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Cubic formula?

  1. Feb 9, 2008 #1
    The quadratic formula is, as you are well aware:

    [tex]ax^2+bx+c=0[/tex] [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

    What is the cubic formula?

    [tex]ax^3+bx^2+cx+d=0[/tex]

    Thanks.
     
  2. jcsd
  3. Feb 9, 2008 #2
    Cardano's formula will get you one root, but you have to transform the general equation a*x^3+b*x^2+cx+d=0 into e*z^3+f*z+g=0 by use of the substitution z=x-b/3. Mathworld explains a bit more: http://mathworld.wolfram.com/CubicFormula.html
     
  4. Feb 9, 2008 #3
  5. Feb 9, 2008 #4
    Thanks.
     
  6. Oct 17, 2010 #5
    could someone just show me with an example on how to work the cardano and Del ferro method from start to finish say using x^3-6x^2+2x-1
     
  7. Oct 17, 2010 #6

    HallsofIvy

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    The first thing you do is "reduce" the equation to the form [itex]x^3+ bx= c[/itex] without any [itex]x^2[/itex] term. To do that, let x= y- a. Then [itex]x^3= (y- a)^3= y^3- 3ay^2+ 3a^2y- a^3[/math] and [itex]x^2= (y- a)^2= y^2- 2ay+ a^2[/math]

    [tex]x^3- 6x^2+ 2x-1= y^3 -ay^2+ a^2y- a^3- 6y^2+ 12ay- 6a^2+ 2y- 2a- 1[/tex][tex]= y^3+ (-a- 6)y^2+ (a^2+ 2)y+ (-a^3- 2a- 1)[/math]

    That will have no "[itex]y^2[/itex]" term is a= -6 and, in that case, the polynomial is
    [itex]y^3+ 38y+ 229[/itex] and so our equation is [itex]y^3+ 38y+ 229= 0[/itex].

    Here's a quick review of Cardano's formula:
    [itex](a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3[/itex]
    [itex]-3ab(a+ b)= -3a^2b- 3ab^3[/itex]
    so that [itex](a+ b)^3+ 3ab(a+ b)= a^3+ b^3[/itex].

    In particular, if we let x= a+b, m= 3ab, and [itex]n= a^3+ b^3[/itex], [itex]x^3+ mx= n[/itex].

    Can we go the other way? That is, given m and n, can we find a and b and so find x?

    Yes, we can. From [itex]m= 3ab[/itex], we have [itex]b= m/(3a)[/itex] so [itex]a^3+ b^3= a^3+ m^3/(3^2a^3)= n[/itex]. Multiplying through by [itex]a^3[/itex], [itex](a^3)^2+ m^3/3^3= na^3[/itex] or [itex](a^3)^2- na^3+ m^3/3^3= 0[/itex].

    We can think of that as a quadratic equation in [itex]a^3[/itex] and solve it with the quadratic formula:
    [tex]a^3= \frac{n\pm\sqrt{n^2- 4\frac{m^3}{3^3}}}{2}[/itex][itex]= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}[/tex]
    and a is the cube root of that.

    From [itex]n= a^3+ b^3[/itex] we have
    [tex]b^3= n- a^3= \frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}[/tex].


    Now, in this problem, [itex]y^3+ 38y+ 229= 0[/itex] or [itex]y^3+ 38y= -229[/itex] so m= 38 and n= -229.
    [tex]\frac{n}{2}= -\frac{229}{2}[/tex]
    and
    [tex]\left(\frac{n}{2}\right)^2= \frac{52441}{4}[/itex]

    [tex]\frac{m}{3}= \frac{38}{3}[/tex]
    and
    [tex]\left(\frac{m}{3}\right)^3= \frac{54872}{27}[/tex]

    So
    [tex]a^3= -\frac{229}{2}\pm\sqrt{\frac{52441}{4}- \frac{54872}{27}}[/tex]
    and
    [tex]b^3= -\frac{229}{2}\mp\sqrt{\frac{52441}{4}- \frac{54872}{27}}[/tex]

    Calculate those numbers, take the cube roots to find a and b and then find x= a+ b.
    Each of those will have 3 cube roots but in the various ways of combining them, some things will cancel so that there will be, at most, 3 roots to the equation.
     
  8. Oct 17, 2010 #7
    thank yo for taking the time to show me this but im still confused, after plugging in y-a into the equation how does a= -6
     
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