if a^2+b^2+2b+4a+5=0 then what is the value of a-b/a+b.

• a^2+b^2+2b+4a+5=0
  => (a^2+4a+4)+(b^2+2b+1)= 0
  => (a+2)^2 + (b+1)^2 = 0
  => a+2 = 0 & b+1 = 0
  => a = -2 & b = -1
  
  a-b/a+b = -2+1 / -2-1 = 1/3
• 10 years agoHelpfull: Yes(71) No(1)
• a^2+b^2+2b+4a+5=0
  => (a^2+4a+4)+(b^2+2b+1)= 0
  => (a+2)^2+(b+1)^2 = 0
  
  when sum of two or more squares is 0 then individual term must be 0
  so a= -2 & b=-1
• 10 years agoHelpfull: Yes(6) No(0)
• a=-2,b=-1.
• 10 years agoHelpfull: Yes(1) No(1)
• 1/3.
  => a^2 + b^2 +2b + 4a + 5 = 0 is to be rewritten in the following way :
  => a^2 + 4a + b^2 + 2b + (4+1) = 0
  => (a^2 + 2.2.a + 4) + b^2 + 2.1.b + 1 = 0
  => (a^2 + 2.2.a + 2^2) + (b^2 + 2.1.b + 1^2) = 0
  => (a+2)^2 + (b+1)^2 = 0 as a^2 + b^2 + 2.a.b = (a+b)^2
  By this we get :
  a = -2 and b = -1
  Putting values of a and b in (a-b)/(a+b) we get 1/3 answer.
• 10 years agoHelpfull: Yes(1) No(1)
• (a^2+4a+4)+(b^2+2b+1)=0
  =>(a+2)^2+(b+1)^2=0
  =>a+2=0&b+1=0
  =>a=-2&b=-1
  =>a-b/a+b=-2+1/-2-1
  =>-1/-3
  =>1/3
• 10 years agoHelpfull: Yes(1) No(0)
• @RAKESH thanks bhai
• 10 years agoHelpfull: Yes(0) No(2)
• a-b/a+b = 1/3
• 10 years agoHelpfull: Yes(0) No(2)
• a^2+4*a+4+b^2+2*b+1=0
  a^2+2*(a)*2+2^2+b^2+2*b+1^2=0
  (a+2)^2+(b+1)^2=0
  
  it means
  (a+2)=0 and (b+1)=0, b=(-1)
  a=(-2)
  
  now a-b=(-2-(-1))=(-2+1)=-1
  and a+b=(-2+(-1))=(-2-1)=-3
  
  now a-b/a+b=(-1)/(-3)=1/3
  
  so 1/3 is the ans
• 9 years agoHelpfull: Yes(0) No(0)