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(B) Given that $\alpha$ and $\beta$ are roots of $2x^2-4x+3=0$. Sum of roots: $\alpha+\beta = -(\frac{-4}{2}) = 2$. Product of roots: $\alpha\beta = \

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$-2$

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(B) Given that $\alpha$ and $\beta$ are roots of $2x^2-4x+3=0$. Sum of roots: $\alpha+\beta = -(\frac{-4}{2}) = 2$. Product of roots: $\alpha\beta = \frac{3}{2}$. Calculate $\alpha^2+\beta^2$: $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (2)^2 - 2(\frac{3}{2}) = 4-3 = 1$. Calculate $\alpha^4+\beta^4$: $(\alpha^2+\beta^2)^2 = \alpha^4+\beta^4+2\alpha^2\beta^2$. $1^2 = \alpha^4+\beta^4 + 2(\frac{3}{2})^2$. $1 = \alpha^4+\beta^4 + 2(\frac{9}{4}) = \alpha^4+\beta^4 + \frac{9}{2}$. $\alpha^4+\beta^4 = 1 - \frac{9}{2} = -\frac{7}{2}$. Now,substitute these values into the expression: $\frac{2(\alpha^4+\beta^4)+3(\alpha^2+\beta^2)}{\alpha+\beta} = \frac{2(-\frac{7}{2}) + 3(1)}{2} = \frac{-7+3}{2} = \frac{-4}{2} = -2$.

If $\alpha$ and $\beta$ are the roots of the equation $2x^2-4x+3=0$,then $\frac{2(\alpha^4+\beta^4)+3(\alpha^2+\beta^2)}{\alpha+\beta} = $

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