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 the

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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 the terms: $(m + 1)x^2 - (m + 1)bx = (m - 1)ax - (m - 1)c$.Rearranging into the standard quadratic form $Ax^2 + Bx + C = 0$:$(m + 1)x^2 - [b(m + 1) + a(m - 1)]x + c(m - 1) = 0$.Expanding the coefficient of $x$: $b(m + 1) + a(m - 1) = bm + b + am - a = m(a + b) - (a - b)$.So,the equation is $(m + 1)x^2 - [m(a + b) - (a - b)]x + c(m - 1) = 0$.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 $-B/A$.Setting the sum of roots to zero: $m(a + b) - (a - b) = 0$.Therefore,$m(a + b) = a - b$,which gives $m = \frac{a - b}{a + b}$.

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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