$I f x^{2} + y^{2} = 29 a n d x y = 2, find the value of$

$(i) x + y$ $(i i) x - y$ $(i i i) x^{4} + y^{4}$

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Solution

$x^{2} + y^{2} = 29, x y = 2 (i) (x + y)^{2} = x^{2} + y^{2} + 2 x y = 29 + 2 \times 2 = 29 + 4 = 33 ∴ x + y = \sqrt[\pm]{33} (i i) (x - y)^{2} = x^{2} + y^{2} - 2 x y = 29 - 2 \times 2 = 29 - 4 = 25 ∴ x - y = \sqrt[\pm]{25} = \pm 5 (i i i) x^{2} + y^{2} = 29 Squaring on both sides, (x^{2} + y^{2})^{2} = (29)^{2} \Rightarrow (x^{2})^{2} + (y^{2})^{2} + 2 x^{2} y^{2} = 841 \Rightarrow x^{4} + y^{4} + 2 (x y)^{2} = 841 \Rightarrow x^{4} + y^{4} + 2 (2)^{2} = 841 (∵ x y = 2) \Rightarrow x^{4} + y^{4} + 2 \times 4 = 841 \Rightarrow x^{4} + y^{4} + 8 = 841 \Rightarrow x^{4} + y^{4} = 841 - 8 = 833 ∴ x^{4} + y^{4} = 833$

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