If α and β are the roots of the equation x2+px+2=0 and 1α and 1β are the roots of the equation 2x2+2qx+1=0, then (α−1α)(β−1β)(α+1β)(β+1α) is equal to

If α and β are the roots of the equation x2+px+2=0 and 1α and 1β a

1.	If α and β are the roots of the equation x2+px+2=0 and 1α and 1β are the roots of the equation 2x2+2qx+1=0, then (α−1α)(β−1β)(α+1β)(β+1α) is equal to
Answer» If $α$ and $β$ are the roots of the equation $x^{2} + p x + 2 = 0$ and $\frac{1}{α}$ and $\frac{1}{β}$ are the roots of the equation $2 x^{2} + 2 q x + 1 = 0$ , then $(α - \frac{1}{α}) (β - \frac{1}{β}) (α + \frac{1}{β}) (β + \frac{1}{α})$ is equal to

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If α and β are the roots of the equation x2+px+2=0 and 1α and 1β are the roots of the equation 2x2+2qx+1=0, then (α−1α)(β−1β)(α+1β)(β+1α) is equal to

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