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    How do you find the solution to x^3-27x-90=0.Thanks for the help in advance.
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    Use this formula where a=1, b=0, c=-27 and d=-90
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    (Original post by xpointx)
    How do you find the solution to x^3-27x-90=0.Thanks for the help in advance.
    If this is A-level, then only graphically or numerically
    However there is a method which leads to the solution of x^3-3ax-b=0 being given by
     u=\sqrt[3]{\frac{1}{2}(b+\sqrt{b^2+4a^3})}  , v=\frac{a}{u} then solution of equation is u+v
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    (Original post by xpointx)
    How do you find the solution to x^3-27x-90=0.Thanks for the help in advance.
    Is this for FP1?

    You can expand the equation:

    (x-\alpha)(x-\beta)(x-\gamma)=0

    And compare it to your cubic to work out the values of \alpha, \: \beta \: \text{and} \: \gamma.

    That should give you the roots with a bit of manipulation I would think.
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    x^3 - 27x - 90 = 0

    x^3 - 27x = 90

    x ( x^2 - 27) = 90

    x ( x^2 - x^3 ) = 90

    :. x ( x - 3 ) ( x - 3 ) = 90

    So the solutions you can work out.

    x=90
    x=93

    Working

    x - 3 = 90
    :. x= 90+3
    x=93
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    (Original post by Sena5)
    x^3 - 27x - 90 = 0

    x^3 - 27x = 90

    x ( x^2 - 27) = 90

    x ( x^2 - x^3 ) = 90

    :. x ( x - 3 ) ( x - 3 ) = 90

    So the solutions you can work out.

    x=90
    x=93


    Working

    x - 3 = 90
    :. x= 90+3
    x=93
    You've made quite a few mistakes here:

    First you changed 27 to x^3 which isn't correct.

    Then you said x^2 - x^3 = (x-3)(x-3) which is also wrong.

    Then you read off solutions to x ( x - 3 ) ( x - 3 ) = 90 but forgot that the right-hand-side is not 0.
 
 
 
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