The Cubic Equation

[Anonymous No.7529602]

The purpose of this thread will be to exposit the general solution (that is, three algebraically distinct roots) of the general cubic equation

<span class="math"> \displaystyle ax^3 + bx^2 + cx + d = 0 [/spoiler]

in the situations where the coefficients a,b,c and d are alternately real or complex. In particular, Wikipedia gives two distinct forms of the general solution, the latter of which I call the Cardano form (which amounts to pic related), which is a bit nicer to write out:

https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots

However, in wiki's current treatment, it is insisted that this latter form has caveats attached where generalizing the coefficients (say, to complex numbers) is concerned - as if to suggest that the the earlier form, which is quite similar in appearance, does not! Owing to their similar appearance, this intimation seems odd, but I haven't thought it all the way through yet, which is why this thread exists. Why (if at all, I'm not even convinced that this is true) is there some loss of generality between the two forms?

Owing to the above, I will first derive the solution with real coefficients, all the while soliciting help to understand what happens when they are variously replaced with complex terms (while thinking through same myself). For the above reasons, pains will be taken to outline the operations entailed in each step, like any good-old-fashioned formal proof, the better to walk it back afterwards. I come up with about four steps where square and cube roots are invoked, at a glance.

I should also simply mention that I have been starting to review the behavior of (fractional) nth roots of complex numbers, but I wanted to develop enough that this could be a halfway decent thread. One thing at a time.

https://en.wikipedia.org/wiki/Nth_root#Complex_roots

Finally, if you like anything in this thread at all or are moved to help... great! If not, you can hide the thread!

[Anonymous No.7529616]

>>7529602
Just popping into your thread to post this:

http://pastebin.com/i1iw2b4S

It's a section of a Python script that I'm currently using. Obviously replace the coefficients with your own equation. I'm sure there's a few of these on the internet, just thought I'd add another.

Cheers.

[Anonymous No.7529617]

The proof strategy with real coefficients, which I will commence after this sketch of same, largely apes wiki's current treatment, but I diverge a bit from wiki in the latter half by making a few different observations. The larger steps are:

-begin with general equation
-depress cubic, get rid of square term (first substitution)
-carry out a second substitution (Viete's substitution)
-name various expressions and establish relationships between them (mostly consistent with wiki's treatment)
-observe that "w+w-" is a solution of the depressed cubic
-invoke the third roots of unity and FTA
-walk everything back
-"declare problem solved, establish general solution".

Pic related is an expanded form of the FORMER (more general?) solution from wiki. Note the algebraic similarity (just with that big mess in the denominator instead).

Now, the cubic with real coefficients.

[Anonymous No.7529626]

>>7529616

Thanks!

"Let a,b,c and d be real numbers. Solve for x."

The next few posts will amount to highly condensed step-exposition, with lines in between indicating the exact manipulations made, purpose of the step, etc.

[Anonymous No.7529658]

<span class="math"> \displaystyle ax^3 + bx^2 + cx + d = 0 [/spoiler]

Premise.

<span class="math"> \displaystyle x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} = 0 [/spoiler]

Division by (necessarily non-zero) a.

<span class="math"> \displaystyle \bigg( t- \frac{b}{3a} \bigg)^3 + \frac{b}{a} \bigg( t- \frac{b}{3a} \bigg)^2 + \frac{c}{a} \bigg( t- \frac{b}{3a} \bigg) + \frac{d}{a} = 0 [/spoiler]

First substitution: x = t - (b/3a)

<span class="math"> \displaystyle t^3 + \frac{3ac-b^2}{3a^2} t + \frac{27a^2 d - 9abc + 2b^3}{27a^3} = t^3 + pt + q = 0 [/spoiler]

+-x/ algebraic manipulation (expansion), definition of liner and constant term as p,q

<span class="math"> \displaystyle t=w - \frac{p}{3w} \rightarrow -3w^2 + 3tw + p = 0 \rightarrow w = \frac{-3t\pm\sqrt{9t^2 +12p}}{-6} [/spoiler]

OBSERVATION: existence of some w, contingent on quadratic formula ( +-x/, and esp! √ )

<span class="math"> \displaystyle t^3 + pt + q = 0 \rightarrow \bigg( w - \frac{p}{3w} \bigg)^3 + p \bigg( w - \frac{p}{3w} \bigg) + q = 0 [/spoiler]

Following feasibility check, second (Viete's) substitution t = w - (p/3w)

<span class="math"> \displaystyle w^3 + q - \frac{p^3}{27w^3} = 0 \rightarrow w^6 + qw^3 - \frac{p^3}{27} = 0 [/spoiler]

+-x/ expansion, /multiplication/ by w^3 to get quadratic in same

(the interesting bits will have to wait, lunchtime)...

[Anonymous No.7529838]

>>7529602
good luck with that anon, take my bump.

[Anonymous No.7529861]

>>7529658

<span class="math"> \displaystyle z=w^3 \rightarrow z^2 + qz - \frac{p^3}{27} = 0 \rightarrow z = w^3 = \frac{-q \pm \sqrt{q^2 + \frac{4}{27} p^3} }{2} [/spoiler]

Definition/substitution: z=w^3, usage of quadratic formula, +-x/, esp √ !

<span class="math"> \displaystyle w = \sqrt[3]{ \frac{-q \pm \sqrt{q^2 + \frac{4}{27} p^3} }{2} } [/spoiler]

First invocation of the cube root.

<span class="math"> \displaystyle \sqrt[3]{z} = w = \sqrt[3]{ \frac{-q + \sqrt{q^2 + \frac{4}{27} p^3} }{2} } \;\;\; ; \;\;\; \sqrt[3]{ \bar{z} } = \bar{w} = \sqrt[3]{ \frac{-q - \sqrt{q^2 + \frac{4}{27} p^3} }{2} } [/spoiler]

Backtracking definitions, to differentiate the choices of quadratic form. This step and the previous are the most amenable to criticism (solicited).

<span class="math"> \displaystyle -(3a^2) p = b^2 - 3ac = \Delta_{0} \;\;\; ; \;\;\; (27a^3)q = 27a^2 d -9abc +2b^3 = \Delta_{1} [/spoiler]

Definitions. Specifications of "discriminant" forms.

<span class="math"> \displaystyle (729a^6)(q^2 + \frac{4}{27} p^3) = ( \Delta_{1}^{2} - 4 \Delta_{0}^{3} ) [/spoiler]

<span class="math"> \displaystyle - \frac{ \Delta_{1}^{2} - 4 \Delta_{0}^{3} }{27a^2} = (-27a^4)(q^2 + \frac{4}{27} p^3) = \Delta [/spoiler]

<span class="math"> \displaystyle \Delta = -27a^2 d^2 + 18abcd - 4ac^3 - 4b^3 d + b^2 c^2 [/spoiler]

+-x/, establishment of relationships of various forms, consistent with wiki's current treatments, helpful to "see"/reconcile their equivalence. (When you expand out the delta stuff and the p-q stuff per wiki's definition of "Large-C", you get similar but distinct results differing by a simple factor(s)...)

Now for an interesing bit, involving my above w and w-bar...

[Anonymous No.7529927]

>>7529861

The next few steps are to 1) critically observe that w+w-bar is precisely a solution of the above DEPRESSED cubic equation t^3 + pt + q = 0, and consequently (eventually) of the original equation! And, 2) once one's suspicions are aroused (if one is fortunate enough to /already/ be aware of complex roots of unity), it turns out that two distinct other forms involving same also check out. And 3) if one is further blessed to be aware of FTA (without having had to prove same), we can invoke same to declare the original problem completely solved---where real a,b,c,d are concerned, remember. That is, we're currently investigating the OP equation-situation when a,b,c,d are real, while admitting of complex x. Now I will slow down a bit because this bit is actually rather nice.

First, consider the (potential) equality

<span class="math"> \displaystyle ( w + \bar{w} )^3 + p( w + \bar{w} ) + q \;\;\; ?=? \;\;\; 0 [/spoiler]

In particular, the person who is furthermore blessedly aware of the Binomial theorem can quickly expand the first cubic term (taken by itself) to

<span class="math"> \displaystyle w^3 + 3w \bar{w} (w + \bar{w}) + \bar{w}^3 [/spoiler]

But the outer two cubic terms of /this/ expression sum to precisely -q, as a quick check with the above development can show, and with similar manipulation, 3ww-bar is precisely equal to -p, so the whole thing vanishes, cancelling the linear and constant terms of the earlier suspect equation, and we have a root!

And this is where I am being a bit cavalier with radical forms. For example, there is occasion in the above to assume things like (x^3)^(1/3) = (x^(1/3))^3 = x, which of course is generally not true, least of all when dealing with complex numbers. Which is really the nub of what I would have to suss out in order to deal with later complex-coefficient situation. For now, we will conclude the real-coefficient case however. The first step is admitting that you have a problem...

[Anonymous No.7529932]

>>7529658
OP I'm (>>7529616) monitoring your thread like a motherfucker, I'm actually working on a problem where it would be great if I could find the differentials of the quintic roots where the a, b, c, d coefficients of my equation contain variables in my model so a simpler expression would be a godsend.

How to I cite this thread in my paper?
>Anonymous (2015), "The Cubic Equation", Journal of /sci/, thread no. >>7529602

[Anonymous No.7529959]

Continuing formally,

<span class="math"> \displaystyle ( w + \bar{w} )^3 + p( w + \bar{w} ) + q = 0 [/spoiler]

observation of w+w solution of /depressed/ cubic, (and below:) invocation of third roots of unity, +-x/, once again, ESPECIALLY NOTE usage of √ 3√ manipulations (suspect for complex-coefficient numbers later on, but we proceed)

<span class="math"> \displaystyle ( uw + \bar{u} \bar{w} )^3 + p( uw + \bar{u} \bar{w} ) + q = 0 [/spoiler]

Here, there is a very nice and simple algebraic derivation, very much without tears. Here, u simply denotes /any/ of the third roots of unity, and u-bar denotes /that/ root's complex conjugate, which itself is always a third root of unity. in particular, for some u, notice that uu-bar is always equal to one (verify for yourself if need be). Similar binomial-theorem expansion and cancellation of various terms (u^3 = u-bar^3 = 1) shows that this is the basis of our general solution.

<span class="math"> \displaystyle - \frac{1}{3a} \bar{w} = v \;\;\; ; \;\;\; - \frac{1}{3a} w = \bar{v} \;\;\; ; \;\;\; x = t - \frac{1}{3a} b [/spoiler]

Extraction of -1/(3a) from all terms, personal book-keeping about final "bar" forms (corresponding to similar radical forms with SUBTRACTIONS in place of additions), recollection of original substitution in preparation for walking it all back

<span class="math"> \displaystyle x = - \frac{1}{3a} \bigg( uv + \bar{u} \bar{v} + b \bigg) [/spoiler]

Explicit statement of general form of solution, "The Cardano form". "u" may be any of the three third roots of unity, giving rise to three algebraically distinct forms (equivalent to the OP picture). Invocation of FTA to declare problem solved (where real coefficients a,b,c,d are concerned). The roots may variously be numerically distinct, or equal, per FTA.

Recall that in addition to giving rise to the OP pic, my treatment is equivalent to the LATTER general solution given at

https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots

where they use a big Capital-C instead of my v.

[Anonymous No.7529978]

>>7529932

OP here. I don't know what "differentials of quintic roots" are, and although depending on what you're doing there might be a "simpler" expression, given this thread's subject I should point out that there does not exist a /similar/, i.e. closed-form expression for quintics, or indeed higher-degree polynomials in general:

https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

Moreover, I am innocent of more complex manipulations of higher-degree equations, hence my exploration of the simpler cubic as a personal enrichment. :^)

Outright mention of 4chan is unwise in any area of polite society, least of all as a citation in an academic paper. You are welcome to save this thread of course. I am not doing anything original ITT (perhaps just simplifying the procedure a bit, so far), so you could search some other source which amounts to the same thing, and cite that.

[Anonymous No.7529995]

>>7529978
>differentials of quintic roots

In essence the results of the cubic roots are some functions that I want to differentiate individually. This is my thread >>7529459 that describes my current sub-problem a bit better; where each of the functions are one of the roots of the cubic and I use the minimum root. It's actually a pretty cool coincidence that you posted your thread.

I was just joking about the citation haha.

[Anonymous No.7530020]

Both in the interest of sheer autism, and also actually of making a few other remarks about the Cardano-form solution, we present a simple-yet-expanded form of the OP picture's solutions, all nice and cleaned up (assuming this post doesn't break jsMath):

<span class="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} [/spoiler]

<span class="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} [/spoiler]

<span class="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} [/spoiler]

If you tape two 8.5"x11" papers together long-ways, it's just enough room to write all this legibly by hand, with some explanation as well.

[Anonymous No.7530052]

Bumping.

Quick question due to me not being a hardcore maths guy myself, is this all your original work? Also, if it is, what exactly are you solve and what effects would this solution have in wider mathematics?

I might just be mistaken and this thread is for editing a wikipedia page or something.

[Anonymous No.7530064]

>>7530020
how do i implement formula?

[Anonymous No.7530118]

>>7530052

No. I am simply synthesizing material (and trying to straight-line a derivation) which has been known for over four-hundred years, as a personal enrichment, and if /sci/ likes, then so much the better. But there's really nothing new here. As I said, most of what I'm doing is simply aping wikipedia's existing page

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

, with a little tweest by way of John Derbyshire (the w-w-bar thing).

In searching the subject and various derivations, some physics guy had occasion to use general cubics for something. General solutions of this exact type are literally a dead-end though, being impossible for degree-five equations, or higher (see >>7529978 for more on this).

>>7530064

Consider a polynomial of the form (OP) with real coefficients a,b,c,d. Then you should be able to simply plug those coefficients into the Beasties in order to find the three roots (which may be real, complex, or have multiplicity). I'm cooking up a very simple example-check atm, and the sticking point seems to be about choice of (radical) roots.

[Anonymous No.7530314]

>>7530020

What can be remarked about these Beasties? Of course, a,b,c and d are (here) real, as we keep harping. And of course a is nonzero, otherwise we can all just go home, the thing wasn't a cubic at all, we divide by zero, john was phone, memes memes etc.

But we are told that a cubic with real coefficients necessarily has at least one real root. But is the /former/ equation necessarily that real root (in the situation where there is exactly one), or might the radical terms cancel in a goofynice way which takes care of the latter "i" terms? Or are the radical forms necessarily positive? The answer to the latter two questions is no, once we recall that a cubic equation with real coefficients may have three real, distinct roots.

I observe that the very friendly (real-domain'd, real-valued) odd function f(x) = x^3 - x (pictured) has the three distinct real roots -1, 0, and 1, as well as a nice graph near the origin. So I had better damn well be able to check vs. the Beastie Boys. It turns out fine, albeit some sleight of hand about choosing the "right" or if you like, "principal" third-root of i.

It goes like this. When you plug in a=1, c=-1, and b=d=0, seven of the eight terms in a large radical form immediately vanish. It turns out that the first equation gives 0, the second gives 1, and the third equation gives -1, which you can work out for yourself (hint: -i is the third root of i that you want to use).

The point is that the smaller square root form is not necessarily positive, and the latter equations may yield a real number. But this is not the same as saying that that WHEN a real-coefficient cubic has EXACTLY ONE real root among its three distinct roots, that the FIRST equation is necessarily the one which gives that root-after all, the other two equations are perfectly capable of giving real roots, as we've already seen! Let's see (if possible) if I can pick a function such that one of the latter two equations gives the unique real answer.

[Anonymous No.7530602]

>>7530314

Although Mathworld anticipates my observation (toward the bottom third of the page just after equation 69):

http://mathworld.wolfram.com/CubicFormula.html

to the effect that the forms of the above general equations do not, of themselves, necessitate that the FIRST such equation, not directly involving an "i" term, is always the one giving the unique real answer (when the other two are not real), I have a suspicion that some qualified answer along these lines is the case. That word, "qualified". I should like to finish this bit with some more qualitative, narrative remarks, then move on to the next part of my project, but first, a simple banality about zero coefficients:

Since a is necessarily non-zero, what is left pertaining to a zero/non-zero dichotomy are seven cases concerning real b,c,d: either exactly one of these is zero (3 cases), exactly two are(another three cases), or all three are, in which case every term vanishes and we are left with the ultimate, trivial banality that x^3 = 0.

In playing at teasing out a complex answer from the first equation of >>7530020 , while at the same time requiring a real answer from one of the latter two (and failing), I began simply, considering polynomials such as x^3 + x^2 and x^3 -1. Throughout this, when you are cognizant of complex numbers, there is a sense of forever stepping outside the "+-" simplicity of the quadratic derivation (which allows you to stay in the intellectually cozy space of the real line): but if you want to fully account for things going forward into higher degrees, then you must always be aware that any non-zero complex number (including of course all non-zero real numbers) has exactly n distinct nth roots, which I've been hinting at throughout the thread. And this is exactly what I don't have a good bead on: the multivariate nature of choice, branch cuts, etc, etc. I may just have to get (Weyl's?) little book about Riemann surfaces for further enrichment.

[Anonymous No.7530635]

>>7530602

Therefore, not finding an example, I leave aside the question of whether (a) unique real-root polynomial can be given by one of the LATTER TWO equations >>7530020 (when the other two are necessarily complex): the internet is suggesting to me that this can't happen, but I don't know one way or the other. I welcome help.

What I must start to do instead, is the second (and un-rehearsed) half of this thread's program: to consider the situation where the COEFFICIENTS a,b,c,d are themselves complex (in addition to admission of complex solutions x).

Is the Cardano-Form >>7530020 in any way substantively less-general than the former general form which wiki first details at

https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots

? I am becoming doubtful, as both seem to entail and imply the same happy "choices" of principal roots in manipulation. By 3√8 we typically mean just 2, for example (and not one of the others, although they exist as well), or by √81 we denote 9 (and not -9), etc.

But this is exactly where I must do some exercise. We recall that the complex numbers form a FIELD,

https://en.wikipedia.org/wiki/Field_%28mathematics%29

which immediately gives us a bonanza of the corresponding properties (esp. commutativity, associativity, distributivity etc), but owing to my personal temperament I want to remind myself of various of these behind the scenes (I know it amounts to real-number manipulation of real and imaginary parts in many cases, but still).

THEN, I must internalize the form of the distinct nth roots of various complex numbers, especially their geometry (haven't found a good quick visual link for this yet!):

https://en.wikipedia.org/wiki/Nth_root#Complex_roots

Since the above derivation of the cubic formulae only involved taking of square and cube roots in its "trickiest" spots, I will then be in a position to appreciate what becomes of the Cardano Form in the situation where a,b,c,d are non-real.

[Anonymous No.7530722]

>>7530635

By way of thinking out loud on this point, I previously re-established (for my own benefit) that the complex numbers are CLOSED under: addition, subtraction, and multiplication, and the complex number less-zero (or, "the punctured plane") is closed under division. Same behavior as the reals.

A quick review reminds me that the complex numbers ARE COMMUTATIVE UNDER ADDITION AND MULTIPLICATION, while they ARE NOT COMMUTATIVE UNDER SUBTRACTION OR DIVISION, which is a bit spurious since because this was already untrue of the reals, the reals, being also complex, spoil same for the larger complex case, but there we are. Of course you can get around this by "adding negatives", but the point is that the operations with the complexes have the same status vis a vis commutativity as with the reals. I would further imagine that associativity and distributivity also check out, but I'll skip that for just now to get to the larger point:

How good is the first-half program,

>>7529658
>>7529861
>>7529927
>>7529959
>>7530020

elucidated step-by-step, when we replace the previous /real/ coefficients with more general /complex/ ones? MOST of the steps appear to carry over just fine, with the sticking point of various /invocations/ of square and cube roots/, which deserve a deeper exposition. And since we will have occasion to take fractional roots of complex numbers, is there an algorithmic expression for how to proceed?

[Anonymous No.7530772]

Why are you posting this?

[Anonymous No.7531555]

>>7530772
>Why are you posting math on /sci/?

Really anon? He's just discussing his thoughts, you can hide his thread if you want to like he said.

[Anonymous No.7531816]

A quick review of De Moivre's theorem, trig/polar forms for complex numbers, and that old special-case chestnut

<span class="math"> \displaystyle e^{i \pi}+1 = 0 [/spoiler]

shows us both the geometry of nth roots of ANY GIVEN (non-zero) complex number, and how to find them.

Indeed, they always rest evenly spaced on some circle in the complex plane, which is always centered on the origin. The radius of this circle is precisely the modulus (absolute value) of every point on the circle, in particular the roots, with which we are concerned. With reference to the above z, this radius/modulus (of the nth roots) is precisely r^(1/n). As for the arguments (being a bit vague now), they are (modular) fraction/multiples of the argument of the original z-that is, you add a given argument to itself n times (multiply by n) to produce the argument of the target. And if you type something like "fifth roots of 1-2i" into wolfram alpha, it gives you a nice picture of what's going on. So much the better that we are only concerned with square and cube roots in this program!

Example: 1+i has an absolute value of √2, and an argument of π/4, or as I will crudely say, 45 degrees. Its three third roots are such that their absolute values are ( √2 )^(1/3) (taken in the principal sense, now, we will only ever deal with positive real numbers in this context so all is well), and the arguments can be multiplied modularly (spun around) three times over so that they end up at the 45 degree ray.

In pic related, notice how the root in the second quadrant is at a nice 135-degree angle. spin it (two more) times to end up at 405-degrees, or 45 degrees. And the radius of the circle is just a titch bigger than 1, which is just what ( √2 )^(1/3) is.

Now, in principle, we can interpret all possibilities for square and cube roots of complex numbers, especially with relation to the Beasties >>7530020 . Principle... I seem to remember something about "principal" roots?

[Anonymous No.7532705]

>>7530772
>Why are you posting math on /sci/?

[Anonymous No.7532851]

Note to self: find the roots of the equation ix^3+ix^2+ix+i=0 using the above equns and see what you come up with.

[I got a head cold which is slowing me down a bit!]

[Anonymous No.7532900]

>>7532851

aaand it seems that the first equation (when you make "convenient/appropriate" choices of square and cube roots, which is the crux of this whole thing) returns x = -1, which obviously checks out.

[Anonymous No.7533106]

>>7532900

Got it, though I punted to a calculator to suss out some real cube-root terms!

The second equation gives up -i, while the third equation >>7530020 gives up i, all three of which check out.

The point being that "The Cardano form" can be made to give up the correct (and distinct, where applicable) answers for some non-real a,b,c,d, but it hinges on appropriate choice of roots. For all that, wiki's language still seems to suggest/insist that there is a distinction between the two solution forms given there (referenced in the OP), but even mathworld's page ends up with what is effectively the Cardano-form.

[Anonymous No.7533675]

>>7532900

Indeed, any time that all four of a,b,c,d are equal, you will have the same solution set of 1, -i, +i due to algebra, which does not require an onerous equation. Which just about exhausts my (self-contained) interest in this problem.

Looking back at the original derivation and its steps involved, I tentatively conclude that the only real sticking point from transition from real to complex coefficients is a need to be more mindful of the various second and third roots which can be admitted of. And so I tentatively conclude that Cardano's formula holds good for complex coefficients, "up to" the appropriate choices of square and cube roots which do in fact lead to solutions of a given cubic equation when checking, which is a bit like what wiki said. Which is hardly a grand finish, but I got done what I wanted to get done with this thread and a bit more, and so I am pleased. :^)

At some later date, now that I know how this goes, I will exposit The White Whale: The Quartic. There is no major intellectual leap from the cubic to the quartic, just another order of magnitude of symbol pushing...

(have a look at wiki's general quartic solution):

https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots