1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reduced cubic equation

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

    Find the solutions of:


    using the Cardano's method.

    2. Relevant equations

    3. The attempt at a solution









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

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








    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
  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


    User Avatar
    Staff Emeritus
    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.
    [tex]u= \frac{-1+ i\sqrt{3}}{2}[/tex]
    [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


    User Avatar
    Staff Emeritus
    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Reduced cubic equation
  1. Cubic equation (Replies: 3)

  2. Cubic equation (Replies: 5)