If x2 + y2 = 29 and xy = 2, find the value of
(i) x + y
(ii) x − y
(iii) x4 + y4
(i) We have:
$(x+y)^{2}=x^{2}+2 x y+y^{2}$
$\Rightarrow(x+y)=\pm \sqrt{x^{2}+2 x y+y^{2}}$
$\Rightarrow(x+y)=\pm \sqrt{29+2 \times 2} \quad\left(\because x^{2}+y^{2}=29\right.$ and $\left.x y=2\right)$
$\Rightarrow(x+y)=\pm \sqrt{29+4}$
$\Rightarrow(x+y)=\pm \sqrt{29+4}$
$\Rightarrow(x+y)=\pm \sqrt{33}$
(ii) We have:
$(x-y)^{2}=x^{2}-2 x y+y^{2}$
$\Rightarrow(x-y)=\pm \sqrt{x^{2}-2 x y+y^{2}}$
$\Rightarrow(x+y)=\pm \sqrt{29-2 \times 2} \quad\left(\because x^{2}+y^{2}=29\right.$ and $\left.x y=2\right)$
$\Rightarrow(x+y)=\pm \sqrt{29-4}$
$\Rightarrow(x+y)=\pm \sqrt{25}$
$\Rightarrow(x+y)=\pm 5$
(iii) We have:
$\left(x^{2}+y^{2}\right)^{2}=x^{4}+2 x^{2} y^{2}+y^{4}$
$\Rightarrow x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}$
$\Rightarrow x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2(x y)^{2}$
$\Rightarrow x^{4}+y^{4}=29^{2}-2(2)^{2} \quad\left(\because x^{2}+y^{2}=29\right.$ and $\left.x y=2\right)$
$\Rightarrow x^{4}+y^{4}=841-8$
$\Rightarrow x^{4}+y^{4}=833$
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