If alpha and beta are the roots of 6x^2 - 6x | vedclass

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(B) Given $\alpha, \beta$ are the roots of $6x^2 - 6x + 1 = 0$.By Vieta's formulas,$\alpha + \beta = \frac{6}{6} = 1$ and $\alpha\beta = \frac{1}{6}$.

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$\frac{a}{1} + \frac{b}{2} + \frac{c}{3} + \frac{d}{4}$

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(B) Given $\alpha, \beta$ are the roots of $6x^2 - 6x + 1 = 0$.By Vieta's formulas,$\alpha + \beta = \frac{6}{6} = 1$ and $\alpha\beta = \frac{1}{6}$.We need to evaluate $S = \frac{1}{2}[a + b\alpha + c\alpha^2 + d\alpha^3] + \frac{1}{2}[a + b\beta + c\beta^2 + d\beta^3]$.$S = a + \frac{b}{2}(\alpha + \beta) + \frac{c}{2}(\alpha^2 + \beta^2) + \frac{d}{2}(\alpha^3 + \beta^3)$.Using $\alpha + \beta = 1$ and $\alpha\beta = \frac{1}{6}$:$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 1^2 - 2(\frac{1}{6}) = 1 - \frac{1}{3} = \frac{2}{3}$.$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 1^3 - 3(\frac{1}{6})(1) = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting these values into $S$:$S = a + \frac{b}{2}(1) + \frac{c}{2}(\frac{2}{3}) + \frac{d}{2}(\frac{1}{2}) = a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4}$.

If $\alpha$ and $\beta$ are the roots of $6x^2 - 6x + 1 = 0$,then the value of $\frac{1}{2}[a + b\alpha + c\alpha^2 + d\alpha^3] + \frac{1}{2}[a + b\beta + c\beta^2 + d\beta^3]$ is

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