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

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In summary: 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 xYes, I know that there are few ways, but I need to solve it with Cardanos. Please help.
  • #1
Physicsissuef
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Homework Statement



Find the solutions of:

[tex]x^3-3x-2=0[/tex]

using the Cardano's method.

Homework Equations





The Attempt at a Solution



[tex]x=u+v[/tex]

[tex](u+v)^3-3(u+v)-2=0[/tex]

[tex]u^3+v^3+(u+v)(3uv-3)-2=0[/tex]

[tex]3uv-3=0[/tex]

[tex]uv=1[/tex]

[tex]u^3v^3=1[/tex]

[tex]u^3+v^3=2[/tex]

[tex]u^3=v^3=1[/tex]

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

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

[tex]u=v=1[/tex]

[tex]u=v=\frac{-1}{2}+i\frac{sqrt{3}}{2}[/tex]

[tex]u=v=\frac{-1}{2}-i\frac{sqrt{3}}{2}[/tex]

x=u+v

[tex]x_1=2[/tex]

[tex]x_2=-1+i\sqrt{3}[/tex]

[tex]x_3=-1-i\sqrt{3}[/tex]

I substitute above and something is wrong. Where is the error?
 
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  • #2
there may be a better way out..observinf that if [tex]f(x)=x^3-3x-2[/tex].
clearly [tex]f(-1)=0[/tex]
satisfies the eqn...hence
[tex](x+1)[/tex] is a factor of [tex]f(x)[/tex].
Now use division and get [tex]f(x)[/tex] in the form of factors and put the other factor(other than [tex]x+1[/tex])
to be zero and solve for
[tex]x[/tex]
 
  • #3
Yes, I know that there are few ways, but I need to solve it with Cardanos. Please help.
 
  • #4
Physicsissuef said:

Homework Statement



Find the solutions of:

[tex]x^3-3x-2=0[/tex]

using the Cardano's method.

Homework Equations





The Attempt at a Solution



[tex]x=u+v[/tex]

[tex](u+v)^3-3(u+v)-2=0[/tex]

[tex]u^3+v^3+(u+v)(3uv-3)-2=0[/tex]

[tex]3uv-3=0[/tex]

[tex]uv=1[/tex]

[tex]u^3v^3=1[/tex]

[tex]u^3+v^3=2[/tex]

[tex]u^3=v^3=1[/tex]
?? 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 [itex]u^3+ 1/u^3= 2[/itex]. 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 [tex]u=v=\sqrt[3]{1}[/tex].
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 [tex]z=\sqrt[3]{1}[/tex] (using complex numbers).

[tex]u=v=1[/tex]

[tex]u=v=\frac{-1}{2}+i\frac{sqrt{3}}{2}[/tex]

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

[tex]x_1=2[/tex]

[tex]x_2=-1+i\sqrt{3}[/tex]

[tex]x_3=-1-i\sqrt{3}[/tex]

I substitute above and something is wrong. Where is the error?
 
  • #5
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?
 
  • #6
Yes, basically, that is what you do.
 
  • #7
Ok, thanks. But why I read that also suming [itex]u_1+v_1=x_1[/tex], will work if I found [itex]v_1[/itex], by using complex algebra?
 

1. How does Cardano's method work for solving cubic equations?

Cardano's method involves using a formula to find the roots of a cubic equation. The formula is complex and involves finding the roots of a depressed cubic equation, which is a simplified version of the original equation. Once the roots of the depressed cubic are found, they can be used to find the roots of the original equation.

2. What are the three cases for solving a cubic equation with Cardano's method?

The three cases for solving a cubic equation with Cardano's method are when the equation has three real roots, when it has one real root and two complex conjugate roots, and when it has three real roots but one is a repeated root. Each case requires a different approach to solving the equation.

3. How accurate is Cardano's method for finding the roots of a cubic equation?

Cardano's method is a highly accurate method for finding the roots of a cubic equation. It is able to find the exact roots of the equation, rather than just approximations like other methods. However, the method can be time-consuming and complex, so it may not be the best choice for every situation.

4. Can Cardano's method be used to solve all cubic equations?

No, Cardano's method is not able to solve all cubic equations. There are certain types of cubic equations, such as those with irrational or non-real coefficients, that cannot be solved with this method. In these cases, other methods such as numerical approximation may be used.

5. Are there any limitations to using Cardano's method for solving cubic equations?

One limitation of Cardano's method is that it can only be used for cubic equations. It cannot be applied to equations of higher degrees. Additionally, the method can become very complex and difficult to use for equations with large coefficients, making it less practical in those situations.

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