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Thread: system of equations

  1. #1
    Senior Member pacman's Avatar
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    system of equations

    (1) x^4 = y - 1
    (2) 3x^2 = y + 1

    find all solutions in (x,y)
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  2. #2
    Member eXist's Avatar
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    Solve each equation for y:

    $\displaystyle [x^4 = y - 1] \Rightarrow [x^4 + 1 = y]$
    $\displaystyle [3x^2 = y + 1] \Rightarrow [3x^2 - 1 = y]$

    Since both equations are equal to y, you can set the two equations equal:

    $\displaystyle x^4 + 1 = 3x^2 - 1$

    Move everything to one side:

    $\displaystyle x^4 - 3x^2 + 2 = 0$

    Factor:

    $\displaystyle (x^2 - 2)(x^2 - 1) = 0$

    Set each equal to 0:

    $\displaystyle (x^2 - 2) = 0$
    $\displaystyle (x^2 - 1) = 0$

    Solve for x:

    $\displaystyle x = \pm\sqrt{2}$
    $\displaystyle x = \pm\sqrt{1}$

    Make it look a little nicer:

    $\displaystyle x = -\sqrt{2}, \sqrt{2}, -\sqrt{1}, \sqrt{1}$

    And finally, you can substitute these back in to get the values for y if you really need them .
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  3. #3
    Senior Member pacman's Avatar
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    Your answer checks with this graph, x^4 - 3x^2 + 2 = 0, thanks.
    Attached Thumbnails Attached Thumbnails system of equations-quartic.gif  
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