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Math Help - Proof In a System of Equations

  1. #1
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    Proof In a System of Equations

    Prove that x=y=z in the equations

    x+y^2+z^4=0
    y+z^2+x^4=0
    z+x^2+y^4=0

    where x, y, and z are real numbers.
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  2. #2
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    Quote Originally Posted by Winding Function View Post
    Prove that x=y=z in the equations

    x+y^2+z^4=0
    y+z^2+x^4=0
    z+x^2+y^4=0

    where x, y, and z are real numbers.
    Can you give us some indication as where this problem comes from?
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    Can you give us some indication as where this problem comes from?
    An Alg2/Trig Text. Is it too easy/difficult?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Winding Function View Post
    Prove that x=y=z in the equations

    x+y^2+z^4=0
    y+z^2+x^4=0
    z+x^2+y^4=0

    where x, y, and z are real numbers.
    I'm sorry, are you saying prove that x=y=z is the only solution?
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  5. #5
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    Dummy bro = plz read books

    x=-(y^2+z^4)

    y=-(z^2+x^4)

    z=-(x^2+y^4)

    let us consider that x=y=z and let us assume a real value which is positive.
    clearly any real number raised to even powers like 2 or 4 must give positive values
    as seen from the equation a positive value cannot be equal to a negative value. so the question is wrong.

    now let us assume a real negative value.
    x=y=z=-a
    clearly LHS of the equations are negative.
    -a raised to powers of 2 or 4 will produce positive values which are much larger or smaller (in case of negative fractions) than -a; then
    due to the negative sign outside bracket RHS become negative values which are much smaller or larger (in case of negative fractions) than -a

    Hence we draw conclusions that the equations are wrong.
    Last edited by mr fantastic; December 27th 2008 at 02:14 AM. Reason: Deleted all the smilies and the capital letters and the exclaimation marks.
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by susrut View Post
    x=-(y^2+z^4)

    y=-(z^2+x^4)

    z=-(x^2+y^4)

    [...] now let us assume a real negative value.
    x=y=z=-a
    clearly LHS of the equations are negative.
    -a raised to powers of 2 or 4 will produce positive values which are much larger or smaller (in case of negative fractions) than -a; then
    due to the negative sign outside bracket RHS become negative values which are much smaller or larger (in case of negative fractions) than -a

    HENCE WE DRAW CONCLUSIONS THAT THE EQUATIONS ARE WRONG!!!!!!
    The equation -a=-(a^2+a^4)\Longleftrightarrow a(1-a-a^3)=0 has at least one nontrivial solution because the polynomial 1-X-X^3 has one real root... The equations are not wrong.
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  7. #7
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    susrut,

    It's very rude to use a title such as "Dummy bro = plz read books".
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    I'm sorry, are you saying prove that x=y=z is the only solution?
    I'm trying to prove that for any set of solutions, (x,y,z), x=y=z.
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  9. #9
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    HELLO FLYINGSQUIRREL

    first of all i thank you

    1-X-X^3=0 has one real root which comes to be NEARLY equal to 0.68
    so x=y=z~0.68 is a solution to the equations
    but it is not a strong enough proof as it is a backdoor approach

    please reply



    SORRY TO ALL WHO FELT MY TITLE ON LAST POST ON THIS PAGE RUDE
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  10. #10
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by susrut View Post
    1-X-X^3=0 has one real root which comes to be NEARLY equal to 0.68
    so x=y=z~0.68 is a solution to the equations
    but it is not a strong enough proof as it is a backdoor approach
    I did not meant to prove that x=y=z, I just wanted to show that the three equations we are given are correct.
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