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

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(D) Given the cubic equation $5x^3 - 3x^2 + 2x - 4 = 0$. Comparing with $ax^3 + bx^2 + cx + d = 0$,we have $a=5, b=-3, c=2, d=-4$. By Vieta's formulas

Vedclass · Accepted Answer

$\frac{-2}{5}$

Vedclass · Answer

(D) Given the cubic equation $5x^3 - 3x^2 + 2x - 4 = 0$. Comparing with $ax^3 + bx^2 + cx + d = 0$,we have $a=5, b=-3, c=2, d=-4$. By Vieta's formulas: $\alpha + \beta + \gamma = -\frac{b}{a} = \frac{3}{5}$ $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{2}{5}$ $\alpha\beta\gamma = -\frac{d}{a} = \frac{4}{5}$ We need to find $\sum \alpha^2 \beta^2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2$. Using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$,let $x=\alpha\beta, y=\beta\gamma, z=\gamma\alpha$. Then $(\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 + 2(\alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta)$. $(\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \sum \alpha^2\beta^2 + 2\alpha\beta\gamma(\alpha + \beta + \gamma)$. Substituting the values: $(\frac{2}{5})^2 = \sum \alpha^2\beta^2 + 2(\frac{4}{5})(\frac{3}{5})$. $\frac{4}{25} = \sum \alpha^2\beta^2 + \frac{24}{25}$. $\sum \alpha^2\beta^2 = \frac{4}{25} - \frac{24}{25} = -\frac{20}{25} = -\frac{4}{5}$.

If $\alpha, \beta, \gamma$ are the roots of the equation $5x^3 - 3x^2 + 2x - 4 = 0$,then find the value of $\sum \alpha^2 \beta^2$.

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