If alpha, beta, gamma are the roots of the eq | vedclass

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Question

(A) Given that $\alpha, \beta, \gamma$ are the roots of the equation $2x^3 - 3x^2 + 5x - 7 = 0$. From Vieta's formulas: $\alpha + \beta + \gamma = \fr

Vedclass · Accepted Answer

$-\frac{17}{4}$

Vedclass · Answer

(A) Given that $\alpha, \beta, \gamma$ are the roots of the equation $2x^3 - 3x^2 + 5x - 7 = 0$. From 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 $\sum \alpha^2 \beta^2 = (\alpha \beta)^2 + (\beta \gamma)^2 + (\gamma \alpha)^2$. Using the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$,let $a = \alpha \beta, b = \beta \gamma, c = \gamma \alpha$: $(\alpha \beta + \beta \gamma + \gamma \alpha)^2 = (\alpha \beta)^2 + (\beta \gamma)^2 + (\gamma \alpha)^2 + 2(\alpha \beta^2 \gamma + \beta \gamma^2 \alpha + \gamma \alpha^2 \beta)$ $(\alpha \beta + \beta \gamma + \gamma \alpha)^2 = \sum \alpha^2 \beta^2 + 2 \alpha \beta \gamma(\beta + \gamma + \alpha)$ Substituting the values: $(\frac{5}{2})^2 = \sum \alpha^2 \beta^2 + 2(\frac{7}{2})(\frac{3}{2})$ $\frac{25}{4} = \sum \alpha^2 \beta^2 + \frac{21}{2}$ $\sum \alpha^2 \beta^2 = \frac{25}{4} - \frac{42}{4} = -\frac{17}{4}$

If $\alpha, \beta, \gamma$ are the roots of the equation $2x^3 - 3x^2 + 5x - 7 = 0$,then $\sum \alpha^2 \beta^2 =$

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