solving quartic equations using Ferrari's method

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tmoria

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Hi,


I created the quartic equation 36x4−72x3−391x2−123x+270=0\displaystyle 36x^4-72x^3-391x^2-123x+270=0


by multiplying out (3x−2)(2x+3)(3x+5)(2x−9)=0\displaystyle (3x-2)(2x+3)(3x+5)(2x-9)=0 so that I know the roots are x=23   x=−32   x=−53   and   x=92\displaystyle x=\dfrac{2}{3}\ \ \ x=-\dfrac{3}{2}\ \ \ x=-\dfrac{5}{3}\ \ \ and\ \ \ x=\dfrac{9}{2}


I then tried following Ferrari's method from http://en.wikipedia.org/wiki/Quartic_formula#Summary_of_Ferrari.27s_method to work from the quartic to get the roots. I'm apparently doing something wrong as I can get nowhere near a solution

I have A=36   B=−72   C=−391   D=−123   E=270\displaystyle A=36\ \ \ B=-72\ \ \ C=-391\ \ \ D=-123\ \ \ E=270

plugging these in I get . . .

α=CA−3B28A2=−44536\displaystyle \alpha=\dfrac{C}{A}-\dfrac{3B^2}{8A^2}=-\dfrac{445}{36}

β=B38A3−BC2A2+DA=−27518\displaystyle \beta=\dfrac{B^3}{8A^3}-\dfrac{BC}{2A^2}+\dfrac{D}{A}=-\dfrac{275}{18}

γ=CB216A3−3B4256A4−BD4A2+EA=269\displaystyle \gamma=\dfrac{CB^2}{16A^3}-\dfrac{3B^4}{256A^4}-\dfrac{BD}{4A^2}+\dfrac{E}{A}=\dfrac{26}{9}

P=−α212−γ=−24295315552\displaystyle P=-\dfrac{\alpha^2}{12}-\gamma=-\dfrac{242953}{15552}

Q=αγ3−β28−α3108=−1188727555038848\displaystyle Q=\dfrac{\alpha\gamma}{3}-\dfrac{\beta^2}{8}-\dfrac{\alpha^3}{108}=-\dfrac{118872755}{5038848}

There's nothing in the above that can cause any problems but the next stage . . .

R=−Q2±Q24+P327\displaystyle R=-\dfrac{Q}{2}\pm\sqrt{\dfrac{Q^2}{4}+\dfrac{P^3}{27}}

this produces a complex R (i.e. the square-root part is negative).

continuing on with the next three steps does not remove this complex element and so doesn't give any of the initial roots.

Am I doing this incorrectly or have I misunderstood the method ?

Many thanks.
 
Complex numbers may result along the way. Assumig your work is correct, It should be cancelled in the end step (where U is added to -P/(3U) and others).
 
Hi Daon, thanks for the reply.

I have checked and rechecked the working shown for errors and can find none.

The complex square-root does not get cancelled on any of the consequent steps.

I have even written a simple BASIC routine that does the calculating and it produces the same results I am getting.
 
tmoria said:
Hi Daon, thanks for the reply.

I have checked and rechecked the working shown for errors and can find none.

The complex square-root does not get cancelled on any of the consequent steps.

I have even written a simple BASIC routine that does the calculating and it produces the same results I am getting.
Click to expand...

I ran the numbers you have and have verfied they do cancel. Here is U - P/(3U)

http://www.wolframalpha.com/input/?...55/10077696+(77441+i)/(31104+sqrt(3)))^(1/3))
 
Last edited:
Solving Depressed Cubic

In your reference there is a history in which Gerolamo Cardano gives credit in the book Ars Magna (1545) to his servant, Lodovico Ferrari for the derivation of the Quartic function (above).


Earlist credit is due for Tartaglia's (1500) contribution, by deriving the depressed cubic formula used by Ferrari.


Let xxx = px + q describe the depressed cubic in x.
Then solutions for x are given by

\(\displaystyle \text{x = (}
\sqrt[3]{\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}{3}\right)^3}+\frac{q}{2}}
\text{ ) + ( }
\sqrt[3]{\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}{3}\right)^3}}
)\)

This can lead to some very simple surprising results. As you have noticed, expressions that seem to be irrational and even complex, may in-fact be rational.
 
Golden Example

It is easy to find two irrational numbers (a and b) that sum to a rational.
You have shown that the cube roots of those two irrational numbers (a and b) can also sum to a rational number. This is only true if (a and b) are carefully chosen!!!!

My favorite is ...

Link: http://math2.org/mmb/thread/44154
 
Welcome tmoria

Denis said:
3 kisses for lookagain? :confused:
Click to expand...

Hey Dennis,

meet tmoria. He is new to FMH. I have read his posts, so far, and I am impressed with his curiosity.

Hi tmoria,

I added the "Golden" post (above) because I read your posts and thought you might like it. FMH has a high volume of posts to get homework help -- that's our purpose. There is a relatively small community of those who provide help. I hope you continue in that role too. Feel free to add your 2 cents to any thread, if you feel that it adds to the benefit of the OP (original poster). We are all here to do what we can to benefit the OP. You are the OP on this thread. Denis referred to "lookagain" who is an Elite Member like Dennis. "lookagain" is very good at catching details and keeping us honest -- we love him dearly.

There is also a "Math Odds & Ends" Forum Category for a less formal exchange.
 
Last edited:
Thanks again Bob for your help. Can I ask what software you used for outputting the solution ?
 
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