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

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

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Given the cubic equation $x^3+2x^2-3x-1=0$,let $\alpha, \beta, \gamma$ be its roots.From 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 from $(i), (ii),$ and $(iii)$:$(-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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