If the roots of the quadratic equation frac{x | vedclass

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(A) Given the equation: $\frac{x - m}{mx + 1} = \frac{x + n}{nx + 1}$.Cross-multiplying,we get: $(x - m)(nx + 1) = (x + n)(mx + 1)$.Expanding both sid

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$n = 0$

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(A) Given the equation: $\frac{x - m}{mx + 1} = \frac{x + n}{nx + 1}$.Cross-multiplying,we get: $(x - m)(nx + 1) = (x + n)(mx + 1)$.Expanding both sides: $nx^2 + x - mnx - m = mx^2 + x + mnx + n$.Rearranging the terms: $(n - m)x^2 - 2mnx - (m + n) = 0$.Multiplying by $-1$: $(m - n)x^2 + 2mnx + (m + n) = 0$.Let the roots be $\alpha$ and $\frac{1}{\alpha}$.The product of the roots for a quadratic equation $ax^2 + bx + c = 0$ is given by $\frac{c}{a}$.Therefore,$\alpha \cdot \frac{1}{\alpha} = \frac{m + n}{m - n}$.$1 = \frac{m + n}{m - n}$.$m - n = m + n$.$-n = n$,which implies $2n = 0$,so $n = 0$.

If the roots of the quadratic equation $\frac{x - m}{mx + 1} = \frac{x + n}{nx + 1}$ are reciprocal to each other,then

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