If a+b = 3, a^2+b^2 = 7 then a^4+b^4?

• a+b=3 s.b.s,
  => (a+b)^2=9 => a^2+b^2+2ab=9
  => 7+2ab=9 (given,a^2+b^2 = 7) =>ab=1.
  now, a^2+b^2 = 7
  s.b.s,
  => a^4+b^4+2a^2*b^2=49
  => a^4+b^4=49-2*1 (a^2*b^2=(ab)^2 and ab=1)
  therefore, a^4+b^4=47
• 9 years agoHelpfull: Yes(10) No(0)
• my ans is 12
• 9 years agoHelpfull: Yes(0) No(2)
• A+b = 3
  so a^2+b^2 = 7 = (a+b)^2-2ab
  implies ab=1
  a^4+b^4 = (a^2+b^2)^2 - 2*a^2*b^2
  solving this the ans is 47
• 9 years agoHelpfull: Yes(0) No(0)
• a^2+b^2=(a+b)^2-2ab =9-2ab
  9-2ab =7,ab=1
  a^4+b^4=49-2=47
• 9 years agoHelpfull: Yes(0) No(0)