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Math Help - Rearranging help?

  1. #1
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    Rearranging help?

    Alright, this has stumped me.

    How can I rearrange:

    2x^3 - x - 4 = 0

    to be:

    x = ROOT(2/x + 1/2) ?
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  2. #2
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    Quote Originally Posted by Drummerdude93 View Post
    Alright, this has stumped me.

    How can I rearrange:

    2x^3 - x - 4 = 0

    to be:

    x = ROOT(2/x + 1/2) ?
    Hi Drummerdude,

    2x^3-x-4=0

    2x^3-x=4

    Factor...

    x\left(2x^2-1\right)=4

    2x^2-1=\frac{4}{x}

    2x^2=\frac{4}{x}+1

    x^2=\frac{2}{x}+\frac{1}{2}

    Finally, take the square root.
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  3. #3
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    Quote Originally Posted by Drummerdude93 View Post
    Alright, this has stumped me.

    How can I rearrange:

    2x^3 - x - 4 = 0

    to be:

    x = ROOT(2/x + 1/2) ?
    note that the transformation is incorrect ... it should be x = \pm \sqrt{\frac{2}{x} + \frac{1}{2}} \, ; x \ne 0

    what problem are you solving that led you to ask about this?
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