The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

A. 20 B. 30
C. 40 D. None of these

• x^+y^2+z^2=138
  xy+yz+zx=131
  as we know.. (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
  so (x+y+z)^2=138+(2*131)
  (x+y+z)^2=400
  so x+y+z=20
• 11 years agoHelpfull: Yes(8) No(1)
• a^2+b^2+c^2=138
  ab+bc+ca=131
  Hence, (a+b+c)^2=138-131=7
  a+b+c=square root of 7
• 13 years agoHelpfull: Yes(4) No(13)
• answer is A. 20
  solution is very simple:-
  apply the identity of (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
  => a+b+c=sqrt(138+2(131))=20 is the answer
• 7 years agoHelpfull: Yes(1) No(0)