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

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Question

(C) Given the cubic equation $x^3+0x^2+2x+5=0$. By Vieta's formulas: $\alpha+\beta+\gamma = 0$ $\alpha\beta+\beta\gamma+\gamma\alpha = 2$ $\alpha\beta

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(C) Given the cubic equation $x^3+0x^2+2x+5=0$. By Vieta's formulas: $\alpha+\beta+\gamma = 0$ $\alpha\beta+\beta\gamma+\gamma\alpha = 2$ $\alpha\beta\gamma = -5$ We need to evaluate $\sum \frac{\beta+\gamma}{\alpha^2}$. Since $\alpha+\beta+\gamma = 0$,we have $\beta+\gamma = -\alpha$. Substituting this into the expression: $\sum \frac{\beta+\gamma}{\alpha^2} = \sum \frac{-\alpha}{\alpha^2} = \sum -\frac{1}{\alpha} = -\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)$ $= -\left(\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}\right)$ $= -\left(\frac{2}{-5}\right) = \frac{2}{5}$

If $\alpha, \beta, \gamma$ are the roots of $x^3+2x+5=0$,then $\sum \frac{\beta+\gamma}{\alpha^2} = $

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