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    How do I derive the dicriminant of a cubic function??
    i.e   {-4b^3d + b^2c^2 - 4ac^3 + 18abcd - 27a^2d^2.} ,
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    What's your definition of "discriminant"?
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    one must think of a cubic
    with graphic properties
    and shape and turning points
    most appropriately
    so I would recommend you derive
    and solve equal to nought
    and with these station'ry values
    seek the solution that was sought
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    (Original post by generalebriety)
    What's your definition of "discriminant"?
    ermm, I dont know what the definition is .. all I know is it can be used to find the nature of stationary points:o: ..
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    (Original post by rbnphlp)
    ermm, I dont know what the definition is .. all I know is it can be used to find the nature of stationary points:o: ..
    Well, that'll be why you can't derive it...

    What level are you at? GCSE / AS / A2?
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    (Original post by generalebriety)
    Well, that'll be why you can't derive it...

    What level are you at? GCSE / AS / A2?
    ok let me rephrase my question How would I solve a cubic finction

    ax^3+bx^2+cx+d ..

    I just finished my Alevels
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    B^2-4ac is the discriminant.
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    (Original post by learner_dancer)
    B^2-4ac is the discriminant.
    not for a cubic function it isnt.:p:
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    (Original post by rbnphlp)
    not for a cubic function it isnt.:p:
    Nope.
    Sigh.
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    (Original post by rbnphlp)
    ok let me rephrase my question How would I solve a cubic finction

    ..

    I just finished my Alevels
    Newton raphson?

    There is a general formula but its very long and impossible to memorise. Your best bet would be to find a factor then do polynomial divison.
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    (Original post by rbnphlp)
    ok let me rephrase my question How would I solve a cubic finction

    ax^3+bx^2+cx+d ..

    I just finished my Alevels
    That's a completely different question. See here.

    For your original question (though be warned that this will require you to do a lot of algebra): take two polynomials P and Q. Consider the product of all (x - y), where P(x) = 0 and Q(y) = 0, i.e. x is a root of P and y is a root of Q. Obviously this product is 0 if any one of its factors is 0, i.e. if any one root of P happens to also be one root of Q, and is non-zero otherwise. Now suppose we want to know whether P has a repeated root, say r. Well, if it does, then P is 0 at r, but P also turns at r, so P' (the derivative of P) is 0 at r. So let Q = P' in the above quantity. But of course we don't know the roots yet - so what can we do? Well, suppose P has roots r, s, t; then by writing P = ax^3 + bx^2 + cx + d = a(x-r)(x-s)(x-t) we can find that -d/a = rst, c/a = rs+rt+st, -b/a = r+s+t; from there on it's all just horrible algebra. (If you're interested in a less hurried explanation of this, search Wikipedia or something for the "resultant", a related quantity.)
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    wiki fail you mean.
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    (Original post by Simplicity)
    Newton raphson?

    There is a general formula but its very long and impossible to memorise. Your best bet would be to find a factor then do polynomial divison.
    I thought of that , but it doesnt work for the equation i wanted to solve

    x^3+x=6
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    (Original post by rbnphlp)
    I thought of that , but it doesnt work for the equation i wanted to solve
    What?
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    (Original post by rbnphlp)
    I thought of that , but it doesnt work for the equation i wanted to solve
    Wolfram alpha gives this
    http://www76.wolframalpha.com/input/...+x%5E3%2Bx%3D6

    P.S. Where did you get the problem. As I'm pretty sure no one would ask you that in a textbook. So you must have made that up.
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    (Original post by rbnphlp)
    I thought of that , but it doesnt work for the equation i wanted to solve

    x^3+x=6
    google depressed cubic.
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    (Original post by Simplicity)
    Wolfram alpha gives this
    http://www76.wolframalpha.com/input/...+x%5E3%2Bx%3D6

    P.S. Where did you get the problem. As I'm pretty sure no one would ask you that in a textbook. So you must have made that up.
    No I i didnt make it up .. I came Across it on " the history of mathematics"
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    Another way to solve it is to introduce a linear shift y = x-p to 'remove' the x^2 term, to then get you an equation of the form hy^2 \pm iy = j. Then introduce another substitution of the form z=qy to transform the above equation into something of the form 4z^3 \pm 3z = k, and then note that a solution to this is given by either cos (arcos k)/3 , cosh (arcosh k)/3, or sinh (arsinh k)/3 depending on whether you have a +, - and whether k is greater or smaller than 1.

    i'm not too sure of the above, but i think the general gist of it is right.
 
 
 
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