a+b=3;a^2+b^2=7;then a^4+b^4=?

• Ans: 47
  Given: a+b=3 a^2+b^2=7 , (a+b)^2=a^2+b^2+2ab therefore,9=7+2ab ab=1;
  Soln:
  a^4+b^4 = (a^2)^2 + (b^2)^2= (a^2+b^2)^2 -2((ab)^2)
  a^4+b^4= 7^2 -2(1^2)
  = 49-2=47
  
• 11 years agoHelpfull: Yes(19) No(0)
• answer is 47
• 11 years agoHelpfull: Yes(2) No(1)
• after solving we get ab=1,
  then simplifying the power of a^4+b^4
  and putting the values we get 47.
• 11 years agoHelpfull: Yes(2) No(1)
• a+b=3
  a^2+b^2=(a+b)^2-2(a*b)
  so on solving,ab=1
  similarly
  a^4+b^4=(a^2+b^2)^2-2(a^a*b^b)
  =7^2-2*(a*b)^2
  =49-2(1)
  =47
  on solving,it results 47
• 11 years agoHelpfull: Yes(0) No(1)
• Answer is 47
  (a+b)^2=a^2+b^2+2*a*b
  9=7+2*a*b
  a*b=1
  we know that, (a^2+b^2)^2 = a^4 + b^4 + 2*a^2*b^2
  so a^4 +b^4 = 47
  
• 11 years agoHelpfull: Yes(0) No(0)
• ANS: 47
  GIVEN:
  a+b=3 -eq1
  a^2+b^2=7 -eq2
  TO FIND:
  a^4+b^4=?
  SOL:
  eq2 can be written as:
  a^2+b^2+2ab-2ab=7
  9-2ab=7 from eq 1 a^2+b^2=9
  ab=1
  we know that:
  (a^2+b^2)^2=a^4+b^4+2(ab)^2
  (7)^2=a^4+b^4+2*1
  a^4+b^4=47
• 11 years agoHelpfull: Yes(0) No(0)