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Solve math equation without assumptions

  • #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
 

Answers and Replies

  • #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".
 
  • #3
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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
 
  • #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.
 
  • #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.
 
  • #6
VietDao29
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Milind_shyani said:
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
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!!!
dextercioby said:
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.
In fact, it can be reduced to a cubic equation. :)
 
Last edited:

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