1. Tardigrade
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  4. If α and β are the roots of the equation x 2+ px +2=0 and (1/α) and (1/β) are the roots of the equation 2 x2+2 q x+1=0, then (α-(1/α))(β-(1/β))(α+(1/β))(β+(1/α)) is equal to:

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Q. If $\alpha$ and $\beta$ are the roots of the equation $x ^{2}+ px +2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the roots of the equation $2 x^{2}+2 q x+1=0,$ then
$\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$ is equal to:

JEE MainJEE Main 2020Complex Numbers and Quadratic Equations

Solution:

$\alpha, \beta$ are roots of $x^{2}+p x+2=0$
$\Rightarrow \alpha^{2}+ p \alpha+2=0 \& \beta^{2}+ p \beta+2=0$
$\Rightarrow \frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $2 x ^{2}+ px +1=0$
But $\frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $2 x ^{2}+2 qx +1=0$
$\Rightarrow \,\,\,\ p=2 q$
Also $\alpha+\beta=-p \,\,\,\, \alpha \beta=2$
$\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$
$=\left(\frac{\alpha^{2}-1}{\alpha}\right)\left(\frac{\beta^{2}-1}{\beta}\right)\left(\frac{\alpha \beta+1}{\beta}\right)\left(\frac{\alpha \beta+1}{\alpha}\right)$
$=\frac{(- p \alpha-3)(- p \beta-3)(\alpha \beta+1)^{2}}{(\alpha \beta)^{2}}$
$=\frac{9}{4}( p \alpha \beta+3 p (\alpha+\beta)+9)$
$=\frac{9}{4}\left(9- p ^{2}\right)=\frac{9}{4}\left(9-4 q ^{2}\right)$