Question

pls solve...............​

Answer 1

Given :

$\mathsf{x=(2+\sqrt{5})^{1/2}+(2-\sqrt{5})^{1/2}}$

Squaring both sides, we get

$\mathsf{x^{2}=2+\sqrt{5}+2(2+\sqrt{5})^{1/2}(2-\sqrt{5})^{1/2}+2-\sqrt{5}}$

$\mathsf{=4+2\sqrt{4-5}=4+2\sqrt{-1}}$

$\&\:\mathsf{y=(2+\sqrt{5})^{1/2}-(2-\sqrt{5})^{1/2}}$

Squaring both sides, we get

$\mathsf{y^{2}=2+\sqrt{5}-2(2+\sqrt{5})^{1/2}(2-\sqrt{5})^{1/2}+2-\sqrt{5}}$

$\mathsf{=4-2\sqrt{4-5}=4-2\sqrt{-1}}$

$\therefore \mathsf{x^{2}+y^{2}=6+2\sqrt{-1}+2-2\sqrt{-1}}$

$\implies \boxed{\mathsf{x^{2}+y^{2}=8}}$