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

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(B) Given that $\alpha$ and $\beta$ are the roots of $6x^2 - 6x + 1 = 0$.From the properties of roots,$\alpha + \beta = -(-6)/6 = 1$ and $\alpha\beta

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

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(B) Given that $\alpha$ and $\beta$ are the roots of $6x^2 - 6x + 1 = 0$.From the properties of roots,$\alpha + \beta = -(-6)/6 = 1$ and $\alpha\beta = 1/6$.The given expression is $\frac{1}{2}[a + b\alpha + c\alpha^2 + d\alpha^3] + \frac{1}{2}[a + b\beta + c\beta^2 + d\beta^3]$.This simplifies to $a + \frac{b}{2}(\alpha + \beta) + \frac{c}{2}(\alpha^2 + \beta^2) + \frac{d}{2}(\alpha^3 + \beta^3)$.Substituting $\alpha + \beta = 1$ and $\alpha\beta = 1/6$:$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 1^2 - 2(1/6) = 1 - 1/3 = 2/3$.$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 1^3 - 3(1/6)(1) = 1 - 1/2 = 1/2$.Substituting these values back into the expression:$= a + \frac{b}{2}(1) + \frac{c}{2}(2/3) + \frac{d}{2}(1/2) = a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4}$.

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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