if (1/a^2 )+ (1/b^2 )= 1/4. then what could be the minimum value of ab ? .(a,b are integers and a>b)

• 1/a^2 + 1/b^2=1/4
  =>(a^2+b^2)=(a.b)^2/4;
  (a+b)^2=a^2+b^2+2.a.b
  =(a.b)^2/4 +2.a.b
  =ab(ab+8)/4
  
  so either ,
  ab=0 or ab=-8;
• 9 years agoHelpfull: Yes(23) No(5)
• pushpak ur ans is wrong bcoz if u r taking b=0 i.e 1/0^2 which is impossible
• 9 years agoHelpfull: Yes(5) No(0)
• ans is 0 assume a=2 b=0 then ans is 1/4 and a*b =0
• 9 years agoHelpfull: Yes(4) No(13)
• ans is -64
  a=8 and b=-8 because condition is a>b
  substitute the values and get the answer is -64
• 9 years agoHelpfull: Yes(3) No(3)
• 1/a^2+1/b^2=1/4
  =>(b^2+a^2)/(ab)^2=1/4
  =>(ab)^2= 4(b^2+a^2)
  =>ab= 2 sqrt(a^2+b^2)
  Let a =1 and b=0 because a>b
  ANS ab=2
  
• 9 years agoHelpfull: Yes(3) No(4)
• Sorry the above Given process is correct but a=2√2 and b=-2√2 so ab=-8 then it satisfies all above conditions because a>b
• 9 years agoHelpfull: Yes(3) No(0)
• Ans a.b= -8
  a=2 sqrt2 and b= -2 sqrt2
  
  1/(2 sqrt2)^2 + 1/(-2 sqrt2)^2= 1/8+1/8=2/8=1/4
• 9 years agoHelpfull: Yes(2) No(1)
• 1/8+1/8=1/4
  so may be ab=8
• 9 years agoHelpfull: Yes(2) No(5)
• AM=>GM ans 8
• 8 years agoHelpfull: Yes(2) No(0)
• (b^2+a^2)/a^2*b^2=1/4
  4(b^2+a^2)=a^2*b^2
  4b^2+4a^2=a^2*b^2
  √(4a^2+4b^2)=ab
  so a should be greater than b
  so consider minimum value a=-2√2,b=-2√2
  then ab=8 The a,b can satisfy all above conditions
  a and b cannot be 0 because 1/0 is undefined
• 9 years agoHelpfull: Yes(0) No(2)
• put the value of a as 4 and b as 3
  so we get the answer as ab=10(2*5)
  
• 8 years agoHelpfull: Yes(0) No(0)
• A to q, a and b are integers so ab must be integer...let a=8 and b=6 then, ab= sqrt root(a^2+b^2)/2,therefore ab is 5.
• 8 years agoHelpfull: Yes(0) No(0)
• Acc to q, a and b are integers so ab must be integer.
  let a=8 and b=6
  therefore ab=(sqrt(a^2+b^2))/2
  put the the value of a and b in above equation so we get
  the minimum value of ab is equal to 5.
• 8 years agoHelpfull: Yes(0) No(0)