|(x+y)2-4|+[(x-y)2-3]2=0
=>(x+y)^2-4 =0 and (x-y)^2-3 =0
=>(x+y)^2 = 4 (1) and
(x-y)^2 = 3 (2)
(1) -(2)
4xy =1
xy =1/4
(1)+(2)
2(x^2+y^2) =7
x^2+y^2 = 7/2