Question

x=1+(root 2) and y=1-root 2
Find value of x^2+y^2​

Answer 1

$\sf{x = 1 + \sqrt{2} } \\ \sf{ y = 1 - \sqrt{2} }$

$\sf {x}^{2} = {(1 + \sqrt{2} )}^{2}$

expand using $\sf{{(a+b)}^{2}}$ formula

$\sf {x}^{2} = { {1}^{2} + ( \sqrt{2} )}^{2} + 2(1)( \sqrt{2} ) \\ \sf {x}^{2} = { 1 + 2} + 2 \sqrt{2} \\ \sf {x}^{2} = 3 + 2\sqrt{2}$

expand using $\sf{{(a-b)}^{2}}$ formula

$\sf {y}^{2} = { ({1} - \sqrt{2}) }^{2} \\ \sf {y}^{2} = { {1}^{2} + ( \sqrt{2} )}^{2} - 2(1)( \sqrt{2} ) \\ \sf{ {y}^{2} = 1 + 2 - 2 \sqrt{2} } \\ \sf{ {y}^{2} = 3 - 2 \sqrt{2} }$

now,

$\sf {x}^{2} + {y}^{2} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2}$

$\fbox{ \sf {x}^{2} + {y}^{2} = 6}$