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Question

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

Vedclass · Accepted Answer

$\frac{27}{16}$

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(A) Given the quadratic equation $7x^{2}-3x-2=0$.From the properties of roots,the sum of roots $\alpha+\beta = -(-3)/7 = 3/7$ and the product of roots $\alpha\beta = -2/7$.We need to evaluate the expression $E = \frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$.$E = \frac{\alpha(1-\beta^{2}) + \beta(1-\alpha^{2})}{(1-\alpha^{2})(1-\beta^{2})} = \frac{\alpha - \alpha\beta^{2} + \beta - \alpha^{2}\beta}{1 - \beta^{2} - \alpha^{2} + \alpha^{2}\beta^{2}} = \frac{(\alpha+\beta) - \alpha\beta(\alpha+\beta)}{1 - (\alpha^{2}+\beta^{2}) + (\alpha\beta)^{2}}$.Using $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta = (3/7)^{2} - 2(-2/7) = 9/49 + 4/7 = 9/49 + 28/49 = 37/49$.Substituting the values:Numerator $= (3/7) - (-2/7)(3/7) = 3/7 + 6/49 = 21/49 + 6/49 = 27/49$.Denominator $= 1 - 37/49 + (-2/7)^{2} = 1 - 37/49 + 4/49 = 49/49 - 33/49 = 16/49$.Therefore,$E = \frac{27/49}{16/49} = \frac{27}{16}$.

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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