Solve the following

Question:

If x2 + y2 = 29 and xy = 2, find the value of

(i) x + y

(ii) x − y

(iii) x4 + y4

Solution:

(i) We have:

$(x+y)^{2}=x^{2}+2 x y+y^{2}$

$\Rightarrow(x+y)=\pm \sqrt{x^{2}+2 x y+y^{2}}$

$\Rightarrow(x+y)=\pm \sqrt{29+2 \times 2} \quad\left(\because x^{2}+y^{2}=29\right.$ 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 $\left.x y=2\right)$

$\Rightarrow x^{4}+y^{4}=841-8$

$\Rightarrow x^{4}+y^{4}=833$