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Cubic equations

  1. Jun 2, 2010 #1
    Hi colleages. can you help me to solve the cubic equation below:
    2N(Ep-En)hp^3(x)-3[M(x,t)(Ep-En)-2NEnh]hp^2(x)-6Enh[M(x,t)+Nh]hp(x)+Enh^2[3M(x,t)+2Nh]=0 notice that all variables in the equation are dependent on x only, except M is dependent on x and t.
    En, Ep, N and h are constants.
    i know the solution when the equation dependent on one variable, but the problem is that M dependent on two variables x and t.
    please help me and i will be very grateful for you...
     
  2. jcsd
  3. Jun 2, 2010 #2
    I can't speak for everyone, but personally I find that presentation rather impenetrable! You will probably get a better response if you write it out in using superscripts etc.
     
  4. Jun 3, 2010 #3
    well mrbohn1 i'll try to write the real equation in the form:
    ax^3-b(x,t)x^2-c(x,t)x+d(x,t)=0
    but don't overlook of the real equation in the first post to know what i mean...
    thanks
     
  5. Jun 3, 2010 #4
    As written it's not generally a cubic equation (I think in either post).

    For example if [itex]a=b(x,t)=c(x,t)=0[/itex] in the second post, you're asking for the solutions of [itex]d(x,t)=0[/itex], and without being more specific I don't think you'll get any sensible answers.

    Similarly in the first post you'd need to be more specific about what [itex]p(x)[/itex] (or is that [itex]h_p(x)[/itex]?) and [itex]M(x,t)[/itex] are.

    As mrbohn1 says you'ld probably get a better response if you made the typesetting clearer in the clarification.
     
  6. Jun 3, 2010 #5
  7. Jun 4, 2010 #6
    My colleages. the cubic equation is as shown below:
    2N(Ep-En)X^3-3[M(x,t)(Ep-En)-2NEnh]X^2-6Enh[M(x,t)+Nh]X+Enh^2[3M(x,t)+2Nh]=0 this is cubic equations. i want to solve this equation and get the roots of it.
    and all the coefficients of the equation are constant except M(x,t).
    By meaning,
    you can write the equation in the form aX^3+bX^2+cX+d=0
    where a, b, c and d=coefficients of the equation and their values as showm in the main equation above.
    but you must notice that M(x,t) is dependent on x and t, indeed this is a real trouble.
    thanks
     
  8. Jun 4, 2010 #7

    Office_Shredder

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    The fact that M(x,t) depends on x means that it's impossible to give a root as is. That' not really a cubic polynomial, and we have no idea what the actual equation you want us to solve is. For example if M(x,t)=x2 then what you really have is a fourth degree polynomial which has no general solution.
     
  9. Jun 6, 2010 #8
    Not correct. A fourth degree polynomial has solutions in radicals - see http://en.wikipedia.org/wiki/Quartic_function" [Broken].

    Which doesn't invalidate the above.

    If for example we set

    [tex]M(x,t)=\frac{2N(E_p-E_n)x^3+6NE_nhx^2-6NE_nh^2x+4NE_nh^3-f(x,t)}{3(E_p-E_n)x^2+6E_nhx-E_nh^2}\text{ (*)}[/tex]

    where [itex]f(x,t)[/itex] is any function of [itex]x[/itex] and [itex]t[/itex], the given equation becomes

    [tex]f(x,t)=0[/tex].

    (I've assumed [itex]x[/itex] and [itex]X[/itex] in the given equation are the same.)

    So if we don't get anything more specific about the form of [itex]M(x,t)[/itex] it's apparent that we're asked, effectively, for the general "solution" to [itex]f(x,t)=0[/itex], which would actually be just the equation itself.

    More specific information might (or might not) allow for the equation to be solved for [itex]x[/itex] in terms of [itex]t[/itex], which is presumably what omarxx84 is asking for when he refers to the roots.

    For instance if [itex]M(x,t)[/itex] were defined by [itex]\text{(*)}[/itex] with [itex]f(x,t)=x^2-(1+t)x+t[/itex] we could say [itex]x=1\text{ or }t[/itex].

    On the other hand if [itex]M(x,t)[/itex] were defined by [itex]\text{(*)}[/itex] with [itex]f(x,t)=p_{\lfloor |x|+1\rfloor}+p_{\lfloor |t|+1\rfloor}[/tex] (where [itex]p_n[/itex] is the [itex]n^{th}[/itex] prime and [itex]\lfloor r\rfloor[/itex] is [itex]floor(r)[/itex]) then nobody is likely to come up with a formula for [itex]x[/itex] in terms of [itex]t[/itex] any more useful than the original equation.
     
    Last edited by a moderator: May 4, 2017
  10. Jun 6, 2010 #9

    Office_Shredder

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    Sorry I meant to make it a fifth degree polynomial. My mistake
     
  11. Jun 7, 2010 #10
    In which case it would still have roots, but not necessarily in radicals. See http://en.wikipedia.org/wiki/Ultraradical.

    The point is that even though we may be able to solve the equation we can't even begin without a more specific description of [itex]M(x,t)[/itex].
     
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