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Solving Cubic Functions

by TheDominis
Tags: cubic, functions, solving
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TheDominis
#1
Nov19-09, 01:15 PM
P: 2
x3 - 12x + 1 = 0

How does one solve for x?
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Mark44
#2
Nov19-09, 01:43 PM
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It's not a trivial process. See this Wikipedia article, http://en.wikipedia.org/wiki/Cubic_function
especially the Summary about halfway down the page.
HallsofIvy
#3
Nov20-09, 05:14 AM
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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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