if a+b=3 and a^2+b^2=7 then a^4+b^4= ?

• (a+b)^2=a^2+b^2+2ab
  3^2=7+2ab
  then ab=1..
  
  a^4+b^4=(a^2)^2 + (b^2)^2
  this in the form of " a^2+b^2 = (a+b)^2-2ab"
  so
  (a^2)^2 + (b^2)^2= ((a^2+b^2)^2)-2*a^2*b^2
  =7^2-2*1
  =49-2
  a^4+b^4 =47
• 9 years agoHelpfull: Yes(33) No(0)
• a+b=3
  a^2+b^2+2ab=9
  as given a^2+b^2=7
  then 2ab=2
  ab=1
  a=1/b,b=1/a
  then from a.a+b.b=7
  by substitution we got two equations
  1+b^4=7b^2
  1+a^4=7a^2
  then
  2+a^4+b^4=7(a^2+b^2)
  2+a^4+b^4=7(7)
  then
  a^4+b^4=49-2
  finally a^4+b^4=47
• 9 years agoHelpfull: Yes(4) No(0)
• (a+b)^2 = a^2 + b^2 + 2ab
  => 9 = 7+2ab
  =>ab = 1
  
  (a^2+b^2)^2 = 7^2= 49 = a^4 +b^4 +2a^2b^2
  a^4+b^4 +2 = 49
  a^4+b^4 =47
  47 is the answer
• 9 years agoHelpfull: Yes(4) No(0)
• a + b=3
  a^2 + b^2=7
  (a+b)^2=a2 + 2ab + b2
  ab=1
  (a + b)^4=(a^4 + 2 a^2 b^2+ b^4)
  
  a^4 + b^4=47
• 9 years agoHelpfull: Yes(2) No(0)
• a^2+b^2=7
  =>(a+b)^2-2ab=7
  =>ab=1
  
  a^4+b^4=(a^2+b^2)^2-2(ab)^2
  =7^2-2*1^2
  =47
• 9 years agoHelpfull: Yes(1) No(0)
• a+b=3
  (a+b)^2=a^2+b^2+2ab
  9=7+2ab
  ab=1
  (a^4+b^4)=(a^2+b^2)^2-2a^2b^2
  =49-2
  =47
• 9 years agoHelpfull: Yes(1) No(0)
• 47
  a^2+b^2=(a+b)^2-2ab
  =>2ab=3^2-7
  =>ab=1
  a^4+b^4=(a^2+b^2)-2(ab)^2
  =49-2=47
  
• 9 years agoHelpfull: Yes(0) No(0)
• (a+b)^2=9
  a^2+b^2+2*a*b=9
  7+2*a*b=9
  a*b=1
  a^4+b^4=(a^2+b^2)^2-2*a^2*b^2
  =49-2
  =47
• 9 years agoHelpfull: Yes(0) No(0)
• first a^4+b^4=(a^2+b^2)^2-2a^2*b^2; ....1
  then put the value of a^2+b^2=7 in equ. 1
  ab=1;
  ans is 49-2*1=47
• 8 years agoHelpfull: Yes(0) No(0)
• a+b=3 ans a^2+b^2=7
  solving above equations
  ab=1
  (a^2+b^2)^2=a^4+b^4+2(ab)^2
  7^2=a^4+b^4+2(1^2)
  a^4+b^4=47
• 8 years agoHelpfull: Yes(0) No(0)