero.Sum of roots $= -\frac{B}{A} = 0$,which implies $B = 0$.Therefore,$b(m + 1) + a(m - 1) = 0$.$bm + b + am - a = 0$.$m(a + b) = a - b$.$m = \frac{a - b}{a + b}$.

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

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