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

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(A) 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 this,w

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

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(A) 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 this,we have $(m + 1)x^2 - (m + 1)bx = (m - 1)ax - (m - 1)c$.Rearranging the terms into the standard quadratic form $Ax^2 + Bx + C = 0$:$(m + 1)x^2 - [b(m + 1) + a(m - 1)]x + c(m - 1) = 0$.Since the roots are equal in magnitude but opposite in sign,their sum must be zero.For a quadratic equation $Ax^2 + Bx + C = 0$,the sum of roots is $-\frac{B}{A}$.Thus,$-\frac{B}{A} = 0$,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 but opposite in sign,then the value of $m$ is

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