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Solve math equation without assumptions

  1. Mar 6, 2006 #1
    One mathematical sum is harassing me since a few days the problem is;:-
    a^2+b^2=25 and a^3+b^3=91. find the value of a and b.
    Now here we cannot take that into consoderation that as a^2+b^2=25 , a=3 and b=4. But we have to find the answer without any assumtions.May i please get the answer as soon as possible
     
  2. jcsd
  3. Mar 6, 2006 #2

    dextercioby

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    In which set of numbers do you search the solution to your system of equations...?

    Daniel.

    P.S. This is NOT "Advanced Physics".
     
  4. Mar 6, 2006 #3
    Forget the location. Let me help you.
    a^3 + b^3 = (a+b)(25-ab)
    Put a + b = x.
    Then ab = [(a+b)^2 - (a^2 + b^2)]/2 = (x^2 - 25)/2
     
  5. Mar 7, 2006 #4

    dextercioby

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    It doesn't really help. A solution is to reduce everything to a 6-th order algebraic equation (either in "a" or in "b") to which you know that 3 and 4 are solutions. You can reduce the degree to 4 then and to a 4-th order algebraic equation, the exact solutions do exist. All 4 of them are complex (with nonzero imaginary part)...

    That's why i asked about the set in which you want to determine the solutions...

    Daniel.
     
  6. Mar 7, 2006 #5

    Physics Monkey

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    Just guess! There are only a few numbers that a and b even have a chance of being. In fact, there is really only one pair of numbers that could work.
     
  7. Mar 7, 2006 #6

    VietDao29

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    Why do I hate people who double post so much!!! :grumpy: :grumpy: :grumpy:
    Why must you post one problem in two different threads in only one board??? Why???
    Maths factorization problem!!!
    In fact, it can be reduced to a cubic equation. :)
     
    Last edited: Mar 7, 2006
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