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Math Help - [SOLVED] Find number of distinct pairs (x,y)?(difficult)

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Find number of distinct pairs (x,y)?(difficult)

    The number of distinct pairs (x,y) of real numbers satisfying x = x^3 + y^4 and y = 2xy is:

    A)5
    B)12
    C)3
    D)7
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    y = 2xy

    y - 2xy = 0

    y(1 - 2x) = 0

    y = 0 or x = \frac{1}{2}

    for y = 0 ...

    x = x^3 + y^4

    x = x^3

    x - x^3 = 0

    x(1 - x^2) = x(1 - x)(1 + x) = 0

    (0,0) , (1,0) and (-1,0)

    for x = \frac{1}{2} ...

    \frac{1}{2} = \frac{1}{8} + y^4

    y^4 - \frac{3}{8} = 0

    \left(y^2 + \sqrt{\frac{3}{8}}\right)\left(y^2 - \sqrt{\frac{3}{8}}\right) = 0

    \left(y^2 + \sqrt{\frac{3}{8}}\right)\left(y - \sqrt[4]{\frac{3}{8}}\right)\left(y + \sqrt[4]{\frac{3}{8}}\right) = 0

    \left(\frac{1}{2}, \sqrt[4]{\frac{3}{8}}\right) and  \left(\frac{1}{2}, -\sqrt[4]{\frac{3}{8}}\right)

    looks like 5 distinct solutions to me ... hope I didn't miss any.
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