Cardano's method of solving cubics

[mathmaniac1]

Solve using cardan's method.

\(\displaystyle x^3-13x+12=0\)

\(\displaystyle x=v+u\)

\(\displaystyle 3uv=-p=13\)

\(\displaystyle v^3+u^3=-q=-12\)

\(\displaystyle 27v^6+324v^3-13=0\)

\(\displaystyle v^3=\frac{-324\pm\sqrt{324^2-27*4*13}}{54}\)

Please solve for x.I know I am asking for too much,but seems like I am not able to get the desired answers even though nothing seems wrong with the method.

I would be so grateful towards anybody who helps.

For the moderator: please change * to x,I don't know how to do that.
Thanks

 

[polygamma]

Cardano's formula states that $x = u+v$ is a solution to $x^{3} +px+q =0$ where $u = \Big(-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3} \right)^{3}} \Big)^{\frac{1}{3}}$

and $v = \Big(-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3} \right)^{3}} \Big)^{\frac{1}{3}}$.

So for $x^{3}-13x+12 = 0$ we have $u = \Big(-6 + \sqrt{36 -\frac{2917}{27}} \Big)^{\frac{1}{3}} = \Big(-6 + \sqrt{-\frac{1225}{27}} \Big)^{\frac{1}{3}}$

$= \Big(-6 + \frac{35 i}{3 \sqrt{3}} \Big)^{\frac{1}{3}}$

which according to Wolfram Alpha equals $\Big( \frac{1}{216} \left( 9 + 5 \sqrt{3}i \right)^{3} \Big)^{\frac{1}{3}} = \frac{1}{6} \left(9 + 5 \sqrt{3} i \right)$

And $v$ is the complex conjugate of $u$, that is $\frac{1}{6} \left(9 - 5 \sqrt{3} i \right)$.

So $ x = u+v = \frac{18}{6} = 3$

 

[mathmaniac1]

Cardano's formula states that $x = u+v$ is a solution to $x^{3} +px+q =0$ where $u = \Big(-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3} \right)^{3}} \Big)^{\frac{1}{3}}$

and $v = \Big(-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^{2} + \left(\frac{p}{3} \right)^{3}} \Big)^{\frac{1}{3}}$.

But if I do it the other way,i.e,thinking that v has 6 values,won't I get 6(v+u) s or is 3 of them surely going to repeat

So for $x^{3}-13x+12 = 0$ we have $u = \Big(-6 + \sqrt{36 -\frac{2917}{27}} \Big)^{\frac{1}{3}} = \Big(-6 + \sqrt{-\frac{1225}{27}} \Big)^{\frac{1}{3}}$

$= \Big(-6 + \frac{35 i}{3 \sqrt{3}} \Big)^{\frac{1}{3}}$

which according to Wolfram Alpha equals $\Big( \frac{1}{216} \left( 9 + 5 \sqrt{3}i \right)^{3} \Big)^{\frac{1}{3}} = \frac{1}{6} \left(9 + 5 \sqrt{3} i \right)$

Is it the only possible cube root?

And $v$ is the complex conjugate of $u$, that is $\frac{1}{6} \left(9 - 5 \sqrt{3} i \right)$.

How can you be sure?

So $ x = u+v = \frac{18}{6} =
3$

What does wolframalpha say about the other cube roots?...and what about the other roots?please show me how you would those too...

Thanks a lot,but I am wondering what happened when I did it.Why is my discriminant not negative like yours?Why don't I get anyone of the desired roots?

 

[chisigma]

It is useful to remember that, given the values of u and v... $$ u= \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}\ v= \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}}\ (1)$$

... if $r_{0}= 1$, $r_{1} = e^{i\ \frac{2}{3}\ \pi}$ and $r_{2} = e^{i\ \frac{4}{3}\ \pi}$ are the 'cubic roots' of 1, then the roots of the equation $x^{3} + p\ x + q=0$ are given by...

$$x_{0}= r_{0}\ u + r_{0}\ v\ ;\ x_{1}= r_{1}\ u + r_{2}\ v\ ;\ x_{3}= r_{2}\ u + r_{1}\ v\ (2)$$

With (1), (2) and a bit of patience You can find that the three roots of $x^{3} - 13\ x + 12 = 0$ are $x_{0}=3$, $x_{1}= 1$ and $x_{2}=-4$... Kind regards $\chi$ $\sigma$

 

[CellCoree]

for your question and interest in Cardano's method for solving cubic equations. As a scientist, my response would be to provide you with the steps to solve the given equation using Cardano's method. Please note that this method can be complex and may require some algebraic manipulation. I will try to simplify the steps as much as possible.

Step 1: Rewrite the equation in the form x^3 + px + q = 0. In this case, p = -13 and q = 12.

Step 2: Let x = u + v, where u and v are unknown variables.

Step 3: Substitute x = u + v into the original equation. This will give you a new equation in terms of u and v.

(u + v)^3 - 13(u + v) + 12 = 0

Step 4: Expand the left side of the equation using the binomial theorem.

u^3 + 3u^2v + 3uv^2 + v^3 - 13u - 13v + 12 = 0

Step 5: Group the terms with u and v together.

(u^3 + 3uv^2 - 13u) + (3u^2v + v^3 - 13v) + 12 = 0

Step 6: Equate each group to zero.

u^3 + 3uv^2 - 13u = 0

3u^2v + v^3 - 13v = 0

Step 7: Solve for v in the first equation.

v = -u^2/3u + 13/3

Step 8: Substitute this value of v into the second equation.

3u^2(-u^2/3u + 13/3) + (-u^2/3u + 13/3)^3 - 13(-u^2/3u + 13/3) = 0

Step 9: Simplify the equation to get a new equation in terms of u.

27u^6 - 13u^2 + 12 = 0

Step 10: Use the quadratic formula to solve for u^2.

u^2 = (13±√(13^2 - 4*27*12))/(2*27)

u^2 = (13±√(169 - 1296