The Quartic Equation and its Solutions.

[Anonymous No.7613239]

In this thread, I'm going to solve the quartic.

It is hoped that a few other people will find this thread interesting. Some of you may find it tedious, or dull. If you're in the latter camp, and are considering writing something rude, or off-topic ... just hide the thread, instead!

[Anonymous No.7613269]

>>7613239
I'm waiting

[Anonymous No.7613276]

First, let's say a bit about WHAT'S (algebra) going to be done, and WHY (motivation), not in that order.

[[[[[ -WHY? ]]]]]

I've recently appreciated the general algebraic solution of the cubic equation, having worked through it myself for my first time. The next-highest polynomial (and last) for which a /general/ algebraic solution exists, is the quartic - Abel, Ruffini, and Galois all variously showed that higher-degree /general/ solutions of univariate polynomials do not exist. So in principle, once I'm done with this thread, I can put one small area of mathematics (of personal interest) to bed, where my own understanding is concerned. The principal motivations for this thread are therefore 1) "because it's there", and 2) "there's an end in sight" - I don't care that it's old, I never knew it before, and writing a thread will help me to internalize the general idea of the derivation, which since that's also something I want, is another motivator.

More interestingly, I think this will help prepare me to learn a little Galois theory and related topics a bit later.

[[[[[ -WHAT? ]]]]]

In order to answer this question, we first need to get some terminology straight.

-a MONIC polynomial has a leading coefficient of one.
-a POLYNOMIAL EQUATION is an equation such that one side is a polynomial, and the other is zero; i.e. the polynomial is set equal to zero, e.g. ax^2 + bx + c = 0 is a POLYNOMIAL EQUATION.
-a UNIVARIATE POLYNOMIAL is one (a polynomial) which only involves one variable; x^2 - 5x is a UNIVARIATE POLYNOMIAL, yx + x - 2y is NOT, where both x and y are variables.

Finally...

-an ALGEBRAIC SOLUTION of a POLYNOMIAL EQUATION is an expression (or family of expressions) which expresses the roots of that polynomial equation in terms of its coefficients, finitely applying only the following SIX arithmetic operations: addition, multiplication, subtraction, division, integer exponentiation, integer root extraction (square root, third root, etc).

[Anonymous No.7613307]

It is the above specific notion of GENERAL ALGEBRAIC SOLUTION to UNIVARIATE POLYNOMIAL EQUATIONS, then, to which the Abel-Ruffini "nope, can't do it for higher-than-a-quintic" theorem applies, THESE such solutions are what may not be found for degree five or higher.

The algebraic solution is also the specific notion which may be applied to the quartic. Thus a PROPOSITION, a little HOW, and then I will get crunching, with a view toward clarity of exposition for any who may care to read, and also with clarification of steps, and also pointing out the bits where I have some uneasiness.

PROPOSITION:

There exists an algebraic solution of the polynomial equation

[math] \displaystyle ax^4 + bx^3 + cx^2 + dx +e = 0 [/math] .

[[[[[ HOW ]]]]]

Much of the rest of this thread will be spent aping FERRARI'S derivation, specifically its treatment at Wikipedia under the "quartic function" article, but at a bit more meandering pace for detail.

I will also hold that the coefficients a,b,c,d,e are themselves real for the moment (and consequently do the HS algebra in view of this); most of the arithmetic where the coefficients are themselves complex carries over, but I haven't closed this loop and therefore I restrict myself to the real-coefficient case for the time being (I stopped short of this part in my derivation of the cubic as well). Of course, the variable x may be complex.

However, solving the quartic by means of elementary algebra really requires "all hands on deck": you need the quadratic and cubic formulas in addition to the other tricks. Therefore I take my first detour: Let's solve a linear equation in a strange fashion.

[Anonymous No.7613415]

WARNING: this is a sidebar. Don't confuse the a's, x's etc outside of their respective contexts.

Dig this. ax + b = 0 is a polynomial equation of degree one. Rather than just solving for x, let's take the long way and make a substitution:

[math] \displaystyle ax+b = 0 \rightarrow x + \frac{b}{a} = 0 \;\;\; ; \;\;\; x = t - \frac{b}{1a} \rightarrow t - \frac{b}{a} + \frac{b}{a} = 0 \rightarrow t = 0 \;\;\; ; \;\;\; x = - \frac{b}{a} [/math]

We made a substitution of the form

[math] \displaystyle x = t - \frac{b}{na} [/math]

where n is the degree of the (univariate) polynomial (equation). Now dig this, following (again) division by leading coefficient, it works a bit less trivially on degree two, an alternative to completing the square (following root extraction and back-substitution):

[math] \displaystyle x^2 + \frac{b}{a} x + \frac{c}{a} = 0 \;\;\; ; \;\;\; x = t - \frac{b}{2a} \rightarrow x = \pm \frac{ \sqrt{b^2 - 4ac} }{2a} - \frac{b}{2a} [/math]

And when deriving the cubic, the next substitution of the form x = t - b/3a may be made, where a and b denote in that context again the leading and next-down coefficient, where that problem is concerned. And, as you might expect (and is actually the case), you can start out the derivation of the /quartic/ with the above method, which is a special case of a /Tschirnhaus transformation/, whose exact context you are free to check out, but I am not interested to comment further on at this time. The idea of such transformations is to simplify polynomials by substitution.

For our purposes, the above special case of a Tschirnhaus transformation, when applied to a univariate polynomial equation, always returns a MONIC, DEPRESSED (univariate) polynomial equation. Depressed just means that the next-down term's coefficient is zero (and vanishes). These are easier to manipulate - and as we've seen, the concept can be ret-conned to the near-trivial examples where algebraic solutions exist.

[Anonymous No.7613454]

This will be a simple comment on proof strategy, or how I went about deriving (aping, re-discovering for myself) the cubic formula. My method deviates from Cardano's (per wiki with help from John Derbyshire), and simply invokes third roots of unity and FTA toward that end, to simplify things:

-divide by leading coefficient
-Tschirnhaus transformation (x = t - 3a); depressed cubic w/ new coeff's
-establish feasibility of some "w"
-Viete's substituion (t = w - 3w)
-usage of Quad. Form. and extraction of third root
-bookkeeping (name things)
-(The good bit): Observation of w+w-bar SOLUTION of DEPRESSED cubic, invocation of complex third roots of unity
-Invocation of FTA etc to establish equivalent "Cardano" form.

This has been simply a personal and relevant note, since the cubic formula is implied in all this - a "resolvent" cubic is necessary to perform this thread's objective, per the Ferrari solution, and one will appear in due course.

Enough, let's go to work.

[Anonymous No.7613557]

First, we reconsider the univariate quartic polynomial equation

[math] \displaystyle ax^4 + bx^3 + cx^2 + dx +e = 0 [math]

and we turn this into a MONIC univariate quartic polynomial equation by division of the (necessarily? Oh yes, I forget to mention the obvious: the leading coefficient is necessarily nonzero, whether real or complex) leading coefficient a.

(division)

Then, we perform the appropriate Tschirnhaus transformation: x = t - b/4a .

(substitution of variable)

Then, we rearrange,

(+-x/):

[math] \displaystyle t^4 + pt^2 + qt + r = 0 [/math] .

This is the DEPRESSED quartic. It is, all at once, a MONIC and a DEPRESSED polynomial equation (depressed meaning that the next-down degree term, three in this case, has a coefficient of zero), equivalent to the original, in the sense that it has the same solution set as the original, and will eventually be amenable to back-substitution by the above (real) line- (complex) plane translation. All of p, q, r, are expressed in /algebraic/ terms of a,b,c,d,e (see wiki).

Okay, fine. So?

So the middle bit is more (+-x/)-tier arithmetic manipulation. The thrust is:

-introduce a perfect-square term,
-introduce a free variable,
-make quadratic formula-tier observations,
-specify a "resolvent cubic",
-use the quadratic formula-tier observations to restrict the new free variable,
-rearrange, rearrange, rearrange.

That was confusing. Let's do our best to unconfuse it. The arithmetic itself is quite simple.

[Anonymous No.7613571]

>>7613307
Ganbatte OP!

I have a question: why do you think the Cardano method may not work for [math]t^3 + pt + q = 0[/math] when p,q are complex?
The only possible stumbling block I can see is the Viete substitution (t = w - p/(3w)) breaking down when w = 0, but this only happens when p = 0 and it should be easy to catch that exception.

For reference (sadly I don't have access to the resources referenced there):
https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots
http://math.stackexchange.com/questions/15865/why-not-write-the-solutions-of-a-cubic-this-way
http://math.stackexchange.com/questions/18867/how-does-one-solve-a-cubic-polynomial-with-complex-coefficients

[Anonymous No.7613580]

First, we reconsider the univariate quartic polynomial equation

[math] ax^4 + bx^3 +cx^2 + dx + e = 0 [/math]

and we turn this into a MONIC univariate quartic polynomial equation by division of the (necessarily? Oh yes, I forget to mention the obvious: the leading coefficient is necessarily nonzero, whether real or complex) leading coefficient a.

(division)

Then, we perform the appropriate Tschirnhaus transformation:

[math] x = t − b/4a [/math] .

(substitution of variable)

Then, we rearrange,

(+−x/):

[math] t^4 + pt^2 + qt + r = 0 [/math] .

This is the DEPRESSED quartic. It is, all at once, a MONIC and a DEPRESSED polynomial equation (depressed meaning that the next-down degree term, three in this case, has a coefficient of zero), equivalent to the original, in the sense that it has the same solution set as the original, and will eventually be amenable to back-substitution by the above (real) line- (complex) plane translation. All of p, q, r, are expressed in /algebraic/ terms of a,b,c,d,e (see wiki).

Okay, fine. So?

So the middle bit is more (+-x/)-tier arithmetic manipulation, getting into the tall weeds. The thrust is:

-introduce a perfect-square term,
-introduce a free variable,
-make quadratic formula-tier observations,
-specify a "resolvent cubic",
-use the quadratic formula-tier observations to restrict the new free variable,
-rearrange, rearrange, rearrange.

That was confusing. Let's do our best to unconfuse it. The arithmetic itself is simple.

[Anonymous No.7613592]

The best thread on /Sci/ right now.

[Anonymous No.7613613]

>>7613571

Thank you very much for your kind encouragement, as well as your careful reading of my concerns up to this point.

In order to learn both the "cubic and quartic", I have mostly followed wiki with a few modifications. My previous /sci/ engagement with the cubic is now archived here (and which I meant to re-introduce into this thread eventually, pertinent response to encouragement is as good an occasion as any):

>>/sci/thread/7529602

The formulae don't display nicely there anymore, but the root of my concern was reading language in the wiki which suggested that there is an accounting issue in transition from real to complex coefficients; the same phrase is present in the very first link which you have so intelligently provided: "However, this formula is applicable without further explanation /only/ when a, b, c, d are real numbers"...

Basically I haven't taken the trouble to re-check just everything needed about complex numbers, day one (preferring this tedium instead). That said, I was moving toward a conclusion of "meh, it's all the same" by the end of the above thread, but I stopped short of a proof of that point, having gotten done what I wanted with the thread at that time.

[Anonymous No.7613636]

>>7613239
Solve Quintic instead pls.

and before you go HURR DURR IT'S IMPOSSIBLE, it's not.

it's only impossible if you only use real/complex numbers.

[Anonymous No.7613640]

>>7613592

Since this nice person is also encouraging me (wew lad), let's start the middle-grind segment.

Consider both (t^2 + p)^2 and its simple three-term expansion. Add the former to the negative latter, and it's zero. Now take this zero, and add this zero to both sides of our above depressed cubic. Then shove the other stuff over to the one side and you get (+-x/)

[math] \displaystyle (t^2 + p)^2 = pt^2 - qt + p^2 - r [/math]

NOW A REALLY IMPORTANT, INNOVATIVE BIT: invoke a brand-new variable, which we shall call y. The idea is to get the above LHS equal to some new perfect square, and the RHS equal to a quadratic. What happens is, the quantity 2t^2 y + 2py + y^2 is added to both sides. The result following (+-x/) rearrangement is the equality of an (algebraic, this bit confused me at first) perfect square, and a quadratic in t:

[math] \displaystyle (t^2 + y + p)^2 = (2y + p)t^2 - qt + (y^2 + 2py + p^2 -r) [/math]

aaand now were are truly in the hairy middle bit. y can be anything at all, it hasn't affected the above, yet. Per the strategy we shall place a restriction on y, and other stuff. I feel a sidebar coming on.

[Anonymous No.7613641]

>>7613636
>What's the point if you don't use real/complex numbers?
>Abel-Ruffini theorem
>Do you just come on this board to make yourself look smart? Let the man work his equation.

[Anonymous No.7613642]

>>7613641
>he doesn't know what UltraRadicals are
fuck off retard.

[Anonymous No.7613645]

>>7613640
I love this thread, is there a reason your TeX equations don't display properly? Can you see the equation, because all I see is the raw text?

[Anonymous No.7613646]

>>7613642
I know what they are, just fuck off this thread. Who the fuck uses fifth-order polynomials in real math anyway?

[Anonymous No.7613649]

>>7613646
>Who the fuck uses fifth-order polynomials in real math anyway?
Anything involving space orbits. And you call me the retard.

[Anonymous No.7613661]

>>7613649
I said real math, and I never called you a retard, retard.

[Anonymous No.7613665]

>>7613649
>Muh elliptic integrals
>Space orbits
How long did you read Wikipedia before you came up with that answer?

[Anonymous No.7613677]

i like the way he is developing the solution, reminiscent of the way i learnt to solve cubics: get rid of 2nd degree term, substitute, solve quadratic. this is just a larger manipulation

[Anonymous No.7613682]

This is legitimately the best thread on /sci/ in years.

The only thread I can think of that topos this was the one where anons calculated the speed of a car from a picture of it crashed into a barn

[Anonymous No.7613687]

>>7613682
I wish I was around for that one. I think I was, but I missed it.

[Anonymous No.7613718]

>>7613645

ayy thx person, I dunno what device you're using. /sci/ has recently made modifications to its LaTeX implementation since new-moot took office, I noticed that days ago.

---

Sidebar (again, don't confuse these a's, b's, etc with the rest): the next step requires an appreciation that such-and-such (a quadratic discriminant) is necessarily equal to zero; thus we provide an argument which goes like this:

"a (quadratic) discriminant is equal to zero IF AND ONLY IF the quadratic polynomial (equation? LHS) is expressible as an algebraic perfect square"

THIS IS ONE PLACE WHERE I REACH OUT FOR HALP. FOR SOME ODD REASON I AM A BIT UNEASY ABOUT THIS, BUT I PROCEED:

Forward: plug and chug, it involves radicals, okay, fine.

REVERSE: (more interesting)

Suppose that a generic quadratic equation's LHS, ax^2 + bx + c (= 0) may be rewritten as a perfect square of the form (α x + ß)^2 (= 0). Expansion of same requires that alpha squared is equal to a, and so on, with the effect that the coefficients have restrictions, meaning the discriminant is zero per the quadratic formula. Thus

[math] \displaystyle x = \frac{ -2 \alpha \beta \pm \sqrt{ 4 \alpha^2 \beta^2 - 4 \alpha^2 \beta^2 } }{2 \alpha^2} [/math]

I just haven't thought through details here, but I think the argument is right.

[Anonymous No.7613735]

>>7613718
Couldn't you just complete the square?

Assuming a != 0
[eqn]ax^2 + bx + c = a \left( x + \frac{b}{2a} \right)^2 + c - \frac{b^2}{4a}[/eqn]

Though there's a potential issue if [math]a<0[/math], depending on your stance on complex numbers..

[Anonymous No.7613748]

>>7613718

So what's the point of this? To buttress an observation about quadratics, and to MAKE AN OBSERVATION ABOUT (the latter equation in) >>7613640 :

"If the quadratic RHS of latter >>7613640 is to be expressible as a perfect square, then we require that its involved discriminant is equal to ZERO".

This is the first extra-hairy detour on this thing. We just set it,

[math] \displaystyle (-q)^2 - 4(2y + p)(y^2 + 2py + p^2 - r) = 0 [/math]

and a re-arrangement (+-x/) (just multiply out and get the y's alone!) gives a MONIC CUBIC POLYNOMIAL EQUATION IN y: ..!

[math] \displaystyle y^3 + \frac{5}{2} py^2 + (2p^2 - r)y + \bigg( \frac{1}{2} p^3 - \frac{1}{2} pr - \frac{1}{8} q^2 \bigg) = 0 [/math]

which per wiki is (a) RESOLVENT CUBIC, though I couldn't verify this earlier. But that's beside the point.

The point here, is that when the free variable y satisfies the above, then it will serve its purpose in our derivation. BUT! Those old Italians, and I, HAVE ALREADY SEEN THIS MOVIE! And here is the true joy of re-discovery: the stuff we developed before will eventually have a payoff, and now we can actually see it, if we have been fortunate enough to derive the cubic. But, some more unsexy grinding has to happen next.

[Anonymous No.7613783]

>>7613665
about 10 minutes.

[Anonymous No.7613808]

>>7613748

Use the DISCRIMINANT-FORM (former-this-post), to observe THIS:

[math] \displaystyle \frac{q^2}{4(2y + p)} = y^2 + 2py + p^2 - r [/math]

What's the point (again)? First, just replace the latter coefficient in (latter RHS) >>7613640 , per the above. But now, the RHS of that form is amenable to re-arrangement /as a perfect square/! It becomes this (+-x/):

[math] \displaystyle (t^2 + y + p)^2 = \bigg( \sqrt{2y + p} \;\; t - \frac{q}{2 \sqrt{2y + p} } \bigg)^2 [/math]

I pause here to consider that a different tactic may involve square root extraction of both sides at around this point (the seemingly obvious thing), something I don't recall having actually done thus far, which once you do it, opens the other complex of worms. Cursory searching of quartic derivations suggests these variants. Here, I will instead hew to wiki's treatment of the Ferrari derivation, which doesn't do this right-now: The next bit is an arithmetic re-arrangement of the previous equation as a sort of psuedo-pair of conjugates.

Then you use zero-cancellation to get at four distinct possibilities per the quadratic formula again, express those, and back-substitute. But I have not exhausted my interest in this by a damn sight, nor have I accounted for the special cases and other caveats.

[Anonymous No.7614599]

>>7613808

Up until this point, we have only required division by the (necessarily non-zero) a. To go down this part, we shall require that 2y + p =/= 0, as we will use division by (square root) of same.

We are going to re-arrange, again (and also back up a bit). Here is the rearrangement (+-x/) :

[math] \displaystyle \bigg( t^2 + y + p + \sqrt{2y + p} \;\; t - \frac{q}{2 \sqrt{2y+p} } \bigg) \bigg( t^2 + y + p - \sqrt{2y + p} \;\; t + \frac{q}{2 \sqrt{2y+p} } \bigg) = 0 [/math]

This is a rearrangement of both of the equivalent equations latter >>7613808 , and particularly of the earlier equivalent latter >>7613640 . I remember finding this presentation at this point odd at first, especially next to those very similar perfect square forms we've just shown, but if the reader is curious to verify, just do this: expand out latter >>7613640 , and set it to one side. Then actually carry out the expansion of this post's equation; there are 25 terms, but what ends up happening is that 12 of the nastier ones cancel immediately, and half of what's left groups up such that the rearrangement is verified (compare with what you've just set aside).

When is this equation true? 1) when one of the factors is equal to zero; 2) when the other factor is equal to zero; or 3) when both are zero. We have thus factored into two quadratic equations in t, each with their own quadratic formula. And, since each of those implies its own two roots, once we marry them up to produce an expression giving four algebraically distinct roots of our fourth-degree polynomial, we will then shout "FTA lol!" and call it done (but then I will spend the rest of the thread going back over details and playing with things).

[Anonymous No.7614713]

This leads to the four algebraically distinct expressions of t, as promised We simply take the two resultant and virtually identical quadratic formulae, and rewrite them as a single expression of t (accounting for the various +- business):

[math] \displaystyle t = \frac{ \pm_{1} \sqrt{ 2y+p } \pm_{2} \sqrt{ -(2y + 3p \pm_{1} \frac{q}{2 \sqrt{2y+p } } ) } }{2} [/math]

Here, the discriminant was cleaned up slightly in the interim, that negative is held outside the discriminant because we require both pm_1's to have the same sign, whichever it may be, in this version of the expression (and holding the negative outside preserves this information); the pm_2, meanwhile, with respect to the pm_1, need not be of same or even opposite parity (as such situations may be vaguely familiar to the reader in the case of trig angle formulae); rather, the expression encapsulates four cases: ++, +-, -+, -- in the appropriate spots. These are precisely our four roots of the equivalent /depressed quartic/, and back-substitution to our original variable of x gives

[math] \displaystyle x = \frac{ \pm_{1} \sqrt{ 2y+p } \pm_{2} \sqrt{ -(2y + 3p \pm_{1} \frac{q}{2 \sqrt{2y+p } } ) } }{2} - \frac{b}{4a} [/math]

FTA, LOL!

Well, not quite. We have the (simpler) case 2y+p = 0 left to investigate. Another confusion that I originally had around this point, went like this: "Wait, I thought that cubic from earlier had three roots, so y is involved here... we have ... seven roots? Twelve?" No; all that we required of the free variable y was /that it satisfy our resolvent cubic/. The expectation, then, is that you can pick any such root of the cubic, plug into the above, and the /same/ four roots of our original quartic will be given, (I bet they get permuted).

And since I'm in the neighborhood, I also want to go back over the definition of a resolvent cubic, /actually test-drive the above/ with a few different items, and mention a result I got about the above Tschirnhaus transforms...

[Anonymous No.7615264]

>>7614599

Alternatively, the factors in this post represent the two quadratic situations, had we done root extraction with latter >>7613808, (this would entail another pm one one side; attempting to square either situation (or factor) by itself to get at our four roots would have resulted in a loss of information, and an inequality with the above quadratic form (the nasty terms that I said vanish when we use the conjugate forms, stay stubbornly put in such a mis-step).

The point being that the comparison of perfect squares, when the square root is taken (and two pm situations arise) lead to the same factors, and ultimately the same roots of the original quartic.

[Anonymous No.7615315]

In the other case where 2y + p = 0, this forces us to conclude that q = 0 as well, per the discriminant form, such that the depressed cubic is a monic quadratic in t^2. ; in this remaining simple case, our expression-of-four roots leads to a significantly different expression, having the form

[math] \displaystyle x = \pm_{1} \sqrt{ \frac{ -p \pm_{2} \sqrt{ p^2 - 4r } }{2} } - \frac{b}{4a} [/math]

Where again, the pm's are uncoupled, ranging over four possibilities.

[Anonymous No.7615318]

>>7613748
Abel died when he was 27??

based abel

[Anonymous No.7615323]

>>7613239
>numerical analysis

[Anonymous No.7615334]

>>7614713
based galois

[Anonymous No.7615371]

[Anonymous No.7615376]

Some notes on the Tschirnhaus transformation, which I used to solve all four polynomials above (proof was sketched for the cubic case):

I've been using a consistent and arbitrary "a,b,c" notation for a given polynomial's coefficients throughout this thread; once I had the idea that the transformation always leads to a monic depressed polynomial, no matter the degree, I wanted to verify this fact out of personal curiosity, which I did in a thread that will age out of the archive in a day or two:

>>7591696

The point was to create a general form for the (by definition monic) depressed polynomial, associated with each degree's general polynomial; the Binomial theorem and some nested series were used, thus:

[math] \displaystyle \sum_{k=0}^{n} A_{k+1} x^{n-k} = \frac{1}{ A_{1} } \; \sum_{r=0}^{n} \Bigg( \sum_{q=0}^{r} \Bigg( A_{q+1} { {n-q \choose r-q} \bigg( - \frac{A_{2}}{n A_{1}} \bigg) ^{r-q} } \Bigg) t^{n-r} \Bigg) = 0 [/math]

where the "A's" denote the nth alphabetical (placeholder) coefficient. This is just a fancy way of saying that division by A_1 followed by the substitution always leads to a depressed equation (the next-down degree term always algebraically vanishes in the middle chunk, regardless of value of n, and none of the latter terms do so, except as numeric accident of choice of coefficients). This formula also gives the associated p's, q's and r's of some depressed polynomial, /in explicit form/ of the original "a,b,c..." coefficients, matching the material that I studied (wiki), which is a nice shortcut.

The point here is that it should be possible to build a flowchart of solution methods for the linear, quadratic, cubic, and quartic equations, that always starts with these two steps, given in explicit form.

[Anonymous No.7615573]

did anyone save the cubic equation thread?
warosu is kill and i can't find it anywhere else

[Anonymous No.7615625]

>>7615573
What do you need?

[Anonymous No.7615628]

>>7615573

Everyone is a step ahead of me. Warosu is working again (for me), and so I pull my derivation of the cubic back into this thread, with an alternative (also correct) pic related; compare both of these with the cubic function wiki section on "general formula of roots" and the commentary there.

Let P(x) = 0 be a polynomial equation of the form ax^3 + bx^2 + cx + d = 0 , where x is some complex number, and the coefficients a,b,c,d are real (and I think complex is fine too, there's maybe just some issue with root extraction, either way I haven't satisfied myself about complex coefficients in either case yet), with a nonzero. Then

[math] \displaystyle x = \bigg( - \frac{1}{3a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( - \frac{1}{3a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]

[math] \displaystyle x = \bigg( \frac{ 1 + \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( \frac{ 1 - \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]

[math] \displaystyle x = \bigg( \frac{ 1 - \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( \frac{ 1 + \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]

Once again, take care not to confuse THESE a,b,c,d's with those of the quartic etc. But this is what is to be plugged in, in an expansion.

[Anonymous No.7615852]

>>7613748

I am currenty struggling to verify that the latter formula (or any multiple of same) in the above post is a "resolvent cubic" of /anything/ as wikipedia claims, per their definition (and wolfram furnishes the same definition):

https://en.wikipedia.org/wiki/Resolvent_cubic

This is odd, since the plug'n'chug should be straightforward. I welcome help with this.

We have three variables floating around in the discussion: x, t, and y. Since the above definition assumes the same variable between polynomials, I'm moving in that check-direction next. I compared both the original and depressed quartic's coefficients with this "y-cubic" form, and didn't get anything that looked right (perhaps since I need to be comparing polynomials with the same variable per the definition!)

[Anonymous No.7616533]

bump

[Anonymous No.7617699]

Now for some bookkeeping.

Following a cursory comparison of the three formulae given at

>>7614713 ,

https://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg , and

http://planetmath.org/QuarticFormula

It appears at-a-glance that the latter two formulations are verbatim copies of each other, and the large-scale layout of terms and pm-branching agree with my ending derivation. The latter formulae, however, correspond to the general /MONIC/ quartic (starting with "a" = 1), and using a,b,c,d on down, which is cosmetically different from my presentation, where I insisted upon carrying the leading coefficient through.

So, once again, take care not to confuse your a,b,c's, depending on context. Here, they are distinct from a layout that I would personally prefer (and that I have used in my own presentation), and they are also not to be confused with my presentation of the cubic: >>7615628 although after scanning a bit, you can start to make out similarities in the forms; something that "looks" very much like our version of the cubic formula is pervasive throughout, as our own y appears not once, not twice, but /three/ times in the derivation, which is why longcat is so very, very long.

Finally, the planetmath link doesn't seem to account for the case when 2y+p = q = 0, which is rather more merciful since y becomes superfluous in this fifth case. You really need this fifth case to call the thing comprehensively solved in terms of an algebraic solution.

Oh yes. I already shouted "FTA, LOL!", but more formally, to the above PROPOSITION: "But this is precisely an algebraic solution of the polynomial equation which was given, being what was to be shown, QED etc etc.

I may amuse myself with a re-write and post of the general formulae in my own terms ITT, but the substantive math is done, apart from two things: complex coefficients (issues with root extraction?), and I still can't get the above Resolvent Cubic thing to verify in any way shape or form.