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Homework Help: Reduced cubic equation

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the solutions of:

    [tex]x^3-3x-2=0[/tex]

    using the Cardano's method.

    2. Relevant equations



    3. The attempt at a solution

    [tex]x=u+v[/tex]

    [tex](u+v)^3-3(u+v)-2=0[/tex]

    [tex]u^3+v^3+(u+v)(3uv-3)-2=0[/tex]

    [tex]3uv-3=0[/tex]

    [tex]uv=1[/tex]

    [tex]u^3v^3=1[/tex]

    [tex]u^3+v^3=2[/tex]

    [tex]u^3=v^3=1[/tex]

    Now [tex]u=v=\sqrt[3]{1}[/tex].

    I found u and v with [tex]z=\sqrt[3]{1}[/tex] (using complex numbers).

    [tex]u=v=1[/tex]

    [tex]u=v=\frac{-1}{2}+i\frac{sqrt{3}}{2}[/tex]

    [tex]u=v=\frac{-1}{2}-i\frac{sqrt{3}}{2}[/tex]

    x=u+v

    [tex]x_1=2[/tex]

    [tex]x_2=-1+i\sqrt{3}[/tex]

    [tex]x_3=-1-i\sqrt{3}[/tex]

    I substitute above and something is wrong. Where is the error?????
     
  2. jcsd
  3. Sep 6, 2008 #2
    there may be a better way out..observinf that if [tex]f(x)=x^3-3x-2[/tex].
    clearly [tex]f(-1)=0[/tex]
    satisfies the eqn...hence
    [tex](x+1)[/tex] is a factor of [tex]f(x)[/tex].
    Now use division and get [tex]f(x)[/tex] in the form of factors and put the other factor(other than [tex]x+1[/tex])
    to be zero and solve for
    [tex]x[/tex]
     
  4. Sep 6, 2008 #3
    Yes, I know that there are few ways, but I need to solve it with Cardanos. Please help.
     
  5. Sep 6, 2008 #4

    HallsofIvy

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    Science Advisor

    ?? I don't see how this follows directly, although it is true. Since you have defined u and v such that uv= 1, you have v= 1/u and so [itex]u^3+ 1/u^3= 2[/itex]. Multiplying both sides by u3, you get the quadratic (in u3) (u3)2- 2(u3)+ 1= 0 which does give u3= 1 or u3= -1 as roots.

    No. This is your error. It does not follow from u3= v3 that u= v. Only that they are both cube roots of 3. And, as you note below, there are 3 of those.

    You do NOT know, as I said, that u= v. You know, rather, that uv= 1 so v= 1/u.
    If
    [tex]u= \frac{-1+ i\sqrt{3}}{2}[/tex]
    then
    [tex]v= \frac{2}{-1+ i\sqrt{3}}[/tex]
    rationalize the denominator by multiplying both numerator and denominator by [itex]-1+ i\sqrt{3}[/itex] and you get
    [itex]\frac{-1- i\sqrt{3}}{2}[/itex]
    the other non-real root of z3= 1.

    Now, x= u+ v gives x= -1 which is correct. The three roots of x3- 3x- 2= 0 are 2, -1, -1.

     
  6. Sep 7, 2008 #5
    So, first I found the 3 solutions of u , and then substitute in uv=1, I get v and then substitute in x=u+v, right?
     
  7. Sep 7, 2008 #6

    HallsofIvy

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    Science Advisor

    Yes, basically, that is what you do.
     
  8. Sep 7, 2008 #7
    Ok, thanks. But why I read that also suming [itex]u_1+v_1=x_1[/tex], will work if I found [itex]v_1[/itex], by using complex algebra?
     
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