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Math Help - Solve this Quadratic Equation

  1. #1
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    Solve this Quadratic Equation

    Is it possible to solve (for x and y) this quadratic equation:

    x^2 + 7xy - 3 = 0
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  2. #2
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    Quote Originally Posted by acevipa View Post
    Is it possible to solve (for x and y) this quadratic equation:

    x^2 + 7xy - 3 = 0
    What sort of solutions are you looking for?
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  3. #3
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    Post not sure exactly

    Hmmm. I really am not sure on this one, I just tried typing it into my fancy calculator and got a strange result so I don't think you can.
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  4. #4
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    Quote Originally Posted by acevipa View Post
    Is it possible to solve (for x and y) this quadratic equation:

    x^2 + 7xy - 3 = 0
    Not without more information:

    x^2 + 7xy - 3 = 0

    7xy  = -x^2 + 3

    y  = \frac{-x^2 + 3}{7x}

    y  = -\frac x7 + \frac 3{7x}

    It should be relatively clear from here that there are an infinite number of solutions.
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  5. #5
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    Quote Originally Posted by angel.white View Post
    Not without more information:

    x^2 + 7xy - 3 = 0

    7xy  = -x^2 + 3

    y  = \frac{-x^2 + 3}{7x}

    y  = -\frac x7 + \frac 3{7x}

    It should be relatively clear from here that there are an infinite number of solutions.
    Alternatively, solving for x using the quadratic formula:

    x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}.

    The interesting question would be to find all integer solutions (x, y).
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  6. #6
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Alternatively, solving for x using the quadratic formula:

    x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}.

    The interesting question would be to find all integer solutions (x, y).
    I'll take a stab at it.
    Premise: x, y are integers

    x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}

    2x = -7y \pm \sqrt{49y^2 + 12}

    setup case 1: -7y + \sqrt{49y^2 + 12} = 2x

    -7y + \sqrt{49y^2 + 12} = 2x

    \sqrt{49y^2 + 12} = 2x+7y

    49y^2 + 12 = 4x^2+28xy+49y^2

    12 = 4x^2+28xy

    3 = x^2+7xy


    setup case 2: -7y - \sqrt{49y^2 + 12} = 2x

    \sqrt{49y^2 + 12} = -2x-7y

    49y^2 + 12 = 4x^2 +28xy+49y^2

    12 = 4x^2 +28xy

    3 = x^2 +7xy


    Continued from 3 = x^2 +7xy
    So it can be seen that regardless of which setup case we choose, positive or negative, we end up evaluating the same equation of 3 = x^2 +7xy

    Note that if x = 0 then 3 = 0, which is false, so x cannot be zero.

    \frac 3x =x + 7y

    Now, because x and y are both integers, x + 7y must be an integer. Which means that 3/x must be an integer.

    So x \in \{ -3, -1, 1, 3\}


    proof case 1: x = -3
    \frac 3x =x + 7y

    \frac 3{-3} =-3 + 7y

    \frac 27 = y

    But this contradicts that y is an integer, so x \not{=} -3


    proof case 2: x = -1
    \frac 3x =x + 7y

    \frac 3{-1} =-1 + 7y

    -\frac 27 = y

    But this contradicts that y is an integer, so x \not{=} -1


    proof case 3: x = 1
    \frac 3x =x + 7y

    \frac 3{1} =1 + 7y

    \frac 27 = y

    But this contradicts that y is an integer, so x \not{=} 1


    proof case 4: x = 3
    \frac 3x =x + 7y

    \frac 3{3} =3 + 7y

    -\frac 27 = y

    But this contradicts that y is an integer, so x \not{=} 3



    Therefore there are no integer values of x for which y can be an integer also.

    So the solution to the problem "find all integer solutions (x, y)" is the empty set.
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  7. #7
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    Quote Originally Posted by angel.white View Post
    I'll take a stab at it.
    Premise: x, y are integers

    x = \frac{-7y \pm \sqrt{49y^2 + 12}}{2}

    2x = -7y \pm \sqrt{49y^2 + 12}

    setup case 1: -7y + \sqrt{49y^2 + 12} = 2x

    -7y + \sqrt{49y^2 + 12} = 2x

    \sqrt{49y^2 + 12} = 2x+7y

    49y^2 + 12 = 4x^2+28xy+49y^2

    12 = 4x^2+28xy

    3 = x^2+7xy


    setup case 2: -7y - \sqrt{49y^2 + 12} = 2x

    \sqrt{49y^2 + 12} = -2x-7y

    49y^2 + 12 = 4x^2 +28xy+49y^2

    12 = 4x^2 +28xy

    3 = x^2 +7xy


    Continued from 3 = x^2 +7xy
    So it can be seen that regardless of which setup case we choose, positive or negative, we end up evaluating the same equation of 3 = x^2 +7xy

    Note that if x = 0 then 3 = 0, which is false, so x cannot be zero.

    \frac 3x =x + 7y

    Now, because x and y are both integers, x + 7y must be an integer. Which means that 3/x must be an integer.

    So x \in \{ -3, -1, 1, 3\}


    proof case 1: x = -3
    \frac 3x =x + 7y

    \frac 3{-3} =-3 + 7y

    \frac 27 = y

    But this contradicts that y is an integer, so x \not{=} -3


    proof case 2: x = -1
    \frac 3x =x + 7y

    \frac 3{-1} =-1 + 7y

    -\frac 27 = y

    But this contradicts that y is an integer, so x \not{=} -1


    proof case 3: x = 1
    \frac 3x =x + 7y

    \frac 3{1} =1 + 7y

    \frac 27 = y

    But this contradicts that y is an integer, so x \not{=} 1


    proof case 4: x = 3
    \frac 3x =x + 7y

    \frac 3{3} =3 + 7y

    -\frac 27 = y

    But this contradicts that y is an integer, so x \not{=} 3



    Therefore there are no integer values of x for which y can be an integer also.

    So the solution to the problem "find all integer solutions (x, y)" is the empty set.
    This might have no solutions with x and y in \mathbb{Z} but it does have four such solutions in \mathbb{Q}.

    (Apply the rational root theorem to get a list of candidates for x, then find the (if it exists) y that gives that root. This also gives a neater proof of no solutions in \mathbb{Z})

    RonL

    (Of course ther is a deliberate mistake here in that there is an implicit assumption that 7y is an integer, correcting the mistake gives an infinite number of rational solutions)
    Last edited by CaptainBlack; April 5th 2008 at 11:14 AM.
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  8. #8
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    Quote Originally Posted by CaptainBlack View Post
    This might have no solutions with x and y in \mathbb{Z} but it does have four such solutions in \mathbb{Q}.

    (Apply the rational root theorem to get a list of candidates for x, then find the (if it exists) y that gives that root. This also gives a neater proof of no solutions in \mathbb{Z})

    RonL
    I think I understand, I looked up the rational root theorem, and got \pm \frac {1, 3}{1, 3}

    So I could have just jumped straight to the 4 proof cases.

    (I assume this is what you were suggesting, because there should be an infinite number of rational solutions to the problem.)
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