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

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(A) Given equation: $\frac{x - m}{mx + 1} = \frac{x + n}{nx + 1}$Cross-multiplying,we get: $(x - m)(nx + 1) = (x + n)(mx + 1)$$nx^2 + x - mnx - m = mx

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$n = 0$

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(A) Given equation: $\frac{x - m}{mx + 1} = \frac{x + n}{nx + 1}$Cross-multiplying,we get: $(x - m)(nx + 1) = (x + n)(mx + 1)$$nx^2 + x - mnx - m = mx^2 + x + mnx + 1$Rearranging the terms: $(n - m)x^2 - 2mnx - (m + 1) = 0$Since the roots are reciprocal,their product is $1$. For a quadratic equation $ax^2 + bx + c = 0$,the product of roots is $\frac{c}{a}$.Thus,$\frac{-(m + 1)}{n - m} = 1$$-(m + 1) = n - m$$-m - 1 = n - m$$n = -1$. Wait,let us re-evaluate the cross multiplication.$(x - m)(nx + 1) = nx^2 + x - mnx - m$$(x + n)(mx + 1) = mx^2 + x + mnx + n$$nx^2 - mnx - m = mx^2 + mnx + n$$(n - m)x^2 - 2mnx - (m + n) = 0$Product of roots = $\frac{-(m + n)}{n - m} = 1$$-(m + n) = n - m$$-m - n = n - m$$-2n = 0 \implies 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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