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    Solve the following equation
     \displaystyle x^4+6x^3+5x^2-12x-5=0 .
    Any ideas ??
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    (Original post by Ano123)
    Solve the following equation
     \displaystyle x^4+6x^3+5x^2-12x-5=0 .
    Any ideas ??
    What sort of thing is this for, the roots aren't very pleasant you're going to need to be creative or use a rather complicated and unpleasant method.
    That is, unless you're permitted to stick it into a computational engine.
    EDIT: Ignore me, I'll go back to sleep
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    (Original post by joostan)
    What sort of thing is this for, the roots aren't very pleasant you're going to need to be creative or use a rather complicated and unpleasant method.
    That is, unless you're permitted to stick it into a computational engine.
    The roots are nice.
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    (Original post by Ano123)
    Solve the following equation
     \displaystyle x^4+6x^3+5x^2-12x-5=0 .
    Any ideas ??
    (x^2+bx+c)(x^2+dx+e)=x^4+6x^3+5x  ^2-12x-5


    (Original post by joostan)
    What sort of thing is this for, the roots aren't very pleasant you're going to need to be creative or use a rather complicated and unpleasant method.
    That is, unless you're permitted to stick it into a computational engine.
    As 5 is prime, can't you infer c and e easily and compare coefficients for the rest?
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    (Original post by Kvothe the arcane)
    (x^2+bx+c)(x^2+dx+e)=x^4+6x^3+5x  ^2-12x-5




    As 5 is prime, can't you infer c and e easily and compare coefficients for the rest?
    Very good point, I have been a wee bit silly .


    (Original post by Ano123)
    The roots are nice.
    Well, I wouldn't go that far, but I guess they could be worse.
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    (Original post by joostan)
    Very good point, I have been a wee bit silly .



    Well, I wouldn't go that far, but I guess they could be worse.
    As far as quartic equations go the roots are nice
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    (Original post by Kvothe the arcane)
    (x^2+bx+c)(x^2+dx+e)=x^4+6x^3+5x  ^2-12x-5




    As 5 is prime, can't you infer c and e easily and compare coefficients for the rest?
    I solved it by completing the square, didn't try factorising it but you're right it factorises nicely into two quadratic factors.
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    (Original post by Kvothe the arcane)
    As 5 is prime, can't you infer c and e easily and compare coefficients for the rest?
    Not sure what you mean by this?

    Why can't you have (x^2 + bx + \sqrt{5})(x^2 + cx - \sqrt{5})?
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    (Original post by Zacken)
    Not sure what you mean by this?

    Why can't you have (x^2 + bx + \sqrt{5})(x^2 + cx - \sqrt{5})?
    Well that's certainly stirred things up...
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    (Original post by Zacken)
    Not sure what you mean by this?

    Why can't you have (x^2 + bx + \sqrt{5})(x^2 + cx - \sqrt{5})?
    For such a quartic to be solvable for an a-level student, I assumed (perhaps wrongly?) that the c coefficient of the quadratics would have to be an integer.
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    (Original post by Kvothe the arcane)
    For such a quartic to be solvable for an a-level student, I assumed (perhaps wrongly?) that the c coefficient of the quadratics would have to be an integer.
    It's not really a fair assumption - it's okay assuming something like "Oh yeah, this is an A-Level quartic so I should try and check whether the first few integers are roots" and then having that not pan out, but making that kind of a mathematical assumption isn't very valid. (imo)
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    (Original post by Zacken)
    It's not really a fair assumption - it's okay assuming something like "Oh yeah, this is an A-Level quartic so I should try and check whether the first few integers are roots" and then having that not pan out, but making that kind of a mathematical assumption isn't very valid. (imo)
    I agree, but it is a very reasonable first attempt to guess a factorisation, and happens to work out, and makes the problem far simpler than I was making it.
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    I stand corrected :getmecoat:.

    Zacken, joostan
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    (Original post by Zacken)
    It's not really a fair assumption - it's okay assuming something like "Oh yeah, this is an A-Level quartic so I should try and check whether the first few integers are roots" and then having that not pan out, but making that kind of a mathematical assumption isn't very valid. (imo)
    And whilst you make a valid point, there are better ways of saying this instead of publicly humiliating users who are genuinely helping. I'm just saying.
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    (Original post by Slowbro93)
    And whilst you make a valid point, there are better ways of saying this instead of publicly humiliating users who are genuinely helping. I'm just saying.
    He didn't really humiliate, did he? Also Kvothe and Zain are good friends (I think)
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    (Original post by Slowbro93)
    And whilst you make a valid point, there are better ways of saying this instead of publicly humiliating users who are genuinely helping. I'm just saying.
    I wasn't trying to come across as humiliating anybody - I'm sorry if that's how it came across, KTA. I rush and end up being more blunt than I'd like sometimes. I hope you know that I wouldn't say anything out of malice to you. But thanks for bringing it up Slowbro, I'll try and look over the tone of my post before posting in the future.
 
 
 

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