Let a, b and c be positive real numbers. If f | vedclass

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

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

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(D) Given the equation: $\frac{x^2-bx}{ax-c} = \frac{m-1}{m+1}$ Cross-multiplying,we get: $(x^2-bx)(m+1) = (m-1)(ax-c)$ Expanding the terms: $x^2(m+1) - bx(m+1) = ax(m-1) - c(m-1)$ Rearranging into the standard quadratic form $Ax^2 + Bx + C = 0$: $x^2(m+1) - x(b(m+1) + a(m-1)) + c(m-1) = 0$ Let the roots be $p$ and $-p$. For a quadratic equation $Ax^2 + Bx + C = 0$,the sum of roots is given by $-\frac{B}{A}$. Since the sum of roots $p + (-p) = 0$,the coefficient of $x$ must be zero: $b(m+1) + a(m-1) = 0$ $bm + b + am - a = 0$ $m(a+b) = a-b$ Therefore,$m = \frac{a-b}{a+b}$

Let $a, b$ and $c$ be positive real numbers. If $\frac{x^2-bx}{ax-c} = \frac{m-1}{m+1}$ has two roots which are numerically equal but opposite in sign,then the value of $m$ is

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