Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding roots to cubics

  1. Oct 16, 2005 #1
    hello

    how can i finding roots to cubics??

    explain by example

    :smile:
     
  2. jcsd
  3. Oct 16, 2005 #2

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Perhaps you have an example in mind?
     
  4. Oct 16, 2005 #3

    mezarashi

    User Avatar
    Homework Helper

    Oh how I just love these open-ended questions ;P
     
  5. Oct 16, 2005 #4

    Fermat

    User Avatar
    Homework Helper

    If you want to find approximate vaues, use newton's method of approximation. (you can google for it).
    For a more analytical approach try here
     
  6. Oct 16, 2005 #5

    x^3+(5*x^2)+(3*x)+9=0
     
  7. Oct 16, 2005 #6

    no i do not want newton's method of approximation :smile:

    i want (kardan) method but i am not sure about the spelling of (kardan)
     
  8. Oct 16, 2005 #7

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    Cardan's formula gives

    [tex]x = -\frac {\left( 179 - 9 \sqrt {345} \right)^{1/3}}{3}-\frac {\left( 179 + 9 \sqrt {345} \right)^{1/3}}{3} - \frac {5}{3}[/tex]

    for the real root. The other two are complex. And, no, I did not do it by hand! :)
     
  9. Oct 16, 2005 #8

    HallsofIvy

    User Avatar
    Science Advisor

    If a and b are any two numbers then
    (a- b)3= a3-3a2b+ 3ab2- b3
    3ab(a-b)= 3a2b- 3ab2

    so (a-b)3+ 3ab(a-b)= a3- b3.

    In particular, if we let x= a-b, m= 3ab, and n= a3- b3, that says that x3+ mx= n. That is, we can pick any two numbers a, b and right down a cubic equation that has x= a- b as a root.
    The question is, can we go the other way: given m and n, can we find a and b so we can write x= a-b as a solution.
    The answer to that question is "Yes, we can"!

    Since m= 3ab, b= m/3a. Putting that int n= a3- b3, we have [tex]n= a^3- \frac{m^3}{3^3a^3}[/tex].
    Multiplying both sides of the equation by a3, we have
    [tex]na^3= a^6- (\frac{m}{3})^3[/tex]
    which looks worse but is just a quadratic equation in a3:
    [tex](a^3)^2- n(a^3)- (\frac{m}{3})^3[/tex].
    Use the quadratic formula to solve that
    [tex]a^3= \frac{n +/- \sqrt{n^2- 4\left(\frac{m}{3}\right)^3}}{2}[/tex]
    [tex]a^3= \frac{n}{2} +/- \sqrt{\left(\frac{n}{2}
    \right)^2- \left(\frac{m}{3}\right)^3}[/tex]

    Since a3- b3= n, solving for b3 gives
    [tex]a^3= -\frac{n}{2} +/- \sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}[/tex]

    Finding the cube root of each of those, then subtracting to get x= a- b gives the formula that Tide cited.

    Warning- applying that formula is really, really hard!
     
    Last edited by a moderator: Oct 16, 2005
  10. Oct 16, 2005 #9
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook