Solving Cubic Equations with Cardano's Method: A Complete Guide

[Physicsissuef]

Homework Statement

Find the solutions of:

x^3-3x-2=0

using the Cardano's method.

Homework Equations

The Attempt at a Solution

x=u+v

(u+v)^3-3(u+v)-2=0

u^3+v^3+(u+v)(3uv-3)-2=0

3uv-3=0

uv=1

u^3v^3=1

u^3+v^3=2

u^3=v^3=1

Now u=v=\sqrt[3]{1}.

I found u and v with z=\sqrt[3]{1} (using complex numbers).

u=v=1

u=v=\frac{-1}{2}+i\frac{sqrt{3}}{2}

u=v=\frac{-1}{2}-i\frac{sqrt{3}}{2}

x=u+v

x_1=2

x_2=-1+i\sqrt{3}

x_3=-1-i\sqrt{3}

I substitute above and something is wrong. Where is the error?

 

[anantchowdhary]

there may be a better way out..observinf that if f(x)=x^3-3x-2.
clearly f(-1)=0
satisfies the eqn...hence
(x+1) is a factor of f(x).
Now use division and get f(x) in the form of factors and put the other factor(other than x+1)
to be zero and solve for
x

 

[Physicsissuef]

Yes, I know that there are few ways, but I need to solve it with Cardanos. Please help.

 

[HallsofIvy]

Homework Statement

Find the solutions of:

x^3-3x-2=0

using the Cardano's method.

Homework Equations

The Attempt at a Solution

x=u+v

(u+v)^3-3(u+v)-2=0

u^3+v^3+(u+v)(3uv-3)-2=0

3uv-3=0

uv=1

u^3v^3=1

u^3+v^3=2

u^3=v^3=1

?? I don't see how this follows directly, although it is true. Since you have defined u and v such that uv= 1, you have v= 1/u and so u^3+ 1/u^3= 2. Multiplying both sides by u³, you get the quadratic (in u³) (u³)²- 2(u³)+ 1= 0 which does give u³= 1 or u³= -1 as roots.

Now u=v=\sqrt[3]{1}.

No. This is your error. It does not follow from u³= v³ that u= v. Only that they are both cube roots of 3. And, as you note below, there are 3 of those.

I found u and v with z=\sqrt[3]{1} (using complex numbers).

u=v=1

u=v=\frac{-1}{2}+i\frac{sqrt{3}}{2}

u=v=\frac{-1}{2}-i\frac{sqrt{3}}{2}

You do NOT know, as I said, that u= v. You know, rather, that uv= 1 so v= 1/u.
If
u= \frac{-1+ i\sqrt{3}}{2}
then
v= \frac{2}{-1+ i\sqrt{3}}
rationalize the denominator by multiplying both numerator and denominator by -1+ i\sqrt{3} and you get
\frac{-1- i\sqrt{3}}{2}
the other non-real root of z³= 1.

Now, x= u+ v gives x= -1 which is correct. The three roots of x³- 3x- 2= 0 are 2, -1, -1.

x=u+v

x_1=2

x_2=-1+i\sqrt{3}

x_3=-1-i\sqrt{3}

I substitute above and something is wrong. Where is the error?

 

[Physicsissuef]

So, first I found the 3 solutions of u , and then substitute in uv=1, I get v and then substitute in x=u+v, right?

 

[HallsofIvy]

Yes, basically, that is what you do.

 

[Physicsissuef]

Ok, thanks. But why I read that also suming u_1+v_1=x_1[/tex], will work if I found v_1, by using complex algebra?