If alpha, beta, gamma are the roots of x^3+2x | vedclass

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(B) Given the cubic equation $x^3+2x^2-3x-1=0$. By Vieta's formulas,we have: $\alpha+\beta+\gamma = -2$ $(i)$ $\alpha\beta+\beta\gamma+\gamma\alpha =

Vedclass · Accepted Answer

$13$

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(B) Given the cubic equation $x^3+2x^2-3x-1=0$. By Vieta's formulas,we have: $\alpha+\beta+\gamma = -2$ $(i)$ $\alpha\beta+\beta\gamma+\gamma\alpha = -3$ $(ii)$ $\alpha\beta\gamma = 1$ $(iii)$ We need to find $\alpha^{-2}+\beta^{-2}+\gamma^{-2} = \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \frac{\beta^2\gamma^2+\alpha^2\gamma^2+\alpha^2\beta^2}{(\alpha\beta\gamma)^2}$. First,calculate $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$ using the identity $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2\alpha\beta\gamma(\alpha+\beta+\gamma)$. Substituting the values: $(-3)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(1)(-2)$ $9 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 - 4$ $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = 9+4 = 13$. Now,$\alpha^{-2}+\beta^{-2}+\gamma^{-2} = \frac{13}{(1)^2} = 13$.

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

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