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Algebraic Identities

Question

Find $4 x^{2} y^{2} + \frac{1}{4 x^{2} y^{2}},$ if $(2 x y + \frac{1}{2 x y}) = 1.6.$

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Solution

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${(2 x y + \frac{1}{2 x y})}^{2} = (2 x y)^{2} + \frac{1}{(2 x y)^{2}} + 2 \frac{1}{2 x y} \cdot 2 x y$
$\Rightarrow (1.6)^{2} = 4 x^{2} y^{2} + \frac{1}{4 x^{2} y^{2}} + 2$
$\Rightarrow 2.56 - 2 = 4 x^{2} y^{2} + \frac{1}{4 x^{2} y^{2}}$
So $4 x^{2} y^{2} + \frac{1}{4 x^{2} y^{2}} = 0.56$ .

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