Question

x=√3-√2÷√3+√2 and y=√3+√2÷√3-√2 find value of (x+y)2

Answer 1

Hope it is helpful...

Answer 2

x= (root3 +root2)/(root3-root2)

x=[(root3+root2)(root3+root2)]/[(root3-root2)(root3+root2)]
x=(root3+root2)^2/[(root3)^2-(root2)^2]
x=(root3+root2)^2
x=(5+2root6) here we (a+b)^2=a^2+b^2+2ab identity
Similarly
y=5-2root6
Therefore
x^2=(5+2root6)^2
=25+24+4root6
=49+4root6
y^2=[5^2-(2root6)^2]
=25+24-4root6
=49-4root6