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Solving Cubic Functions

  1. Nov 19, 2009 #1
    x3 - 12x + 1 = 0

    How does one solve for x?
     
  2. jcsd
  3. Nov 19, 2009 #2

    Mark44

    Staff: Mentor

  4. Nov 20, 2009 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Let x= a- b. Then [itex]a^3= (a-b)^3= a^3- 3a^2b+ 3ab^2- b^3[/itex].

    Also [itex]3abx= 3ab(a- b)= 3a^2b- 3ab^2[/itex].

    So [itex]x^3+ 3abx= a^3- b^3[/itex]. Letting m= 3ab and [itex]n= a^3- b^3[/itex], then x= a-b satisfies [itex]x^3+ mx= n[/itex].

    Suppose we know m and n- can we "recover" a and b and so find x?

    If m= 3ab, then b= m/3a and [itex]n= a^3- m^3/3^3a^3[/itex]. Multiplying through by [itex]a^3[/itex] we get [itex]na^3= (a^3)^2- m^3/3^3[/itex] which we can think of as a quadratic equation for [itex]a^3[/itex]: [itex](a^3)^2- na^3- m^3/3^3= 0[/itex] and solve by the quadratic formula:
    [tex]a^3= \frac{n\pm\sqrt{n^2+ 4\frac{m^3}{m^3}}}{2}[/tex][tex]= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/tex]
    so that
    [tex]a= \sqrt[3]{\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}}}[/tex]

    Since [itex]a^3- b^3= n[/itex], [itex]b^3= a^3- n[/itex] so
    [tex]b^3= -\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/tex]
    and
    [tex]b= -\sqrt[3]{\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}}}[/tex]
    and, of course, x= a- b.
     
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