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More systems of quadratics

  1. Jan 27, 2005 #1
    Sorry guyz...one last question...

    x^2 - 5y^2 = -44
    xy = -24
    divide by x: y = -24/x
    x^2 - 5(-24/x)^2 = -44
    multiply: x^2 - 2880/x = -44

    Where do I go from here?? :uhh:
  2. jcsd
  3. Jan 27, 2005 #2


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    Wrong,u should get a biquadratic.
    [tex] x^{2}-\frac{2880}{x^{2}}=-44 [/tex]

    Can u solve this type of equations...??
    [tex] x^{2}=t [/tex]

    And solve the quadratic equation...

  4. Jan 27, 2005 #3


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    This is wrong.

    As a further technicality, your logic thus far shatters when you consider the possibility that x = 0: you have to do one of three things:

    (1) avoid dividing by x
    (2) handle x = 0 as a special case
    (3) prove x = 0 cannot happen
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