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Can someone please help me with this equation? Watch

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    its

    6x4+11x2+4x

    thanks in advance
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    (Original post by boner in jeans)
    well what do you want done with it? solve, factorise?

    and postgraduate, really? :rofl:

    lol i'm new so i just put in a label..i'm sorry..uhm and i need to solve it..find x ..sorry if my english is kind of bad
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    i eventually reach

    x3+11/6x+2/3 = 0

    how do i continue ?
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    The question is slightly underestimated by the above posts.

    Notice that x = 0 is a solution.

    Hence, to solve 6x^3 + 11x + 4 = 0 you can use either Cardano's method or Viète's substitution.


    That is, by Viète, you can turn x^3 + ax + b = 0 into

    \displaystyle t^6 + bt^3 - \frac{a^3}{27} = 0

    by letting \displaystyle x = t - \frac{a}{3t}. Solve the quadratic.
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    well that was embarrassing listen to jack :getmecoat:
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    (Original post by jack.hadamard)
    The question is slightly underestimated by the above posts.

    Notice that x = 0 is a solution.

    Hence, to solve 6x^3 + 11x + 4 = 0 you can use either Cardano's method or Viète's substitution.


    That is, by Viète, you can turn x^3 + ax + b = 0 into

    \displaystyle t^6 + bt^3 - \frac{a^3}{27} = 0

    by letting \displaystyle x = t - \frac{a}{3t}. Solve the quadratic.


    Can you explain it please ?
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    (Original post by tjava)
    Can you explain it please ?
    Which part?
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    (Original post by jack.hadamard)
    Which part?

    Viète's substitution
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    (Original post by tjava)
    Viète's substitution
    Substitute \displaystyle x = t - \frac{a}{3t} into x^3 + ax + b = 0 and simplify.
    You can solve the resulting quadratic to find three values of t. The you find x.
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    (Original post by jack.hadamard)
    The question is slightly underestimated by the above posts.

    Notice that x = 0 is a solution.

    Hence, to solve 6x^3 + 11x + 4 = 0 you can use either Cardano's method or Viète's substitution.


    That is, by Viète, you can turn x^3 + ax + b = 0 into

    \displaystyle t^6 + bt^3 - \frac{a^3}{27} = 0

    by letting \displaystyle x = t - \frac{a}{3t}. Solve the quadratic.
    Wow, I've never heard of this method for solving cubics before. Think I'm gonna have to read up on this, looks like it could be pretty handy!
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    (Original post by tjava)
    6x4+11x2+4x
    Just a thought. Is this original equation correct. I ask as, if it was 6x^4+11x^2+4 then it factorises nicely.
 
 
 
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