If alpha and beta are the roots of the equati | vedclass

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(C) Given $\alpha, \beta$ are roots of $x^{2}+px+2=0$. Thus,$\alpha+\beta = -p$ and $\alpha\beta = 2$.Since $\frac{1}{\alpha}, \frac{1}{\beta}$ are ro

Vedclass · Accepted Answer

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

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(C) Given $\alpha, \beta$ are roots of $x^{2}+px+2=0$. Thus,$\alpha+\beta = -p$ and $\alpha\beta = 2$.Since $\frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $2x^{2}+2qx+1=0$,they are also roots of $x^{2}+qx+\frac{1}{2}=0$.Since $\frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $x^{2}+px+2=0$ transformed by $x 	o 1/x$,we have $2(1/x)^{2} + p(1/x) + 1 = 0$,which simplifies to $2+px+x^{2}=0$,or $x^{2}+px+2=0$. Comparing $2x^{2}+2qx+1=0$ with $x^{2}+\frac{p}{2}x+\frac{1}{2}=0$,we get $q = p/2$,so $p = 2q$.Now,consider the expression $E = \left(\frac{\alpha^{2}-1}{\alpha}ight)\left(\frac{\beta^{2}-1}{\beta}ight)\left(\frac{\alpha\beta+1}{\beta}ight)\left(\frac{\alpha\beta+1}{\alpha}ight)$.Since $\alpha^{2}+p\alpha+2=0$,$\alpha^{2}-1 = -p\alpha-3$. Similarly,$\beta^{2}-1 = -p\beta-3$.$E = \frac{(-p\alpha-3)(-p\beta-3)(\alpha\beta+1)^{2}}{(\alpha\beta)^{2}} = \frac{(p\alpha+3)(p\beta+3)(2+1)^{2}}{2^{2}} = \frac{9}{4}(p^{2}\alpha\beta + 3p(\alpha+\beta) + 9)$.Substituting $\alpha\beta=2$ and $\alpha+\beta=-p$:$E = \frac{9}{4}(2p^{2} - 3p^{2} + 9) = \frac{9}{4}(9-p^{2})$.

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