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(A) Given the equation $18x^3-15x^2-4x+4=0$. Let the roots be $\alpha, \alpha, \gamma$ where $\alpha > \gamma$. From Vieta's formulas: $2\alpha + \gam

Vedclass · Accepted Answer

$\frac{71}{72}$

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(A) Given the equation $18x^3-15x^2-4x+4=0$. Let the roots be $\alpha, \alpha, \gamma$ where $\alpha > \gamma$. From Vieta's formulas: $2\alpha + \gamma = \frac{15}{18} = \frac{5}{6}$ $(i)$ $\alpha^2 + 2\alpha\gamma = \frac{-4}{18} = -\frac{2}{9}$ (ii) $\alpha^2\gamma = -\frac{4}{18} = -\frac{2}{9}$ (iii) From (ii) and (iii),$\alpha^2 + 2\alpha\gamma = \alpha^2\gamma$. Since $\alpha 
eq 0$,we have $\alpha + 2\gamma = \alpha\gamma$. From $(i)$,$\gamma = \frac{5}{6} - 2\alpha$. Substituting into $\alpha + 2\gamma = \alpha\gamma$: $\alpha + 2(\frac{5}{6} - 2\alpha) = \alpha(\frac{5}{6} - 2\alpha)$ $\alpha + \frac{5}{3} - 4\alpha = \frac{5\alpha}{6} - 2\alpha^2$ $2\alpha^2 - 3\alpha + \frac{5}{3} = \frac{5\alpha}{6}$ $12\alpha^2 - 18\alpha + 10 = 5\alpha$ $12\alpha^2 - 23\alpha + 10 = 0$ $(3\alpha - 2)(4\alpha - 5) = 0$. So,$\alpha = \frac{2}{3}$ or $\alpha = \frac{5}{4}$. If $\alpha = \frac{2}{3}$,then $\gamma = \frac{5}{6} - 2(\frac{2}{3}) = \frac{5}{6} - \frac{4}{3} = -\frac{1}{2}$. Since $\alpha > \gamma$ $(\frac{2}{3} > -\frac{1}{2})$,this is a valid solution. Then $\alpha + \beta^2 + \gamma^3 = \frac{2}{3} + (\frac{2}{3})^2 + (-\frac{1}{2})^3 = \frac{2}{3} + \frac{4}{9} - \frac{1}{8} = \frac{48 + 32 - 9}{72} = \frac{71}{72}$.

If $\alpha, \beta, \gamma$ are the real roots of the equation $18x^3-15x^2-4x+4=0$ such that $\alpha=\beta$ and $\alpha>\gamma$,then $\alpha+\beta^2+\gamma^3=$

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