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Equation problem 2

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem is a 2+b 2+c 2=2(a-b-c)-3
    Find a+b+c


    2. Relevant equations
    none


    3. The attempt at a solution
    well i tried to add 2(ab+ac+cb) to make the a 2+b 2+c 2=(a+b+c) 2
    so the other side also has a 2(ab+ac+cb)
    :. it becomes( if brackets are opened)
    2a-2b-2c+2ab+2ac+2cb-2-1
    =2a+2ab+2ac-2b-2c-2+2cb-1
    =2a(1+b+c)-2(1+b+c) +2cb-1
    = (1+b+c)(2a-2)+2cb-1
    Hey i even close to any proper solution
    Plz help me out!!
     
  2. jcsd
  3. Nov 27, 2009 #2

    Mark44

    Staff: Mentor

    a2 + b2 + c2 = 2a - 2b - 2c - 3
    <==> a2 - 2a + b2 + 2b + c2 + 2c = -3

    Now complete the square in the a terms, b terms and c terms. You should be able to solve for a, b, and c after that.
     
  4. Nov 27, 2009 #3
    I am sorry but I dont't get it.
    Could you please be more elobrate.It would be rather better if you could just work out only 1 more step.
     
  5. Nov 27, 2009 #4

    Mark44

    Staff: Mentor

    (a2 -2a + ?) + (b2 + 2b + ?) + (c2 + 2c + ?) = -3
    Complete the square in each parenthesized group. Whatever you add on the left side, make sure you add the same amount on the right side.
     
  6. Nov 28, 2009 #5
    I've tried it like this
    (a 2-2a+1)+(b2 + 2b + 1) +(c2 + 2c + 1) = -3 +1+1+1
    <==>(a-1)2+(b+1)2+(c+1)2=0
    Is that ok??or is it wrong??:uhh:
    So what do I do after this??:uhh:
     
  7. Nov 28, 2009 #6

    Mark44

    Staff: Mentor

    Yes, that's right, but you're not quite done. The square of any real number is >= 0. So for example, (a - 1)^2 >= 0, and if (a - 1)^2 > 0, what does that say about the other two terms on the left side of the equation?
     
  8. Nov 28, 2009 #7
    So does that mean
    (b+1)>0 and (c+1)>0(??)
    But what does that conclude?
     
  9. Nov 28, 2009 #8

    ideasrule

    User Avatar
    Homework Helper

    Look at the equation (a-1)2+(b+1)2+(c+1)2=0: it says the sum of three squares is 0. None of the squares can be negative. If even one of the squares is positive, what happens?
     
  10. Nov 28, 2009 #9
    Well i cant guess it any more(SORRY!!)
    plz work out the next step
     
  11. Nov 28, 2009 #10
    Do you mean to say :eek:
    (a-1)^2=0
    =>a-1=0
    =>a=1
    Similarly b=-1 c=-1
    and so a+b+c= -1
    But I still have a doubt why can't we consider (a-1)^2= 3or 45 or 1or any positive integer
    cuz you said
    "The square of any real number is >= 0":confused:
    And ideasrule also makes a point.
    I 'm a bit confused !!
    I would be happy if you help!
     
  12. Nov 28, 2009 #11

    Borek

    User Avatar

    Staff: Mentor

    If (a-1)^2 = 3 you would need one of the other expressions to be negative for the sum to be 0 - but squares can be only positive...
     
  13. Nov 28, 2009 #12
    Thanks!!
     
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