Question

$let \: \alpha \: \beta are \: roots \: of \: {x}^{2} + px + 2 = 0 \: and \: \frac{1}{ \alpha } \frac{1}{ \beta } \: are \: the \: roots \: of \: {2x}^{2} - 2qx + 1 = 0. \: then \: find \: the \: value \: of \: ( \alpha + \frac{1}{ \beta } )( \beta + \frac{1}{ \alpha }) ( \alpha - \frac{1}{ \alpha } )( \beta - \frac{1}{ \beta } )$
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Answer 1

Let α and β are roots of x² + px + 2 = 0 and 1/α , 1/β are the roots od 2x² - 2qx + 1 = 0

We have to find the value of (α + 1/β)(β + 1/α)(α - 1/α)(β - 1/β)

solution : α and β are roots of x² + px + 2 = 0

so, α + β = -p , αβ = 2

and 1/α and 1/β are roots of 2x² - 2qx + 1 =0

so, 1/α + 1/β = 2q/2 = q , 1/αβ = 1/2

now (α + 1/β)(β + 1/α)(α - 1/α)(β - 1/β)

= (αβ + 1 + 1 + 1/αβ)(αβ - α/β - β/α + 1/αβ)

= (2 + 2 + 1/2))(2 - α/β - β/α + 1/2)

= 9/2[5/2 - (α² + β²)/αβ ]

= 9/2 [5/2 - {(α + β)² - 2αβ}/αβ ]

= 9/2 [5/2 - {(-p)² - 2(2)}/2]

= 9/2 [5/2 - (p² - 4)/2 ]

= 9/4 (9 - p²)

Therefore the value of (α + 1/β)(β + 1/α)(α - 1/α)(β - 1/β) is 9/4(9 - p²)