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(D) Given the cubic equation $x^3-5x^2-2x+24=0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha+\beta+\gamma = 5$ $\alpha\beta+\

Vedclass · Accepted Answer

$\frac{-61}{6}$

Vedclass · Answer

(D) Given the cubic equation $x^3-5x^2-2x+24=0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha+\beta+\gamma = 5$ $\alpha\beta+\beta\gamma+\gamma\alpha = -2$ $\alpha\beta\gamma = -24$ We need to evaluate the expression: $E = \frac{\beta\gamma}{\alpha}+\frac{\gamma\alpha}{\beta}+\frac{\alpha\beta}{\gamma}$ $E = \frac{(\beta\gamma)^2 + (\gamma\alpha)^2 + (\alpha\beta)^2}{\alpha\beta\gamma}$ Using the identity $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)$,where $a=\alpha\beta, b=\beta\gamma, c=\gamma\alpha$: $E = \frac{(\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2(\alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta)}{\alpha\beta\gamma}$ $E = \frac{(\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\beta+\gamma+\alpha)}{\alpha\beta\gamma}$ Substituting the values: $E = \frac{(-2)^2 - 2(-24)(5)}{-24}$ $E = \frac{4 + 240}{-24} = \frac{244}{-24} = -\frac{61}{6}$

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-5x^2-2x+24=0$,then $\frac{\beta\gamma}{\alpha}+\frac{\gamma\alpha}{\beta}+\frac{\alpha\beta}{\gamma}=$

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