What is the formula for solving third degree equations?

Question

The quadratic formula gives two correct solutions for any second degree equation. I'm told there is a similar, if more complex, method for solving any third degree equation. In other words, if I say Ax^3+Bx^2+Cx+D=0, can you give me the three solutions for x in terms of A, B, C, and D?

Answer 1

Tartaglia the stutterer came up with a general solution to

Ax^3 + Bx^2 + Cx + D = 0 by making a substitution that will always change the polynomial into the form

ax'3 + px' + q = 0. This is done by substituting x' = x - k and simplifying. You always get k from -b/3a.

Now, this rewritten form will always have solutions given by:
x' = u - v,
where
u = [ -q/2 + (q2/4 + p3/27)1/2]1/3
and
v = [ q/2 + (q2/4 + p3/27)1/2]1/3

Ugly, huh?

Answer 2

Solving Third Degree Equations

Answer 3

The proceedure is as follows:

First divide by A to get an equation of the form x^3 +b x^2 +c x+d=0.
Now let x=y -b/3. When you plug this in, expand out, and simplify, you get an equation of the form
y^3 +Ey +F=0.
Now choose u and v with
uv=(-E/3)^3 and
u+v=-F.
This leads to a quadratic equation in u (or v).
Then y=u^(1/3) +v^(1/3).
There are three complex cube roots of u and v and you have to select the cube roots so that u^(1/3)v^(1/3)=-E/3.
Now, x=y+b/3.

The problem with this method is that you have to take cube roots of complex numbers even if the original equation has three real roots. This truens out to be unavoidable.

Answer 4

Yes, there is a formula for finding at least one solution of a cubic equation. However, I do not have it with me (the only copy I had was an image). Trust me, it is VERY ugly and would take much more calculation than you are willing to do. Try searching for it if you want to find it. That's how I found it, and I don't remember it being particularly difficult to find. Good luck!

Answer 5

Calculate

[1a] ... p = (3AC-B^2) / (9A^2)
[1b] ... q = (2B^3 - 9ABC + 27A^2 D) / (27 A^3)

[Now y^3 + 3py + q = 0 with x = y - b/(3A).]

Calculate the discriminant

[2] ... d = q^2 + 4p^3

If d is positive, the equation has one real solution. Calculate

[3a] ... a1 = (-q + sqrt d) / 2
[3b] ... a2 = (-q - sqrt d) / 2
[3c] ... y = a1^(1/3) - a2^(1/3)
[3d] ... x = y - b/(3A)

This is the real solution; the imaginary solutions are

[4] ... y' = [1+sqrt(-3)]/2 a1^(1/3) - [1-sqrt(-3)]/2 a2^(1/3)

and its complex conjugate. (Use formula 3d to find the solutions of the original problem.)

If d is negative, there are three real solutions, which can be found using trigonometry.

Calculate

[5a] ... r = sqrt(-p^3)
[5b] ... t = inv cos (-q / 2r)

The solutions are

[6a] ... y1 = 2 * r^(1/3) * cos (t/3)
[6b] ... y2 = 2 * r^(1/3) * cos (t/3 + 120 deg)
[6c] ... y3 = 2 * r^(1/3) * cos (t/3 - 120 deg)

Again, you find the x values using

[6d] ... x1 = y1 - b/(3A)

and so forth.

Answer 6

Divide the polynomial by A and you will get

x^3 + b*x^2 + c*x + d = 0 ... (1)

where b=B/A, c=C/A, and d=D/A. I posted the solutions of (1) in:

http://www.megaupload.com/?d=N0JSAA8I

The expressions are very long, so you will have to zoom-in the PDF file to be able to read them.

I hope this helps.

Answer 7

What you are looking for is Tartaglia's solution of the general cubic equation. I don't see any point in reproducing it here. it uses a lot of subscripts and powers which are not easy for me to write in tyis format.

I got it from "College Algebra" by Raymond W. Brink published in 1933. It also has Ferrari's solution of the general quartic equation.
You will probably have to find an older algebra book. They don't do much of this kind of thing in college algebra any more.

Answer 8

You can take cube roots of complex numbers if you convert the complex number to polar form and back. The results will still be a nasty looking complex number.

Answer 9

you have to divide the equation by small factor to get a factor and a second degree equation.

do you know the remainder theorem?

you have to use that.