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(A) Given the quadratic equation $7x^{2}-3x-2=0$.From the properties of roots,we have $\alpha+\beta = \frac{3}{7}$ and $\alpha\beta = \frac{-2}{7}$.We

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$\frac{27}{16}$

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(A) Given the quadratic equation $7x^{2}-3x-2=0$.From the properties of roots,we have $\alpha+\beta = \frac{3}{7}$ and $\alpha\beta = \frac{-2}{7}$.We need to evaluate $S = \frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$.$S = \frac{\alpha(1-\beta^{2})+\beta(1-\alpha^{2})}{(1-\alpha^{2})(1-\beta^{2})} = \frac{\alpha-\alpha\beta^{2}+\beta-\alpha^{2}\beta}{1-(\alpha^{2}+\beta^{2})+\alpha^{2}\beta^{2}}$.$S = \frac{(\alpha+\beta)-\alpha\beta(\alpha+\beta)}{1-((\alpha+\beta)^{2}-2\alpha\beta)+(\alpha\beta)^{2}}$.Substituting the values: $\alpha+\beta = \frac{3}{7}$ and $\alpha\beta = \frac{-2}{7}$.Numerator: $\frac{3}{7} - (\frac{-2}{7})(\frac{3}{7}) = \frac{3}{7} + \frac{6}{49} = \frac{21+6}{49} = \frac{27}{49}$.Denominator: $1 - ((\frac{3}{7})^{2} - 2(\frac{-2}{7})) + (\frac{-2}{7})^{2} = 1 - (\frac{9}{49} + \frac{4}{7}) + \frac{4}{49} = 1 - (\frac{9+28}{49}) + \frac{4}{49} = 1 - \frac{37}{49} + \frac{4}{49} = \frac{49-37+4}{49} = \frac{16}{49}$.Thus,$S = \frac{27/49}{16/49} = \frac{27}{16}$.

If $\alpha$ and $\beta$ are the roots of the equation $7x^{2}-3x-2=0$,then the value of $\frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$ is equal to

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