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How do you find the zeroes of the 1/2x^3-x? I factoried it so it would be 1/2x(x^2)-x. The one zero would then be zero, which matches with my book's answer. But I have no idea how my book got +/- the sq. root of two for the other one. Can you demonstrate, please?

  • Algebra -

    The way you factored makes no sense to me>
    Your factors should be
    x((1/2)x^2 - 1)
    so x = 0 or
    (1/2)x^2 - 1 = 0
    x^2 = 2
    then x = ±√2

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