If alpha, beta are the roots of x^2+ax+2=0 an | vedclass

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(A) Given $x^2+ax+2=0$,we have $\alpha+\beta = -a$ and $\alpha\beta = 2$. For $x^2-bx+c=0$,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$,so $

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$\frac{9}{4}(9-a^2)$

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(A) Given $x^2+ax+2=0$,we have $\alpha+\beta = -a$ and $\alpha\beta = 2$. For $x^2-bx+c=0$,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$,so $\frac{1}{\alpha}+\frac{1}{\beta} = b$ and $\frac{1}{\alpha\beta} = c = \frac{1}{2}$. Now,consider the expression: $E = \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)$ $E = \left(\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}\right) \left(\alpha\beta - 1 - 1 + \frac{1}{\alpha\beta}\right)$ $E = \left(\alpha\beta + \frac{1}{\alpha\beta} + 2\right) \left(\alpha\beta + \frac{1}{\alpha\beta} - 2\right)$ $E = \left(\alpha\beta + \frac{1}{\alpha\beta}\right)^2 - 4$ Substituting $\alpha\beta = 2$: $E = \left(2 + \frac{1}{2}\right)^2 - 4 = \left(\frac{5}{2}\right)^2 - 4 = \frac{25}{4} - 4 = \frac{9}{4}$. Wait,let us re-evaluate the expression expansion: $E = \left(\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}\right) \left(\alpha\beta - \frac{\alpha}{\beta} - \frac{\beta}{\alpha} + \frac{1}{\alpha\beta}\right)$ $E = \left(2 + 2 + \frac{1}{2}\right) \left(2 + \frac{1}{2} - \frac{\alpha^2+\beta^2}{\alpha\beta}\right) = \frac{9}{2} \left(\frac{5}{2} - \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\right)$ $E = \frac{9}{2} \left(\frac{5}{2} - \frac{a^2-4}{2}\right) = \frac{9}{2} \left(\frac{5-a^2+4}{2}\right) = \frac{9}{4}(9-a^2)$.

If $\alpha, \beta$ are the roots of $x^2+ax+2=0$ and $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots of $x^2-bx+c=0$,then $\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right) = $

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