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Math Help - Changing the subject

  1. #1
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    Changing the subject

    (x+a)^3 / x = c

    Express x in terms of a and c .

    How do you do it?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by hitmen View Post
    (x+a)^3 / x = c

    Express x in terms of a and c .

    How do you do it?
    With an enormous amount of difficulty. Solving a general cubic equation is a serious pain. Are you sure that exponent isn't a 2, not a 3?

    -Dan
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  3. #3
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    Well, this isnt a textbook question. I cant remember where I saw this. If it is difficult can you just give me the appropriate steps to be taken. Thanks. I should be able to work it out myself.
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  4. #4
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    Hi !

    It's not easy for general formulae...
    If you know one of the roots, then you will easily be able to find the others (polynomial factorisation).

    Now, I know there is a general formula for x^3+px+q=0

    ------------------
    To refer to your exercise, you're looking for :

    (x+a)^3=cx \Longleftrightarrow (x+a)^3-cx=0
    you can change x+a for y :

    y^3-cy+ca=0

    ------------------

    If 4p^3+27q^2 \ge 0, then there is a real solution (& 2 complex ones) :

    x=\sqrt[3]{-\frac q2+\sqrt{\left(\frac q2\right)^2+\left(\frac p3\right)^3}}+\sqrt[3]{-\frac q2-\sqrt{\left(\frac q2\right)^2+\left(\frac p3\right)^3}}

    Unfortunately, I can't remember who discovered it (he was an Italian guy, in the XVI th century if my memory is correct)... I had to give back the book to the library.



    Edit : his name is Bombelli.

    ---------------------

    I've just seen you can use the Cardan method or trigonometric method... Google for it, I don't have enough knowledge to talk about it.
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  5. #5
    Forum Admin topsquark's Avatar
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    See Cardano's method.

    -Dan
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  6. #6
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    solving cubic equation

    even i do substitution, what is next?

    y=x+a
    y^3 = cy-ca
    y^3 -cy +ca = 0

    how to factorise or solve?

    BTW, how do you enter square roots, power and other symbols?
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  7. #7
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    now this is the only thing I dont understand.

    Setting 3uv + p = 0

    Why?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by hitmen View Post
    now this is the only thing I dont understand.

    Setting 3uv + p = 0

    Why?
    If you look at the general form as you go along you will note that the derivation adds one extra variable to the solution, giving us the freedom to choose a single condition that is convenient for us. This is the condition that we choose. Once this substitution is made the equation falls into something like the form
    p^3 + \frac{n}{p^3} = m

    (p^3)^2 - mp^3 + n = 0

    This is a "tri-quadratic" equation, I guess you'd call it. We can easily solve this for p by using the quadratic formula to solve for p^3 then simply taking the cube root to find p. Then you can back substitute to get your answer in terms of the original coefficients.

    You can get a tutorial on LaTeX here, in the first few posts of the sticky thread at the top of the list.

    -Dan
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