If x^2 +y^2-4x-4y+8=0, then the value of x-y is

• @ SAUMITRA DAY :
  
  U have made a little mistake.
  
  In ur solution we can find
  (x-2)^2 + (y-2)^2=0
  
  now, if x,y are real quantities
  then only we can write now if the sum of 2 or more than 2 square terms of real quantities is zero then the terms are individually zero.
  
  So,(x-2)^2=0 and (y-2)^2=0
  x=2
  y=2,so x-y=0..
  
  So,(x-y)=0.
  
• 12 years agoHelpfull: Yes(10) No(1)
• x^2 +y^2-4x-4y+8=0
  x^2 -4x+4 +y^2-4y+4=0
  (x-2)^2 + (y-2)^2=0
  
  I think there is some mistake signs of some terms in qn.
• 12 years agoHelpfull: Yes(2) No(8)
• x^2 +y^2-4x-4y+8=0
  x^2 -4x+4 +y^2-4y+4=0
  (x-2)^2 + (y-2)^2=0
  now if the sum of 2 or more than 2 is zero then this are individually zero.
  now (x-2)^2=0 and (y-2)^2=0
  so x = +2 or -2 and y = +2 or -2
  so the value of x-y = 0 or +4 or -4
• 12 years agoHelpfull: Yes(0) No(8)