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(D) Given the cubic equation $2x^3+3x^2-5x-7=0$. Let the roots be $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -\frac{3}{2}$ $

Vedclass · Accepted Answer

$\frac{67}{49}$

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(D) Given the cubic equation $2x^3+3x^2-5x-7=0$. Let the roots be $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -\frac{3}{2}$ $\alpha\beta+\beta\gamma+\gamma\alpha = -\frac{5}{2}$ $\alpha\beta\gamma = \frac{7}{2}$ We need to find $\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \frac{\beta^2\gamma^2+\alpha^2\gamma^2+\alpha^2\beta^2}{(\alpha\beta\gamma)^2}$. First,calculate $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma)$. Substituting the values: $= (-\frac{5}{2})^2 - 2(\frac{7}{2})(-\frac{3}{2}) = \frac{25}{4} + \frac{21}{2} = \frac{25+42}{4} = \frac{67}{4}$. Now,$(\alpha\beta\gamma)^2 = (\frac{7}{2})^2 = \frac{49}{4}$. Therefore,$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \frac{67/4}{49/4} = \frac{67}{49}$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $2x^3+3x^2-5x-7=0$,then $\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=$

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