How can I get help with solving cubic equations for my homework?

[mikee]

Homework Statement

I was just wondering if anyone could give me a quick lesson on solving cubic equations(Ax^3+Bx^2+Cx+D=0), or a good place to go to, if anyone can help thanks alot

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

here's the way I would argue it:

First, convert from the general cubic to the "reduced cubic" (without the x² term). If we replace x by y- u, then
$$Ax^3+ Bx^2+ Cx+ D= A(y-u)^3+ B(y- u)^2+ C(y- u)+ D= Ay^3- 3Auy^2+ 3Au^2y- Au^3+ By^2- 2Buy+ Bu^2+ Cy- Cu+ D$$
[tex]= Ay^3+ (-3Au+ B)y^2+ (3Au^2- 2Bu+ C)y+ (-Au^3+ Bu^2- Cu)+ D[/itex]
Choose u so that the coefficient of y² is 0: That is, choose u= B/3A so that the equation for y has no y² term. Solve that equation for y, and then x= y+ u.


Now to solve that reduced equation:
If a and b are any two numbers, then
$$(a+b)^3= a^3+ 3a^2b+ 3ab^2+ b^3$$
and
$$3ab(a+b)= 3a^2b+ 3ab^2$$
so
$$(a+ b)^3- 3ab(a+b)= a^3+ b^3$$
or, letting x= a+b, m= 3ab, and n=a³+ b³,
$$x^3- mx= n$$

Given any "reduced" equation Ax³+ Bx+ C= 0, We can divide through by A to get x³+ (B/A)x+ C= 0 or x³- (-B/A)x= -C. m= -B/A and n= -C/A.

Given m and n, can we solve for a and b, and so find x?

Of course, we can! From m= 3ab, b= m/3a. Replacing b by that in n= a³+ b³, n= a³+ m³/(3³a³).l

Multiply through by a³ to get na³= a⁶+ m³/3 <sup>which we can write as a quadratic equation for a3:
(a3)3- na3+ m3/33 and solve that with the quadratic formula:

$$a^3= \frac{n\pm\sqrt{n^2- 4m^2/3^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2-\left(\frac{m}{3}\right)^3}$$

Of course, once you have found a, b= m/3a and x= a+ b.

Using that is a heck of a lot of work which is why it is not normally taught in basic algebra!

Here's a website where they just give the formula:
http://www.math.vanderbilt.edu/~schectex/courses/cubic/</sup>

 

[mikee]

hmm that's a little bit more complicated than i thought but thank you