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

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(B) Given the cubic equation $x^3-2x^2+3x-4=0$ with roots $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha+\beta+\gamma = 2$ $\alpha\beta+\beta

Vedclass · Accepted Answer

$-6$

Vedclass · Answer

(B) Given the cubic equation $x^3-2x^2+3x-4=0$ with roots $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha+\beta+\gamma = 2$ $\alpha\beta+\beta\gamma+\gamma\alpha = 3$ $\alpha\beta\gamma = 4$ We need to evaluate $\sum \alpha\beta(\alpha+\beta) = \alpha\beta(\alpha+\beta) + \beta\gamma(\beta+\gamma) + \gamma\alpha(\gamma+\alpha)$. Since $\alpha+\beta+\gamma = 2$,we have $\alpha+\beta = 2-\gamma$,$\beta+\gamma = 2-\alpha$,and $\gamma+\alpha = 2-\beta$. Substituting these: $\sum \alpha\beta(\alpha+\beta) = \alpha\beta(2-\gamma) + \beta\gamma(2-\alpha) + \gamma\alpha(2-\beta)$ $= 2(\alpha\beta+\beta\gamma+\gamma\alpha) - 3(\alpha\beta\gamma)$ $= 2(3) - 3(4)$ $= 6 - 12 = -6$.

If $\alpha, \beta, \gamma$ are the roots of $x^3-2x^2+3x-4=0$,then find $\sum \alpha \beta(\alpha+\beta)$.

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