$I f x + y = 4 a n d x y = 2, find the value of x^{2} + y^{2} .$

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Solution

$x + y = 4 Squaring on both sides, (x + y)^{2} = (4)^{2} \Rightarrow x^{2} + y^{2} + 2 x y = 16 \Rightarrow x^{2} + y^{2} + 2 \times 2 = 16 (∵ x y = 2) \Rightarrow x^{2} + y^{2} + 4 = 16 \Rightarrow x^{2} + y^{2} = 16 - 4 = 12 ∴ x^{2} + y^{2} = 12$

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