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Cardano's method of solving cubics

mathmaniac

Active member
Mar 4, 2013
188
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
 
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Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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$
 
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mathmaniac

Active member
Mar 4, 2013
188
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 gonna 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 any one of the desired roots?
 
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chisigma

Well-known member
Feb 13, 2012
1,704
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


$\chi$ $\sigma$
 
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