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 2x^2+2qx+1=0, then (α - 1 α )(β - 1 β )(α + 1 β )(β…

$\alpha ,\beta$α,β are roots of ${x}^{2}+px+2=0$x2+px+2=0

$$\Rightarrow \alpha +\beta =-p,\alpha \beta =2$$⇒α+β=−p,αβ=2

$\frac{1}{\alpha },\frac{1}{\beta }$α1​,β1​ are roots of $2{x}^{2}+2qx+1=0$2x2+2qx+1=0

$\therefore \alpha ,\beta$∴α,β are roots of ${x}^{2}+2qx+2=0$x2+2qx+2=0

$$\Rightarrow \alpha +\beta =-2q,\alpha \beta =2$$⇒α+β=−2q,αβ=2

$$\therefore p=2q...(i)$$∴p=2q...(i)

$$\Rightarrow (\alpha -\frac{1}{\alpha })(\beta -\frac{1}{\beta })(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })={\alpha \beta -\frac{\alpha }{\beta }-\frac{\beta }{\alpha }+\frac{1}{\alpha \beta }}\cdot {\alpha \beta +\frac{1}{\alpha \beta }+2}$$⇒(α−α1​)(β−β1​)(α+β1​)(β+α1​)=αβ−βα​−αβ​+αβ1​⋅αβ+αβ1​+2

$$\Rightarrow {2-(\frac{{(\alpha +\beta )}^{2}-2\alpha \beta }{\alpha \beta })+\frac{1}{2}}\cdot {2+\frac{1}{2}+2}$$⇒2−(αβ(α+β)2−2αβ​)+21​⋅2+21​+2

$\Rightarrow {\frac{5}{2}-(\frac{{p}^{2}-4}{2})}\cdot {\frac{9}{2}}=\frac{9}{4}(9-{p}^{2})$⇒25​−(2p2−4​)⋅29​=49​(9−p2) or, $\frac{9}{4}(9-4{q}^{2})$49​(9−4q2)