Talk:Quartic equation

Can somebody explain me, how in Quartic_equation#Reduction_to_a_biquadradic the formula $y^{4}...$ is created, please? Somehow I do not see, if it is substitution, and if yes, what by what. Thx. --Riddick 13:11, 6 Oct 2004 (UTC)

You can compute it by using a resultant, and eliminating x between

x^{4}+cx^{2}+dx+e=0

and y - x² - px - q = 0, where p and q are as given. User:Gene Ward Smith

Is it possible to predict which combination of solution for the long and short y-equations leads to a valid solution for equation (1) (in Quartic_equation#Reduction_to_a_biquadradic)? Thx. --Riddick 16:14, 6 Oct 2004 (UTC)

It would be if instead we used y = (ax+b)/(cx+d) but I couldn't get that to work, though it seems it should. User:Gene Ward Smith

I was just wondering how in the summary, x (the solution) gives only two values when we are solving a quartic which should have four solutions. Thanks! ([email protected])

The signs were a little bit tricky, but I think, I fixed it... --Riddick 13:11, 6 Oct 2004 (UTC)

A certain Arne noted on Wikipedia:Reference desk that there seems to be a mistake in the solution given in the article. I had a short look and I think he is right. The formula for v, which solves equation (5), seems suspect. -- Jitse Niesen 16:08, 28 Sep 2004 (UTC)

It was me (Arne) :-)) --Riddick 16:42, 28 Sep 2004 (UTC)

Now it looks to me like the formula for v (after equation (5)) is correct for real v (I tried it with some example numbers).

But: If v is a complex number, then the formula for v does not bring values that solve (5). --Riddick 18:25, 28 Sep 2004 (UTC)

With the coefficients A=1, B=0, C=6, D=-60 and E=36 the formula for x delivers just two solutions x1=~3.100 and x2=~0.6444 (if I toggle the +/- sign; but indeed toggling the sign of the first square root (where no +/- sign can be found), results in invalid values for x) with v=9.0097912 (which is approximately a real root of (5) but not the result of the given formula for v). The remaining two solutions can be found by the procedure in the last section. But that does not look so smooth... (I strongly believe, that is because v is a real number in this example; maybe I use the wrong calculator). When I use the formulas from the summary I get (with + instead of +/-) x=~2.275-~1.347i and v=~6.304-~1.562i (both values are no roots of their corresponding equations)

on Quartic Equation -- from MathWorld I have found another description of Ferrari's solution (I just tested it, and its works fine (with coefficients from Ferrari's first one): x1=~3.100, x2=~0.6444, x3=~-1.872+~3.810i and x4=~-1.872-~3.810i=x3* with y1=~20.019582457; it sounds a little bit different at some lines (e. g. they demand a real root of the cubic equation)). --Riddick 21:47, 28 Sep 2004 (UTC)

Should I change something, although I do not know, where the mistake is? --Riddick 21:47, 28 Sep 2004 (UTC)

Arne says: Yes. --Riddick 20:38, 1 Oct 2004 (UTC)

I took two formulae from Cubic equation:

u={\sqrt[{3}]{{q \over 2}\pm {\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}

and

v={p \over 3u}

and inserted them at the right(?) place in this article. --Riddick 20:00, 1 Oct 2004 (UTC)

experimental results: the sign patterns +++ = x1, -+- = x2, --- = x3, +-+ = x4 are good sign patterns (the others are not)

Hereby I recommend further tests and/or proves. Thx. --Riddick 20:38, 1 Oct 2004 (UTC)

A=1, B=0, C=6, D=-60, E=36: check
A=1, B=10, C=-6, D=60, E=36: check
A=1, B=-28, C=294, D=-1372, E=2401: check (remark: (x-7)^4 (it is a little bit tricky due to many zeros as inverse (mult)))

It is too long: may be is adapt for a math book not an encyclopedia

Perhaps this article would be better in a mathematical book (wikibook?). It is long, complex and difficult to read. It would be better if the article was shorter. In my opinion, all the demostration should be summarize up. A link to the full explanation of the methods could be given instead. Historic and modern view of the problem could be expand instead. AnyFile 16:57, 24 Oct 2004 (UTC)