If x² + y² = 29 and xy = 2, find the value of
(i) x + y
(ii) x − y
(iii) x⁴ + y⁴

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Solution

(i) We have:

${(x + y)}^{2} = x^{2} + 2 x y + y^{2} \Rightarrow (x + y) = \pm \sqrt{x^{2} + 2 x y + y^{2}} \Rightarrow (x + y) = \pm \sqrt{29 + 2 \times 2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow (x + y) = \pm \sqrt{29 + 4} \Rightarrow (x + y) = \pm \sqrt{33}$

(ii) We have:

${(x - y)}^{2} = x^{2} - 2 x y + y^{2} \Rightarrow (x - y) = \pm \sqrt{x^{2} - 2 x y + y^{2}} \Rightarrow (x + y) = \pm \sqrt{29 - 2 \times 2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow (x + y) = \pm \sqrt{29 - 4} \Rightarrow (x + y) = \pm \sqrt{25} \Rightarrow (x + y) = \pm 5$

(iii) We have:

${(x^{2} + y^{2})}^{2} = x^{4} + 2 x^{2} y^{2} + y^{4} \Rightarrow x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 x^{2} y^{2} \Rightarrow x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 {(x y)}^{2} \Rightarrow x^{4} + y^{4} = 29^{2} - 2 {(2)}^{2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow x^{4} + y^{4} = 841 - 8 \Rightarrow x^{4} + y^{4} = 833$

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