Question

(x-y)^2 + (x+y)^2 = 0 ; Find the value of 'y'.

Answer 1

(x-y)² + (x+y)² = 0
x² + y² - 2xy + x² + y² + 2xy = 0 
2(x² + y²) = 0
x² + y² = 0
y² = -x² 

y =  x sqrt(-1)

y = i x

Answer 2

(x-y)² + (x+y)² = 0
x² + y² - 2xy + x² + y² + 2xy = 0 
2(x² + y²) = 0
x² + y² = 0
y² = -x² 

as squrt can not be negative therfor no real value of y is possible but it can be imaginery 
as negative roots only leds to imaginary parts 

ans y=squart root -x^2 
as -1=i^2 therfor y=ix