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Math Help - Need some help on the workings.

  1. #1
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    Need some help on the workings.

    Given  x+y=2 and  2xy-z^2=1 solve for the unknowns.

    The answer for all of them is all 1. I can get it by substituting 1 into them (with some guessing), but is there any working for getting the answer without guessing? Some help here please?
    Last edited by saw235; January 16th 2010 at 04:08 AM.
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  2. #2
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    Quote Originally Posted by saw235 View Post
    Given [tex] x+y=2 /math] and  2xy-z^2=1 solve for the unknowns.

    The answer for all of them is all 1. I can get it by substituting 1 into them (with some guessing), but is there any working for getting the answer without guessing? Some help here please?
    Since you have more unknowns than equations, the best you can do is to write everything in terms of one variable.
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  3. #3
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    Quote Originally Posted by saw235 View Post
    Given  x+y=2 and  2xy-z^2=1 solve for the unknowns.

    The answer for all of them is all 1. I can get it by substituting 1 into them (with some guessing), but is there any working for getting the answer without guessing? Some help here please?
    x= y= z= 1 is one solution. There are an infinite number of solutions as Prove it said.

    If x+ y= 2, y= 2-x so 2xy- z^2= 2x(2-x)- z^2= 4x- 2x^2- z^2= 1. z^2= 4x- 2x^2- 1 so z= \pm\sqrt{4x-2x^2-1}. That z only exists when \frac{4- 2\sqrt{2}}{4}< x< \frac{4+ 2\sqrt{2}}{4}.

    In that range, the roots are x, y= 2- x and z= \pm\sqrt{4x-2x^2-1} for every value of x. It is true that x= 1 is in that range and then y= 2-1= 1 and z= \sqrt{4- 2-1}= 1 so that (1, 1, 1) is one solution. x= 1, y= 1, z= -\sqrt{4-2-1}= -1 or (1, 1, -1) is also a solution. Those may be the only integer solutions.
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  4. #4
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    Sorry, I don't really understand what do you mean by

    "In that range, the roots are x, y= 2- x and z= for every value of x."

    By roots, which roots are you talking about? Please explain about the above in details and on how do you come to the conclusion that the roots are x, y= 2- x and z = . A little lost here
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