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(A) Given the quadratic equation $x^2+ax+b=0$,the sum of roots is $\alpha+\beta = -a = \frac{1}{2}$,which implies $a = -\frac{1}{2}$.Using the identit

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$\frac{-1}{6}$

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(A) Given the quadratic equation $x^2+ax+b=0$,the sum of roots is $\alpha+\beta = -a = \frac{1}{2}$,which implies $a = -\frac{1}{2}$.Using the identity $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$,we substitute the known values:$\frac{37}{8} = (\frac{1}{2})^3 - 3b(\frac{1}{2})$$\frac{37}{8} = \frac{1}{8} - \frac{3b}{2}$$\frac{36}{8} = -\frac{3b}{2}$$\frac{9}{2} = -\frac{3b}{2} \Rightarrow b = -3$.Finally,calculating $a-\frac{1}{b}$:$a-\frac{1}{b} = -\frac{1}{2} - (\frac{1}{-3}) = -\frac{1}{2} + \frac{1}{3} = \frac{-3+2}{6} = -\frac{1}{6}$.

If $\alpha$ and $\beta$ are the real roots of the equation $x^2+ax+b=0$,where $\alpha+\beta=\frac{1}{2}$ and $\alpha^3+\beta^3=\frac{37}{8}$,then find the value of $a-\frac{1}{b}$.

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