Talk:Cubic equation

I think the Cardano's Formula section describing negative discriminants needs to be revised and wording cleaned up.

spanish version

There are two examples on the spanish version. The method is slightly different. If you copy them, please quote the author.

General cubic formula unclear ?

Hi, I find the general cubic formula rather difficult to follow and in particular the expression "any square roots" and "any cube roots" too vague : I think these points in particular should be clarified. Doe sit mean that possibly complex numbers arise, for instance ? Sorry, I am not a math major at all and cannot help much more... (2A01:E0A:2D1:7190:9959:CCC5:7AC0:707F (talk) 09:50, 30 September 2021 (UTC))

I agree that it is unclear, but i cannot help any more than you...

Good luck finding someone available. Garethphua (talk) 03:44, 25 April 2024 (UTC)

It does indeed mean that complex numbers may arise. I have added a note which I hope clarifies that. JBW (talk) 16:18, 25 April 2024 (UTC)

Suitability of page for general audience

I'm a bit unhappy with this page as it stands. The standard trigonometric substitutions are not given; was there some reason for this? Shouldn't the Chebyshev cube root be given its own page? Also, it ought to be pointed out that in practice these equations are solved by iteration methods. Xanthoxyl 16:31, 23 June 2006 (UTC)
I agree, moreover I did few unreveiled contributions making this matter very simple as shown at
6. All complete roots ... Mladen Stambuk 89.111.250.119 12:29, 5 April 2007 (UTC)

History

Didn't Cardano learn the *formulas* from Tartaglia, rather than a method that shows that the solution is correct? (Cardano came up with the method discussed in this page.)

Accuracy Dispute

This page doesn't seem to be correct. Consider the equation:

$x^{3}-x=x(x^{2}-1)=x(x-1)(x+1)$

This has three distinct real roots (0, 1, and -1). However, according to the page, it has $a_{0}=1$, $a_{2}=-1$, and $a_{1}=a_{3}=0$, and thus $q=4$ and $r=1$ and thus $s=4$ and the equation should have "two real roots, one of which is a double root."

It may be that the problem is just in the definition of the cubic — it is usually the convention that $a_{n}$ is the coefficient of $x^{n}$ (the article reverses the order). However, since obviously no one has ever checked this article, I can't be confident in any of the other equations. (And I don't have time now to check them myself.)

—Steven G. Johnson

This accuracy dispute has been sitting here for over a month with no activity related to it either here or on the main article. Is anyone paying attention to it? Bryan 08:22, 21 Jan 2005 (UTC)

I happened to stumble on the summary of Bryan's edit. The original poster is right in that there is a mistake in the article. -- Jitse Niesen 19:48, 21 Jan 2005 (UTC)

It turned out that the correct definition for t, mentioned in the previous article, is

$t={\frac {(2a_{1}^{3}-9a_{0}a_{1}a_{2}+27a_{0}^{2}a_{3})^{2}}{(a_{1}^{2}-3a_{0}a_{2})^{3}}}$

instead of

$t={\frac {(2a_{2}^{3}-9a_{1}a_{2}a_{3}+27a_{0}^{2}a_{3})^{2}}{(a_{2}^{2}-3a_{1}a_{3})^{3}}}.$

So Steven was right that part of the problem is the definition of the cubic. But even after fixing this there is still the issue of what happens if the denominator vanishes. So I used the discriminant instead.

This would not have been a problem if someone had not changed the cubic polynomial back to $a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$ without changing the definition of t. This is not the standard way to write a cubic polynomial in the theory of equations or algebra in general. This is because if you write it the other way, the coefficents $a_{n}$ become homogenous polynomials of degree n in the roots. You can see in the above expressions for t that in the first form, the numerator and denominator are both homogenous of degree six, and hence t has weight zero in the roots. This fact is not apparent in the second form.

I'm inclined to change it back to conform to the usage of mathematicians who work in this area. Any comment? Gene Ward Smith 22:04, 23 Jan 2005 (UTC)

I also added a short section on history; more can be written here as it is a nice story. I also removed the paragraph on which branch of the cube root to take, as I couldn't see its relevance. Furthermore, I made some more chances throughout and checked everything, except for (5) and the section on Chebyshev radicals. However, I still have a couple of problems with the page:

• I don't understand the section Factorization. It states that we can find the real root by extracting cube roots only of positive quantities, but how?
• The definition of $C_{\frac {1}{3}}$ is confusing: we have $C_{\frac {1}{3}}(0)=2\operatorname {arccosh} {\frac {1}{3}}$ which is not a real number, but both the power series and the statement that this function solves $x^{3}-3x=t$ suggest that it should be real. Furthermore, the definition of p in this section differs from the p in the rest. Does somebody have a reference for these Chebyshev radicals?

If we have one real root, we can find it by extracting the cube roots of real numbers. If for some reason we wanted only to take roots of positive numbers, and had negative ones, we could substitute -x for x, find the roots of that, and take the negative of the result. Gene Ward Smith 22:17, 23 Jan 2005 (UTC)

I understand that. The problem is, how do we get the number from which we need to take the cube root? I took the text in the article to mean that there is a way to do this, but I am not sure what formula archieves this. -- Jitse Niesen 13:38, 24 Jan 2005 (UTC)

The definition given is in terms of the power series and its analytic continuation, and the power series gives ${\sqrt {3}}$ for $C_{\frac {1}{3}}(0)$. I think I'll change it so that it is clearer that the definition is not in terms of the arccosh function, but it says so as it stands. The equations given for the roots give ${\sqrt {3}}$, $-{\sqrt {3}}$, and 0 for the three roots, which is correct. Gene Ward Smith 22:04, 23 Jan 2005 (UTC)

I still don't understand what the cosh is doing there. Are you sure it shouldn't be cos? That would make everything real on the interval [-2,2]. And please, do give a reference; it would improve the article and be a big help to us checking it (at least personally, I find I often make mistakes with cubic and quartic equations!). -- Jitse Niesen 13:38, 24 Jan 2005 (UTC)

It is possible to use either cosh or cos; it's essentially equivalent. Using cos, we can see this relates to the solution of the cubic using trigonometic functions; however putting it that way disguises the fact that it really comes down to an entirely algebraic solution, in terms of an algebraic function, and hence is exactly analogous to using radicals.

Here's an explanation of the trigonometric method:

http://www.sosmath.com/algebra/factor/fac111/fac111.html

Here's a discussion of Omar Khayyam's geometric method, which is closely related to this business:

http://jwilson.coe.uga.edu/emt669/Student.Folders/Jones.June/omar/omarpaper.html

There's a long and convoluted history involving this approach. So far as the mathematical accuracy of what I wrote, I'm a mathematician who specializes in this area, so I come in that way self-referenced. Gene Ward Smith 21:52, 24 Jan 2005 (UTC)

So I didn't remove the accuracy dispute tag. -- Jitse Niesen 19:55, 23 Jan 2005 (UTC)

I would appreciate a comment of June Jones and of Gene Ward Smith to 6. All complete root ... where at last (?) DETERMINANTS for Cubic & Quartic are introduced as well as NORMALIZED (instead depressed) CUBIC that enables GRAPHICAL RESOLVING much simpler than Omar Khayyam's geometric method.
89.111.254.251 12:52, 22 March 2007 (UTC)Mladen Stambuk

Continued accuracy dispute

I notice that the disputed tag was removed. However, the page still isn't correct for the simple example I gave above:

$x^{3}-x=x(x^{2}-1)=x(x-1)(x+1)$

which has roots 0, -1, and 1. According to the revised page's notation, this has: $\alpha _{3}=1$, $\alpha _{1}=-1$, and $\alpha _{2}=\alpha _{0}=0$. Then, according to the page, $\Delta =-4<0$, and the equation should have "one real roots and a pair of complex conjugate roots."

That's my mistake. I was confused by the fact that MathWorld uses different sign conventions for the discriminant on their web site. Sorry. -- Jitse Niesen 13:38, 24 Jan 2005 (UTC)

Please don't remove the disputed tag until you at least check all the equations against the roots of a few simple examples. (I don't care what convention you use for the coefficient numbering, but please be consistent!) —Steven G. Johnson 04:31, Jan 24, 2005 (UTC)

What would you say to my putting the coefficients the other way around, to correspond to how algebraists most often write them, putting the sign of the discriminant to be what algebraists usually want, checking everything, and removing the disputed tag? Gene Ward Smith 00:03, 25 Jan 2005 (UTC)

Please go ahead. I'm a bit surprised that you want to put the coefficients the other way around, but I'm sure you know best. If you prefer to use the t from Chebyshev radicals in the first section instead of the determinant, that's fine with me as well. When I asked you to provide some references, I meant the old-fashioned kind: books and papers. I did not want to question your knowledge, it's just that I'm a mathematician myself and I have often quite a bit of trouble getting all the signs right; I know many of my colleagues have similar problems and I presumed the same goes for you. I hope this clarifies the matter. -- Jitse Niesen 13:10, 25 Jan 2005 (UTC)

I've removed the disputed tag since I've checked at least through the Cardano's method section for a few sample polynomials and it seems to work (modulo a couple of special cases which I've now noted in the article). —Steven G. Johnson 19:53, Jun 5, 2005 (UTC)

Complete formula

For the form:

x³ + bx² + cx + d = 0.

$x={\frac {-{\Bigg (}{\sqrt[{3}]{(6{\sqrt {3(4b^{3}d-b^{2}c^{2}-18bcd+4c^{3}+27d^{2})}}+4b^{3}-18bc+54d)^{2}}}+2b{\sqrt[{3}]{3{\sqrt {3(4b^{3}d-b^{2}c^{2}-18bcd+4c^{3}+27d^{2})}}+2b^{3}-9bc+27d}}+2(b^{2}-3c){\sqrt[{3}]{2}}{\Bigg )}}{\sqrt[{3}]{3{\sqrt {3(4b^{3}d-b^{2}c^{2}-18bcd+4c^{3}+27d^{2})}}+2b^{3}-9bc+27d}}}$

For the form:

ax³ + bx² + cx + d = 0.

$x={\frac {-{\Bigg (}{\sqrt[{3}]{(6{\sqrt {3({\frac {4b^{3}d-b^{2}c^{2}}{a^{4}}}-{\frac {18bcd+4c^{3}}{a^{3}}}+{\frac {27d^{2}}{a^{2}}})}}+{\frac {4b^{3}}{a^{3}}}-{\frac {18bc}{a^{2}}}+{\frac {54d}{a}})^{2}}}+{\frac {2b}{a}}{\sqrt[{3}]{3{\sqrt {3({\frac {4b^{3}d}{a^{4}}}-{\frac {b^{2}c^{2}}{a^{4}}}-{\frac {18bcd}{a^{3}}}+{\frac {4c^{3}}{a^{3}}}+{\frac {27d^{2}}{a^{2}}})}}+2{\frac {b^{3}}{a^{3}}}-{\frac {9bc}{a^{2}}}+{\frac {27d}{a}}}}+2({\frac {b^{2}}{a^{2}}}-{\frac {3c}{a}}){\sqrt[{3}]{2}}{\Bigg )}}{6{\sqrt[{3}]{3{\sqrt {3({\frac {4b^{3}d}{a^{4}}}-{\frac {b^{2}c^{2}}{a^{4}}}-{\frac {18bcd}{a^{3}}}+{\frac {4c^{3}}{a^{3}}}+{\frac {27d^{2}}{a^{2}}})}}+{\frac {2b^{3}}{a^{3}}}-{\frac {9bc}{a^{2}}}+{\frac {27d}{a}}}}}}$

These were based on the formulas given on the page and were tested on a Ti-89... They seemed to work, however there may have been typos during the tex conversion process ... 71.0.202.231 21:38, 4 Apr 2005 (UTC)

Great! I would think these would be too complicated to insert in the article. Wonder what others think. Oleg Alexandrov 23:01, 4 Apr 2005 (UTC)

I don't think it's too useful in practice unless you eliminate the common subexpressions. —Steven G. Johnson 05:51, Apr 5, 2005 (UTC)

I strongly disagree that these formulas are too complicated to insert in the article. When I was looking for information about cubic and quartic equations I was confused why we didn't have explicit formulas like the quadratic equation. 67.71.20.89 02:57, 11 October 2006 (UTC) Jordan

At Cubic function the formulas are given in a usable form.

P.S. I can't get the formula given here to work, and the denominators are way too complicated, and can be rationalized, or at least they can be if you use the method indicated in the Cubic function article. --MathMan64 21:18, 17 October 2006 (UTC)

Would anyone know how to solve an equation like this:

(a+c)x^3 +(-2a+b+2c+d)x^2+(2a-2b+2c+2d)x+2b+2d is congruent (always equal to is what i mean) to 1. b and d are quite easy but how would get values for a and c?

All complete roots in terms of the determinants for
Cubic (along with Graphical resolving) and Quartic

The schoolbook lecture containing few unrevealed contributions is prepared in order
to facilitate understanding and memorizing of Cubic & Quartic equation.
Just opposite to Complete formula above, there is no need for any accuracy dispute since
all denominators (3A & F for Cubic i.e. 8a & P for Quartic) per definition differ from zero.
Instead enormous number of magic substitutions four determinants – E along with
its sub determinant F for Cubic and Q, R & S for Quartic – should be memorized and
inserted into two simple formulae (2) & (4) as presented at the abstracts below:
1. Cubic equation (extended abstract)
$E={1 \over 2}{\begin{vmatrix}0&3A&2B\\3A&B&C\\B&C&3D\end{vmatrix}}={9ABC-27A^{2}D-2B^{3} \over 2}{\mbox{ and its sub-determinant }}$
$F=-{\begin{vmatrix}3A&B\\B&C\end{vmatrix}}={\begin{vmatrix}B&3A\\C&B\end{vmatrix}}=B^{2}-3AC{\mbox{ form Depressed Cubic for }}3AX+B:$
$(3AX+B)^{3}-3F(3AX+B)-2E=0={\Big (}AX^{3}+BX^{2}+CX+D{\Big )}*3^{3}A^{2}\quad (1)$
$(t_{1}+t_{2})^{3}-3t_{1}*t_{2}(t_{1}+t_{2})-{\Big (}t_{1}^{3}+t_{2}^{3}{\Big )}=0;{\mbox{ if }}3AX+B=t_{1}+t_{2}\Rightarrow F=t_{1}*t_{2}{\mbox{ and }}$
$2E=t_{1}^{3}+t_{2}^{3}{\mbox{ along with }}F^{3}=t_{1}^{3}*t_{2}^{3}{\mbox{ form Quadratic for }}t^{3}:$
${\Big (}t^{3}{\Big )}^{2}-2Et^{3}+F^{3}=0\Rightarrow (t_{1,2})_{k}={\sqrt[{3}]{E\pm \ {\sqrt {E^{2}-F^{3}}}}}*e^{\pm \ i{2k\pi \over 3}}$
${\mbox{ where k = 0; 1; 2 and }}e^{\pm \ i{2k\pi \over 3}}=\operatorname {cos} {2k\pi \over 3}\pm \ i\operatorname {sin} {2k\pi \over 3}={\sqrt[{3}]{1}}{\mbox{ (Euler formula) }}$
$X_{k}={\frac {-B+{\sqrt[{3}]{E+{\sqrt {E^{2}-F^{3}}}}}*e^{+i{2k\pi \over 3}}+{\sqrt[{3}]{E-{\sqrt {E^{2}-F^{3}}}}}*e^{-i{2k\pi \over 3}}}{3A}}\quad (2)$
$S={B \over 3A}\Rightarrow X_{0}=-S+{\frac {{\sqrt[{3}]{E+{\sqrt {E^{2}-F^{3}}}}}+{\sqrt[{3}]{E-{\sqrt {E^{2}-F^{3}}}}}}{3A}}{\mbox{ real root and }}$
$X_{1,2}=-S-{t_{1}+t_{2}\pm {\sqrt {-3}}(t_{1}-t_{2}) \over 6A}=-S-{X_{0}+S \over 2}\pm \ {\sqrt {{\frac {F}{3A^{2}}}-3{\Bigg (}{X_{0}+S \over 2}{\Bigg )}^{2}}}$
$0\leq E^{2}\gtreqless F^{3}\gtreqless 0$ discriminates 8 cases; X₁ & X₂ nature in 5 Marginal ones is:
${\mbox{I. }}F=0{\mbox{ and }}E=0\Rightarrow X_{0}=X_{1}=X_{2}=-S{\mbox{ (all real and equal) }}$
${\mbox{ II. }}F=0\Rightarrow X_{0}=-S+{{\sqrt[{3}]{2E}} \over 3A}{\mbox{ and }}X_{1,2}=-S-{{\sqrt[{3}]{2E}} \over 6A}{\Big (}1\pm \ i{\sqrt {3}}{\Big )}{\mbox{ (conjugate) }}$
${\mbox{ III. }}F^{3}=E^{2}\Rightarrow X_{0}=-S+{2{\sqrt[{3}]{E}} \over 3A}{\mbox{ and }}X_{1}=X_{2}=-S-{{\sqrt[{3}]{E}} \over 3A}{\mbox{ (real and equal) }}$
${\mbox{ IV. }}E=0{\mbox{ and }}F<0\Rightarrow X_{0}=-S{\mbox{ and }}X_{1,2}=-S\pm \ i{{\sqrt {-3F}} \over 3A}{\mbox{ (conjugate) }}$
${\mbox{ V. }}E=0{\mbox{ and }}F>0\Rightarrow X_{0}=-S{\mbox{ and }}X_{1,2}=-S\pm \ {{\sqrt {3F}} \over 3A}{\mbox{ (real) }}$
As shown above X_k-formula is suitable for practical calculation if E×F×(E² – F³) = 0.
If F → 0 Cubic is converging into Primitive one (3AX + B)³ = B³ – 27A²D (see I. & II.).

---

In remaining cases its modification is recommended introducing R ≠ 0, x and z = z(g).
$R={2E \over 3|E|A}{\sqrt {f*F}}\gtrless 0{\mbox{ if }}E\gtrless 0{\mbox{ where }}f={\frac {F}{|F|}}=\pm \ 1\Rightarrow f*F>0$
$x={X+S \over R}\Rightarrow 3AX+B=3ARx={2Ex \over |E|}{\sqrt {f*F}}{\mbox{ inserted into (1) gives }}$
${\mbox{ Primeval Cubic: }}4x^{3}\mp \ 3x={\cfrac {|E|}{\sqrt {f*F^{3}}}}=g={\frac {e^{3z}\pm \ e^{-3z}}{2}}\geq 0\quad (3)\Rightarrow$
$x_{k}={\frac {{\sqrt[{3}]{g+{\sqrt {g^{2}\mp \ 1}}}}*e^{i{2k\pi \over 3}}+{\sqrt[{3}]{g-{\sqrt {g^{2}\mp \ 1}}}}*e^{-i{2k\pi \over 3}}}{2}}={\frac {e^{z+i{2k\pi \over 3}}\pm \ e^{-z-i{2k\pi \over 3}}}{2}}$
Identical structure of 4x³ ± 3x = g (Primeval Cubic) and of tigonometric & hyperbolic
formulae for triple argument leads up to either g = sinh3u or g = cosh3v i.e. g = cos3w.
VI. F < 0 i.e. f = – 1, g = sinh3u = 4sinh³u + 3sinhu → z(g) = u = arcsinh(⅓g),
x₀= sinhu, X₀= Rsinhu – S, X_1,2= – R(sinhu ± i 3^½coshu)/2 – S; E = 0 = g → IV.
VII. F > 0 i.e. f = + 1 ≤ g = cosh3v = 4cosh³v – 3coshv → z(g) = v = arccosh(⅓g),
x₀ = coshv, X₀ = Rcoshv – S, X_1,2 = – R(coshv ± i 3^½sinhv)/2 – S; g = 1 → III.
VIII. F > 0 i.e. f = + 1 ≥ g = cos3w = 4cos³w – 3cosw → z(g) = i w = i arccos(⅓g),
x_k = cos(w + k×120°), X_k = Rcos(w + k×120°) – S; E = 0 = g → V.
Conclusion: in all of 8 cases X₀ is real number, but X₁ & X₂ are conjugate if E² > F³.
Besides, Primeval Cubic enables Semi-Graphical resolving (see C below) and
Geometrical Interpretation (Roots Flowchart): ± (y² – x²) = 1, y² + x² = 1 & y = ± x.

2. Quatric equation (abstract)
$Q={\begin{vmatrix}b&4a&4a&3b\\0&0&b&4a\\4e&b&4d&4c\\b&4a&0&0\end{vmatrix}},\quad R={\begin{vmatrix}0&2a&b\\2a&b&c\\b&c&2e\end{vmatrix}},\quad S={\begin{vmatrix}2a&3b\\b&4c\end{vmatrix}}\,$ $(4ax+b)^{4}+2S(4ax+b)^{2}+8R(4ax+b)+Q={\Big (}ax^{4}+bx^{3}+cx^{2}+dx+e{\Big )}*4^{4}a^{3}=$
$={\Big [}(4ax+b)^{2}-P(4ax+b)+Q_{1}{\Big ]}*{\Big [}(4ax+b)^{2}+P(4ax+b)+Q_{2}{\Big ]}=0\Rightarrow$ $P^{2}+2S=Q_{1}+Q_{2},\,{8R \over P}=Q_{1}-Q_{2}\Rightarrow (P^{2}+2S)^{2}-{(8R)^{2} \over P^{2}}=4Q_{1}Q_{2}=4Q$ ${\Big [}P^{6}+4SP^{4}+4(S^{2}-Q)P^{2}-(8R)^{2}{\Big ]}*3^{3}=0=(3P^{2}+4S)^{3}-3F(3P^{2}+4S)-2E$ $E={1 \over 2}{\begin{vmatrix}0&3&8S\\3&4S&4(S^{2}-Q)\\4S&4(S^{2}-Q)&-3(8R)^{2}\end{vmatrix}}=8(S^{3}-9QS+108R^{2})\Rightarrow F=4(S^{2}+3Q)$
$P_{k}^{2}={\frac {{\sqrt[{3}]{E+{\sqrt {E^{2}-F^{3}}}}}*e^{+i{2k\pi \over 3}}+{\sqrt[{3}]{E-{\sqrt {E^{2}-F^{3}}}}}*e^{-i{2k\pi \over 3}}-4S}{3}}>0$
$x={P_{k}-2b \over 8a}\pm \ {1 \over 8a}{\sqrt {{\frac {16R}{P_{k}}}-4S-P_{k}^{2}}}\quad (4)$
Positive & negative value of P_k (k satisfying P_k² > 0 should be chosen) results into
2 (two) symmetric pairs of either real or conjugate (if 16R/P_k < P_k² + 4S) roots.

More history

This sounds completely wrong: <<In the early 16th century, the Italian mathematician Scipione del Ferro found a method for solving a class of cubic equations, namely those x3 + mx = n. In fact, all cubic equations can be reduced to this form if we allow m and n to be negative, but negative numbers were not known at that time.>> You can't reduce a cubic with x^2 in it to the form "x^3 + mx = n". Also, is this saying that negative numbers weren't known at the time??!! The next paragraph talks about square roots of negative numbers. I'd fix this right away, but it's been here a while -- I'm not sure what we're trying to say. Morcheeba 01:37, 10 March 2006 (UTC)

If you have the cubic equation x^3+ax^2+bx+c=0, you can perform the substitution x=y-a/3 to get an equation on the form y^3+my+n=0.

Negative numbers were in fact, if not unknown, then not accepted as "real" numbers in 16th century Europe. Negative numbers have a brief section on the history. Rasmus (talk) 07:17, 10 March 2006 (UTC)

So, given that negative numbers were not considered as proper numbers, did Cardano really notice that "Tartaglia's method sometimes required him to extract the square root of a negative number" (quote from the article), or is this an anachronism and did Cardona notice something else (for instance, that the quadratic equation for u^3 has no solution). -- Jitse Niesen (talk) 09:08, 10 March 2006 (UTC)

TEX

I'm getting these TeX errors, dunno if its because of the power failure but some of the other <math>s seem to work. 203.218.37.45 15:20, 19 April 2006 (UTC)

Failed to parse (Can't write to or create math output directory): \Delta = 4\alpha_1^3\alpha_3 - \alpha_1^2\alpha_2^2 + 4\alpha_0\alpha_2^3 - 18\alpha_0\alpha_1\alpha_2\alpha_3 + 27\alpha_0^2\alpha_3^2.

References Section

The reference to "Cardano solution calculator as java applet" needs to be removed. The applet is a piece of rubbish that doesn't appear to work very well at all.

Resultant

I've tested the nature of the roots against a simple cubic equation $x^{3}-2x^{2}$ according to these discriminants the equation is supposed to only have 1 real root, while it has 2. Why does this discriminant lack an "ab" section? -T. Stokke

The formula for Δ₂ indeed seems to be wrong. When I (or, to be honest, Maple) compute the resultant, I get

$\Delta _{2}=\mathrm {res} (f,f'')=-216a^{3}d+72a^{2}bc-16ab^{3}.$

I'm not too comfortable with putting this in, so I simply removed the old formula. -- Jitse Niesen (talk) 15:18, 31 December 2006 (UTC)

Well the old one worked, if you didnt forget that if $d=0$ then the two roots that coincide (usually called x₂ and x₃) also equal 0

Merger proposal

---

Proposal
The idea is to replace existing articles with unique Cubic & Quartic equation as follows:

A. Merged history paragraphs giving a credit for cubic to Tartaglia (Cardano’s contribution
is suspicious, at least exaggerated) as well as for quartic to Ferrari.
B. Tartaglia’s – Ferrari’s method for resolving cubic and quartic equations containing mentioned
lecture should replace all the others (Lagrange resolvents and Chebyshev radicals are included under trigonometric and hyperbolic substitutions, depressed cubic is replaced with Primeval one).
C. The graph actualy presents Marginal case V: 4Y = (X + 1)³ – 9(X + 1) = O.
Therefore it should be replaced with more representative one given in Primeval form
4y = 4x³ ± 3x – g = 0 (containing all off cubics where F ≠ 0) that can be splited into:
Basic parabola (y = x³) and
either Green (if F > 0) or Red (if F < 0) Straight line (4y = ± 3x) shifted up for ¼g.
Real solution(s) is x-coordinate of the cut(s) of these functions.
Two characteristic examples are to be chosen (common S = 1.5 & R² = ± 6²).
8X³ + 36X² – 162X – 729 = 0 → S = 1.5, E = + 373248, F = + 5184, R = + 6,
g = 1 = 4x³ – 3x or x³ = y = (1 + 3x)/4 → x₀ = 1, x₁ = x₂ = – 0.5,
X₀ = 6×1 – 1.5 = 4.5 & X₁ = X₂ = 6×(– 0.5) – 1.5 = – 4.5
8X³ + 36X² + 270X – 1350 = 0 → S = 1.5, E = + 1469664, F = – 5184, R = + 6,
g = 3.9375 = 4x³ + 3x or x³ = y = (3.9375 – 3x)/4 → x₀ = 0.75,
X₀ = 6×0.75 – 1.5 = 3 but X₁ and X₂ are conjugate numbers.
x-Drawing Ratio: 6cm = 1, 4cm = 1 for y. The conversion from x & y to X & Y coordinates:
6cm = R (1cm = 1 for upper examples due to R = 6) should be taken as X-Drawing Ratio,
Y-axis shifted either right if S > 0 or left if S < 0 (1.5cm right for upper examples),
X-ais mirrored if E < 0.

---

The lecture is prepared by means of Microsoft Equation 3.0 and Word Drawing tool. Therefore it was impossible to paste here neither this drawing nor Geometrical interpretation. I apologize for this inconvenience to Wikipedians asking them to assist. If you (Wikipedia editorial) find it interesting don’t hesitate to contact me for details like:
${\mbox{ Normalized form of Primeval Cubic: }}{\cfrac {x^{3}}{{\tfrac {1}{4}}g}}\mp \ {\cfrac {x}{{\tfrac {1}{3}}g}}=1$
89.111.252.148 16:54, 17 March 2007 (UTC)Mladen Stambuk
ok, I have absolutely no idea what's going on, but as the person who created the image, I'm going to say I chose those values for their aesthetic appearance - it makes a nice looking cubic. enochlau (talk) 23:02, 28 March 2007 (UTC)
[Meanwhile I added on few formulae hoping it to be understood finally. In my opinion aesthetic appearance should not prevail mathematical reasons. Regards Mladen]
89.111.250.119 11:26, 5 April 2007 (UTC)

I noticed that this page has a merger tag, but no corresponding section in the talk page. I decided to add one, instead of taking away the merger tag. Daniel 22:02, 26 October 2007 (UTC)

Yes, a merge of cubic function into cubic equation seems to be appropriate to me. There is not much information in cubic function. It should be easily combined. --MathMan64 00:43, 10 November 2007 (UTC)

As it is now, the cubic function page doesn't have much additional information. It would be nice, however, to add to the cubic function page a discussion of various cases, i.e. how the function looks depending on a,b,c values. Xenonice (talk) 22:50, 26 November 2007 (UTC)

The external link reference:

Dave Auckly, Solving the quartic with a pencil American Math Monthly 114:1 (2007) 29--39

is about quartics and not cubics.

Before the "Lagrange resolvents" section it states:

"...or two real roots (a single and a double root.)"

What does this mean? Id D=0 and there are 2 roots [aside that there should be 3 roots for a cubic] how can there be 2 real roots of a single and double root?

I think this statement needs clarification.

cited reference and D section

The external link reference:

Dave Auckly, Solving the quartic with a pencil American Math Monthly 114:1 (2007) 29--39

is about quartics and not cubics.

Before the "Lagrange resolvents" section it states:

"...or two real roots (a single and a double root.)"

What does this mean? Id D=0 and there are 2 roots [aside that there should be 3 roots for a cubic] how can there be 2 real roots of a single and double root?

I think this statement needs clarification. —Preceding unsigned comment added by 212.20.240.70 (talk) 16:22, 20 January 2009 (UTC)

Could you explain the technical usage of "useless"?

@D.Lazard: Re: positive square root, mea culpa; I was thinking only of real coefficients which is of course silly. (The purpose of the edit was to remove the duplication between talking about "any square root" and "±" which are redundant with each other.)

The connection to the discriminant Δ I thought was valuable; why use the symbols Δ₀ and Δ₁ if there isn't one? The thing that prompted me was the discussion in Discriminant#Degree 3 of the use of the discriminant in the cubic formula. That wasn't mentioned in this article, so I wanted to add it.

But what really confuses me is this bit about preferring the word "useless". That seems directly contrary to the lay usage. "Useless" carries an implication of not-usable, non-functional, will not work. But the text is clear that the general equation does work, it's just more complex than necessary. I agree that "useless" doesn't denote that, but the usage sounded to me like the sort of almost-but-not-quite word choice commonly found in translated text.

I thought "unnecessary" conveyed the intended meaning better. As an example of the distinction, when a motorway option exists, travelling via side streets is unnecessary; travelling via a cul-de-sac is useless. When a window is dirty only on the outside, washing the inside is unnecessary; washing only the inside is useless.

But you say there's a technical mathematical usage of the word "useless"? I've never heard of that. Can you point me to an example? (Searching for "mathematics" and "useless" leads to a large number of essays on pure mathematics and school curricula.)

I know how mathematicians like to redefine common words (one might hope they'd at least limit themselves to nouns, adjectives, verbs and adverbs; but "into" and "onto" are prepositions!), so it wouldn't surprise me if "useless" had a technical meaning in a narrow field, but I really haven't heard of a widely-used technical meaning.

I'm leery of a meaning which actually conflicts with the lay meaning of the word. WP:TECHNICAL and all that. 64.44.80.252 (talk) 00:05, 13 December 2020 (UTC)

It is not "useless" that has a technical meaning, it is "unecessary" (see, for example, necessary condition). So, care must be taken for using "necessary" and "unnecessary" in a mathematical context. Here, the intended meaning of "useless" is clearly "not useful". If you have a better word for this meaning, the word could be changed, but it must be clear that this makes the meaning of the sentence clearer. D.Lazard (talk) 10:42, 13 December 2020 (UTC)

Ah, that makes more sense! I'm sorry for misinterpreting your edit comment!

There are certainly other synonyms like "not required", "excessive", and "superfluous" which don't carry the implication (literary, not formal logic sense!) of "does not lead to a solution" that "useless" does.

But also, there's nothing wrong with "unnecessary" in this context. Even though this article is not in the field of formal logic, the usage is consistent with the formal logical meaning: when simpler methods are sufficient, the general method, although also sufficient, is not necessary.

That is why I didn't twig to your reference to the formal meaning of "necessary": the way I was using it is completely compatible with the that meaning.

With regard to the modal logic usage of "necessary", things get a bit stranger. I suppose if you took "the general formula holds" as a proposition, then it is a necessary proposition. So calling it unnecessary could be interpreted as making the false claim that its negation is possible; i.e. there are circumstances in which it does not hold. But that's a ridiculously contrived reading; the reader is inferring a modal logical statement not explicitly in the text while ignoring the explicit statement that the formula does hold in general. Clearly the more common classical logic usage is intended.

Note that even modal logic textbooks use "necessary" (pp. 253, 255, 268, 284, 285, 304, 320), "unnecessary" (p. 253) and "necessarily" (pp. 258, 309) in the informal sense. Mixed in with formal usage: "is necessarily an isomorphism" (p. 276), "necessary condition" (p. 282), "necessary but not sufficient conditions" (p. 302), "not necessarily ordered structures" (p. 321).

The same is true (albeit less frequently) of Blackburn's standard text "Modal Logic" (available from libgen.rs). §2.2 (p. 71): "But as it turns out, most of this work is unnecessary." §5.3 (p. 286): "We now have all the necessary material to prove Stone’s Theorem." §6.6 (p. 375): "(If necessary, consult Appendix C for further discussion.)" §6.7 (p. 388): "While the size of the entire model may be exponential in |φ| it is not necessary to keep track of all this information."

If modal logicians can handle informal usage of the words "necessary" and "unnecessary" in their specialized texts, then Wikipedia is not going to confuse them.

Considering the available options, "unnecessary" still seems simplest and most natural, but "not needed" isn't much worse. (The main problem is the lack of an adverb form). Which do you prefer? 64.44.80.252 (talk) 16:06, 13 December 2020 (UTC)

This is an article of mathematics, not of logic. Moreover, modal logic is a non-classical logic that is sometimes (in fact rarely) used in mathematical logic, but never used in other parts of mathematics. So, all your references to modal logic are definitely irrelevant here. Moreover, the article talks about a formula that is proved true. The sentence says that in the listed cases the formula has no interest because there are other formulas that are much simpler, and equivalent in this case, and thus there is no reason to use this formula. If one would want using "unnecessary" here, one should write "in these cases, for solving the equation, it is unnecessary to use the general equation. But this is not the formula that is "unnecessary", it is its use. Thus "useless" or "not useful" has a stronger meaning, which is: "although the formula is true, it not useful in any application". This is not logic but plain English. D.Lazard (talk) 17:21, 13 December 2020 (UTC)

@D.Lazard: I was searching for every sense of the term "necessary" I could find in mathematics (sensu lato, including logic) to find the problematic "different technical meaning" to which you referred. The classical logical meaning is not a problem because it applies here. The modal logical meaning is not a problem because it's only used in a specialized field, (we appear to agree on this point!) and even a reader familiar with that field would never mistake the usage here. If you're excluding the field of logic, there must be some third "different technical meaning" to which you were referring. What is it?

In common plain English, "useless" means complete futility. Not merely inefficient (little use), but unable to ever reach the goal (no use at all). Plenty of reductions return their own input in some circumstances; the reduction is useless in those situations. Duplicate factors, and factors of x, in a CRC polynomial are useless for error detection. Linearly dependent equations are useless for solving a system of equations. Naïve floating-point solutions to ill-conditioned systems of equations produce useless answers. Measuring the moments of a sample to estimate the moments of an underlying distribution is useless if the distribution is Cauchy. And so on.

It's not uncommonly used hyperbolically in situations which only approximate that, but the dictionary definition is zero use:

• of no use; not working or not achieving what is needed: "Without fuel, the vehicles will become useless for moving supplies." "It's useless to speculate without more information."[1]
• having or being of no use: (a) ineffectual "a useless attempt" (b) not able to give service or aid: inept[2]
• (1) of no use; not serving the purpose or any purpose; unavailing or futile: "It is useless to reason with him." (2) without useful qualities; of no practical good: "a useless person"; "a useless gadget."[3]
• (1)(a) Being or having no beneficial use; ineffective: "This pen is useless because it's out of ink." See Synonyms at futile. (b) Having no purpose or reason; pointless; to no avail: "It's useless to argue over matters of taste." (2) Incapable of acting or functioning effectively; ineffectual or inept: "He panics easily and is useless in an emergency."[4]
• (1) Cannot be used: "He realised that their money was useless in this country." "Computers would be useless without software writers." (2) does not achieve anything helpful or good: "She knew it was useless to protest." "...a useless punishment which fails to stop people trafficking." (3) no good at all [informal]: "Their education system is useless." "He was useless at any game with a ball." (4) unable to help someone or achieve anything: "She sits at home all day, watching TV and feeling useless."[5]
• (1) Without use or possibility to be used: "This fork has prongs that are bent. It's useless now." (2) Unhelpful, not useful; pointless (of an action): "I think it's useless to keep this discussion going. It's like talking to a wall." "I tried my best to make him quit smoking, but my efforts were useless. He now smokes six packs a day." (3) (derogatory, of a person) Good-for-nothing; not dependable. "Bill never mows the lawn, takes out the trash or anything. He's useless, but I love him anyways." (4) (colloquial, of a person) Unable to do well at a particular task or thing. Useless is mildly insulting: "My brother is useless at most computer games, but he is an awesome PS2 player." "Why do you keep trying? You're obviously useless at it."[6]

You say that the general formula is "not useful for any application" if a special case applies. I disagree and WP:CHALLENGE that claim.

The general cubic formula is useful in any application which does not want to bother checking if the special case applies. To minimize code size in an implementation, for example. Or a constant-time implementation defending against a timing attack. Or because the expected probability of the special case occurring, while not zero, is so low that the cost of the test outweighs the expected saving from the special case.

The word "useless" implies that the general case fails to find a solution and it is necessary for correctness to check for the special case and use alternate techniques in that case. That implication is false, and I want to change the wording to eliminate it.

More generally, neither of us knows a reader's utility function. There are some algorithms which are worse by any sensible utility function, such as bogosort and galactic algorithms, so I don't object to calling them "useless", but that doesn't apply in this case. (I know that I use "inefficient" algorithms all the time, simply because one is conveniently at hand and will get me the answer I want faster than searching for a more "efficient" algorithm. Brute-force search can do a lot on modern computers.)

Actually, that suggests another possible solution: change every instance of "useless" to "inefficient". Does that seem more palatable to you? 64.44.80.252 (talk) 19:58, 13 December 2020 (UTC)

Naming discriminants

AchzehnWeg, I have reverted your edit several times for two reasons. The first one, is that it is not useful, and often confusing to introduce in a section some symbols that are not used in the same section. The second reason, is that it is very confusing to give the same name (here $\Delta$) to two different things (here the discriminants of two different cubics). However, reading your explanation, it appeared that the formulation of the beginning ot the next section was incorrect: the use of "the polynomial" and "the discriminant" seemed to refer to things previously defined and was therefore ambiguous, since there are two polynomials and two discriminants defined in the previous section. So, I have put "a polynomial" and "its discriminant". With this modification, $\Delta$ is perfectly defined by the sentence, and I hope that will convince you. D.Lazard (talk) 15:51, 19 April 2021 (UTC)

Case of real roots

I think it is useful to mention in this article something elementary (hence, no need to have citations), also hinting to deeper structure, but mainly useful when it comes to inequalities for three real numbers. Namely, explicit bounds on the product of three real numbers in terms of the symmetric functions built from these three numbers. This could also trigger search for similar constructions for higher degree polynomials. For $ax^{3}+bx^{2}+cx+d$, writing $\delta ={\sqrt {(b^{2}-3ac)/(9a^{2})}}$ (it is always possible when the cubic polynomial has real roots) and, just for convenience, $m=b/(3a)$, the inequality is: $-2\delta ^{3}<(x_{1}-m)(x_{2}-m)(x_{3}-m)<2\delta ^{3}$. It could be written as $-2\delta ^{3}<x_{1}x_{2}x_{3}+m^{3}+m\delta ^{2}<2\delta ^{3}$, and next: $(m+\delta )^{2}(m-2\delta )<x_{1}x_{2}x_{3}<(m-\delta )^{2}(m+2\delta )$. We may go back (not very illuminating though) to explicit symmetric functions: $(x_{1}+x_{2}+x_{3}+(x_{1}+x_{2}+x_{3}+{\sqrt {x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2}-x_{2}x_{3}-x_{3}x_{1}}})^{2}(x_{1}+x_{2}+x_{3}-2{\sqrt {x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2}-x_{2}x_{3}-x_{3}x_{1}}})<27x_{1}x_{2}x_{3}$ $27x_{1}x_{2}x_{3}<(x_{1}+x_{2}+x_{3}-{\sqrt {x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2}-x_{2}x_{3}-x_{3}x_{1}}})^{2}(x_{1}+x_{2}+x_{3}+2{\sqrt {x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2}-x_{2}x_{3}-x_{3}x_{1}}})$.

This is the best upper/lower bound of the product $x_{1}x_{2}x_{3}$ of any three real numbers in terms of symmetric functions. One may consider this is trivial and ridiculous to worth mentioning in the article, but this is debatable: something simple is not necessarily irrelevant. These inequalities (necessary and sufficient) could be trivially upgraded for positive numbers, when in addition to the above bounds one has the obvious $m>0,\ \delta >m,-d/a>0$, or just $x_{1}+x_{2}+x_{3}>0,\ x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}>0$, and $x_{1}x_{2}x_{3}>0$.

I would appreciate comments from those erasing my post, explaining their view. Thank you in advance. 0ctavte0. — Preceding unsigned comment added by 78.97.139.142 (talk • contribs) 22:38, 5 May 2021 (UTC)

Please, sign your posts in talk pages with four tildes (~~~~).

Wikipedia is not a forum for discussing personal research. There are specialized sites for such discussions, such as Mathematics Stack Exchange (https://math.stackexchange.com). Your post is better suited for such a forum.

About your results, my first comment is the following. When working on a subject that has been deeply studied for century, this is unbelievable to find something something really new. This is the case here. So, the first thing to do, before trying to publish your results is to search for them in the literature, and, if you do not find them, to try to understand why it is not published.

In your case, the notations that you use appear in the Wikipedia article: $\delta ^{2}$ is essentially the coefficient p of the associated depressed cubic. So, your first inequality is an unusual way to express that, if the three roots are reals, the discriminant of the depressed cubic is positive. The huge final inequality may be trivially interpreted as follows: The discriminant is a quadratic function of $-d=x_{1}x_{1}x_{3},$ and the inequality is the standard condition for a quadratic function to be positive.

A last comment: one may derive hundreds formulas from those appearing in the article. For adding one of them to the article, it must not only having been published elsewhere, but it must also having been shown to be useful for some explicit purpose. As far as I know, your reverted addition does not satisfy either of these conditions. D.Lazard (talk) 07:40, 6 May 2021 (UTC).

It is OK. Of course, like all results in the article, it might have had appeared somewhere before. It could have been useful to have the best bounds for the product of three real numbers explicitly mentioned in this extensive article, in one way or the other. Case of real roots needs to be treated separately, with more results included, anyway. (78.97.139.142 (talk) 13:15, 6 May 2021 (UTC))

Algorithm

Resolved

May I append and algorithm written in Python or is it inappropriate? I use the General cubic formula, so I would insert it at the end of section 'General cubic formula'. — Preceding unsigned comment added by Betaminus (talk • contribs) 09:49, 26 May 2021 (UTC)

A program written in Python would certainly not useful in this article: not all readers knows Python. Readers who know Python are supposed to be able to write simple programs in this language. Translating the formula into a Python program is an easy exercise for beginners in Python programming. This is of no value in this article that is not about Python. D.Lazard (talk) 11:37, 26 May 2021 (UTC)

Thank you for your guidance. Betaminus (talk) 16:25, 26 May 2021 (UTC)

Characteristic 2 and 3​

In section Cubic equation#Characteristic 2 and 3 one can read «This allows computing the multiple root». OK, if the polynomial has a multiple root. But NOT OK, if it does not have. E.g. if the derivative is a constant. What then? Or let $f(X):=X^{3}+tX+q\in \mathbb {F} _{2}(t)[X]$ with non-constant derivative $X^{2}+t$ ? I assume there is no multiple root of $f(X)=0$ . What does this knowledge help me to find the non-multiple roots ? –Nomen4Omen (talk) 15:29, 10 February 2022 (UTC)

Bad behavior by user D.Lazard

User D.Lazard has exhibited a persistent pattern of aggressively suppressing even minor changes that do improve the accuracy or readability of this article. I'm not sure what measures to take, but it appears that D.Lazard is not able to appropriately tolerate other editors' changes to this imperfect article, and is treating it like private property. Astro-Tom-ical (talk) 03:16, 18 February 2022 (UTC)

To editor Astro-Tom-ical: It seems that you are a WP:SOCKPUPPET, since your two edits on Cubic equation are signed 166.205.91.21 and BigEmD. If you continue to use several accounts without good and explicit reasons, your accounts will be blocked for editing.

Also, this talk page is not for judging the behavior of other editors, it is for discussing the content of the associated article.

About the content of your post, it seems that you ignore the main rules of Wikipedia. In this case, that is in the case of a reverted edit, the correct procedure is extensively described in WP:BRD. It consists of opening a discussion in this talk page as soon as you disagree with a revert, for explaining the reasons of your edit and why you think that it improves the article. Then, all editors interested in the question can give their opinion and search for a consensus. Before a consensus, the previous stable version prevails. Therefore, I'll restore it. Do not start WP:edit warring, as this may lead to an edit block. D.Lazard (talk) 10:29, 18 February 2022 (UTC)

@D.Lazard Continuing this conversation, I would like to understand why you reverted my edit. In particular I don't understand what "either of the cube roots" could refer to as we are discussing a case of only one real root. Yehoshua2 (talk) 21:42, 1 November 2022 (UTC)

@Yehoshua2: Your edit "the real root" was not correct with this singular.–Nomen4Omen (talk) 08:51, 2 November 2022 (UTC)

@Nomen4Omen Could you elaborate please? Yehoshua2 (talk) 14:15, 2 November 2022 (UTC)

I thought that I did in the new version of § Cardano's formula. But nevertheless:

There are the two real roots

1. ${\sqrt[{3}]{u_{1}}}$   and
2. ${\sqrt[{3}]{u_{2}}}$,

each of which has to be multiplied by a root of unity.

So that finally, there are the three roots (of the cubic equation):

1. ${\sqrt[{3}]{u_{1}}}+{\sqrt[{3}]{u_{2}}}$   and
2. $\varepsilon _{1}{\sqrt[{3}]{u_{1}}}+\varepsilon _{2}{\sqrt[{3}]{u_{2}}}$   and
3. $\varepsilon _{2}{\sqrt[{3}]{u_{1}}}+\varepsilon _{1}{\sqrt[{3}]{u_{2}}}$.

If this isn't sufficiently clear, please improve!

–Nomen4Omen (talk) 14:54, 2 November 2022 (UTC)

Possible pedagogical point: some real cubics have real roots that require cube roots of complex numbers

Here is a possible pedagogical point about the cubic formula that I vaguely recall but I do not have a source for. Consider x^3 - 3x - 1 = 0, with 3 real roots. One (all 3?) of the roots is a = t + 1 / t where t = cbrt((1 + i sqrt(3)) / 2). Suppose that a is given by the principle cube root. In order to compute a, a real number, I have to consider cube roots of complex numbers. If I want to find the real and imaginary parts of t, I still have to consider cube roots of complex numbers. I believe that there does not exist any closed form for a (radicals and "i" allowed) that does not involve taking the cube root of a complex number. (This may be a deep theorem). In this sense, there is something unsatisfying about the cubic formula.

Edit: I missed that this is "casus irreducibilis," which I ctrl+f'ed 3 times once I knew the formal term for it.

This just can't be right

This article includes the phrase, "Babylonian (20th to 16th centuries BC)". That just can't be right, but I don't know what it should be. AWCzarnik (talk) — Preceding undated comment added 04:58, 17 July 2022 (UTC)

38.73.247.37 (talk) 21:42, 27 May 2022 (UTC)

Application of Lagrange resolvent roots

Equations (85) - (91) of Wolfram MathWorld's CubicFormula webpage develop the roots of the reduced cubic using Lagranges theory. I checked the algebra of equation (91). A Fourier transform is not referenced, just algebra. Janet Heine Barrett, in her article "The Roots of Early Group Theory in the Works of Lagrange" (2017) [7], references the algebra needed to transform a reduced cubic equation into a Lagrange resolvent equation. [op. cit., page 7]. As suggested, I checked the algebra. 72.89.23.208 (talk) 15:17, 21 November 2022 (UTC)

There is wrong use of algebraically in the second paragraph

So the problem is algebraically is not only in radicals. Algebraic number is any root of any polinomial, not only those that is in radicals, i.e. a solvable root. Remove the word "algebraically". Valery Zapolodov (talk) 05:01, 15 April 2023 (UTC)

Correct:algebraically ⊋ radicals.

Abel–Ruffini theorem is mentioned further down. −Nomen4Omen (talk) 06:49, 15 April 2023 (UTC)

This is not the use of the word which is wrong, but the article to which it is linked. Here,"algebraically" means "using only algebra". So, I have changed to "Algebra" the target of the wikilink. D.Lazard (talk) 08:20, 15 April 2023 (UTC)

I agree with Doctor Lazard Hu741f4 (talk) 16:14, 15 April 2023 (UTC)

But as User:Valery Zapolodov did point out: For users without deep algebraic pre-knowledge, the formulation suggests the equivalence of:

algebraical ⇔ coefficients, the four basic arithmetic operations, radicals

This impression is made a little bit weaker by the wording "more precisely" instead of "that is" (as it was until 18:35, April 14, 2023‎). User:D.Lazard is right in saying: «"algebraically" means "using only algebra"». –Nomen4Omen (talk) 16:50, 15 April 2023 (UTC)

Cubic formula requires some algebra knowledge, we are not in 20th century where it was tought very early. It means only that algebra that is needed to write out the solution is more complex, Bring radical or hypergeometric functions, and that is before factoring and finding which (first found numerically) roots are solvable and what are not (using splitting field to find the class/ Galois group of it). It is very hard, see e.g.: https://mathematica.stackexchange.com/search?q=solvable+quintic Even when it is solvable using irrational coefficients complicates everything and in fact just in 2022 new formula was invented: https://math.stackexchange.com/questions/4439694/can-a-quintic-equation-be-solved-without-needing-a-piece-of-paper-the-size-of-a/4442541#4442541 Valery Zapolodov (talk) 18:38, 15 April 2023 (UTC)

Sign of the discriminant

There is still an error in the sign of the discriminant in the Cardano section. In the section defining the discriminant, the discriminant for the depressed cubic is asserted to be

$-\left(4\,p^{3}+27\,q^{2}\right).$

However, in the section on Cardano's formula, the discriminant $\Delta$ has the opposite sign. It is important to make the signs consistent because of the section relating the sign of the discriminant $\Delta$ to the number of real roots.Thomas Hales (talk) 00:20, 24 May 2023 (UTC)

In § Cardano's formula, it is not said that $\Delta$ is the discriminant. Nevertheless, I have added a footnote for clarifying this. D.Lazard (talk) 08:12, 24 May 2023 (UTC)

Why was my change reversed by D.Lazard?

What did I do wrong? 121.210.212.82 (talk) 10:47, 2 September 2023 (UTC)

This is explained in the edit summary: Awfully formatted, wrong in the case of three real roots, poor duplication of proofs given later in the article. D.Lazard (talk) 11:31, 2 September 2023 (UTC)

I worked so hard on it! 121.210.212.82 (talk) 12:27, 2 September 2023 (UTC)

Doesn't matter. If it's wrong, it's wrong; if it's badly formatted, it's badly formatted; if it's poorly written, it's poorly written. Please stop trying to insert it.--Jasper Deng (talk) 08:20, 3 September 2023 (UTC)

Use of multiple fonts for the same symbols

To: D.Lazard 18:10, 1 April 2024

You have twice reverted a valid correction I made to this article, for no justifiable reason. The notations v and v and $v$ produce different fonts for what is supposed to be the same symbol, and all three of these were being used previously, depending on where the symbol was appearing: in the text, in an inline expression, or in an offset equation.

And the same occurred for several other math symbols. This is confusing for the reader, and obscures what is being stated. I made the effort to carefully correct these in order to make the notation consistent. But for no justifiable reason, you have reverted these edits, twice, and also incorrectly stated that there was an identification made between the symbols "A" , "B" and "C", when there was not (at least, not by me in this correction).

It is not a matter of preferring one "style" over another, but of making the article more clearly readable by not using different fonts, in different locations, for the same symbol. Please undo your reversions, and be more careful about undoing valid revisions made by others, who are only trying to improve the article by correcting notational inconsistencies. You have twice reverted to previous, erroneous versions that need to be corrected all over again. KarlJacobi (talk) 22:31, 1 April 2024 (UTC)

I have already answered to a similar post on my talk page (at User talk:D.Lazard#About the Cubic equation article). Please, be more careful about calling "valid revisions", revisions that contain blatant errors, and go against the Manual of Style. D.Lazard (talk) 10:19, 2 April 2024 (UTC)

You clearly have no interest in seeing the article's inconsistent use of various

fonts for the same symbols corrected, nor allowing any constructive edits by

other editors. You apparently wish to "guard" the editing of this article as though

it were your own private territory, and view contributions by other editors,

however well founded, as "alien incursions". There were no "blatant errors" in any

of the revisions I made; they were all valid corrections of inconsistencies in notation or

improvements of unclearly worded passages. I am not prepared to waste more time pursuing

this sort of trivial quarrel in the face of such absurd territoriality. Let the article

remain as you insist it remain - with all its errors, inconsistencies, imprecision

and poor English well guarded. KarlJacobi (talk) 15:28, 2 April 2024 (UTC)

Cardano's Formula and "correct solution"

In the Cardano's Formula section there is a paragraph discussing the solution in the form of t=C-p/(3C) and states that this is the "correct solution". First off, there is no difference between the solution of t=u + v and t=u-p/(3u). Secondly since there is more than adequate discussion before these statements regarding the function u there is no need to confuse people by introducing a new term "C". All of this is about getting the 3 solutions for the equation t^3 + pt + q=0. To further the point, it is interesting to note that for p=- 4 and q= 2 the solution immediately turns to complex functions but results in 3 real solutions whether one uses t=u + v or t=u - p/(3u), t = -2.214, 0.539 and 1.675. Take out the discussion involving "C". Chettery (talk) 11:51, 17 June 2024 (UTC)

What is said in the article is that the solution t=u - p/(3u) is alway correct, while the solution t=u+v is correct only when p and q are real numbers and $4p^{3}+27q^{2}>0.$ Indeed, t and u are cube roots, and if $4p^{3}+27q^{2}>0,$ it is supposed that they are real cube roots. In all the other cases, t and u are cube roots of a two complex numbers that have 3 complex cube roots each. This gives a total of 9 values for u+v, such that only three of them are roots. D.Lazard (talk) 13:05, 17 June 2024 (UTC)

I suppose that is a typo when you refer to t=u instead of t=u + v. t=u is not a solution of the cubic equation. Now back to your claim concerning 4 p^3 +27 q^2 > 0. I included in my above submittal the solutions for 4 p^2+27q^2 < 0 using both t= u + v and t=u-p/(3u). Chettery (talk) 14:54, 17 June 2024 (UTC)

I have fixed my typo. In the above example, your solutions are only approximations, since you give rational numbers while the roots are irrational numbers. Moreover, u and v are complex numbers, and, given u, you do not say how you chose the corrresponding v among the three possible values. D.Lazard (talk) 17:02, 17 June 2024 (UTC)

The solutions I provided were -2.214, 0.539 and 1.675. These are only approximate as to have been rounded off to 3 decimal places. The solutions to 8 decimal places are actually 1.67513087, 0.539188873, and -2.21431974. If you do not believe me try inputting one at a time into the equation with p= - 4 and q=2. The solutions u + v were the three out of the nine that satisfied the original equation and so were selected. Chettery (talk) 17:36, 17 June 2024 (UTC)

That is exactly the point. The method which you advocate gives 6 incorrect solutions along with the correct ones; the other method gives only the correct solutions. JBW (talk) 18:45, 17 June 2024 (UTC)

Not so. It does not appear that you have actually done any of these calculations. Chettery (talk) 18:50, 17 June 2024 (UTC)

I am disappointed to see that after sufficient time to do your homework you have still not made the necessary correction to eliminate C-p/(3C) and the corresponding statements. I suspect that an assumption has been made that because the equation now has a single function C, that that function will take the same value at both locations in the expression at the same time. This is not true. The second position has u now representing v and to get all three solutions requires that u may have to take a differing 1 of the 3 values at each location. The final proof for all this is to get 3 solutions that satisfy the original governing equation. If you used a reference for your statement on C, maybe I should take this issue to the reference, please state the reference. But if this is totally your assumption and you printed it without first proving it, then I am disappointed. Chettery (talk) 15:08, 20 June 2024 (UTC)

Your statement "What is said in the article is that the solution t=u - p/(3u) is always correct" assumes that u in both places, u and 1/u, will take the same value of the three values available for each of the three solutions. In fact, only for the first solution is this true. For the second solution, u will take the 2nd value while the u in 1/u term takes the third value and for the third solution this is reversed. Do the calculations instead of making these incorrect statements, eliminate the incorrect statement "always correct". Chettery (talk) 18:45, 28 June 2024 (UTC)

Possible Addition to Cubic Equation

Not sure how to put this in the present document. Looking for suggestions.

Restricting to real or complex numbers, the general cubic equation can always be reduced to the depressed equation with the substitution of x = t-b/3.

The Cardano formula of this depressed cubic equation appears to be able to handle most every combination of the equation coefficients p and q.  For example, with the depressed equation if the term (q^2/4+p^3/27) is negative, the use of complex numbers is most interesting as it offers the complete solution only requiring sorting through the produced offerings to find the answer.  The Cardano solution in the above case offers the sum of two functions each taken to the 1/3 power. If we expressed each function using complex exponentials, we arrive at the form (r^(1/3) [cos⁡(θ/3+2πk/3)+i sin⁡(θ/3+2πk/3) ],k=0,1,2) for each of the summed functions.  As can be seen by varying “k”, there will be 3 terms for each function offering 9 possible combinations. It is interesting to note that of the 9 possible combinations, at least 3 combinations will always render the complete solution of the governing equation. The first solution is usually with k=0 in both functions. The second and third solutions are typically with k=2 in one function and k=3 in the other and vice versa for the third and final solution.

When (q^2/4+p^3/27) is positive requires no further discussion. Chettery (talk) 13:01, 26 June 2024 (UTC)

All of this is more accurately discussed in the article. D.Lazard (talk) 13:26, 26 June 2024 (UTC)

Strange, I see no reference to the exponential form in the article. Did you actually read my comments? Chettery (talk) 14:00, 26 June 2024 (UTC)

Polar form of complex numbers is not mentioned because it is of no help for finding and discussing the solutions. The sentence "All of this is more accurately discussed in the article" is a polite form of "All the correct part of your post is more accurately discussed in the article". In particular, "The first solution is usually with k=0 in both functions" is wrong when all solutions are real. D.Lazard (talk) 15:14, 26 June 2024 (UTC)

With the depressed equation t^3+pt+q=0 and the quantity q^2/4+p^3/27 being negative, the solution starts with complex numbers, but the final result are all real numbers. For instance, for p= - 4 and q=2 we end up with the expression (r^(1/3) [cos⁡(θ/3+2πk/3)+i sin⁡(θ/3+2πk/3)] where θ equals 130.5 degrees for u and -130.5 degrees for v. With r= 1.539 the above expressions yield a final real number because the imaginary portions cancel and is equal to 0.67513 with k=0. I am only trying to improve on your document so I am stumped by the resistance. Chettery (talk) 17:20, 26 June 2024 (UTC)

Sum and product of roots

I think it would be useful to state somewhere that the sum of the three roots always equals $-{\frac {b}{a}}$ and the product of the three roots always equals $-{\frac {d}{a}}$, but I don't know which section this information would best belong in. Grendon84 (talk) 23:18, 15 July 2024 (UTC)

These are Vieta's formulas, which are mentioned and used at several places. Nevertheless, it could be useful to mention these explicitely. Maybe in section § Discriminant, since both the discriminant and Vieta's formulas are relations between the roots and the coefficients. D.Lazard (talk) 08:15, 16 July 2024 (UTC)