# Talk:Cubic equation

## spanish version

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

## 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

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

This has three distinct real roots (0, 1, and -1). However, according to the page, it has ${\displaystyle a_{0}=1}$, ${\displaystyle a_{2}=-1}$, and ${\displaystyle a_{1}=a_{3}=0}$, and thus ${\displaystyle q=4}$ and ${\displaystyle r=1}$ and thus ${\displaystyle s=4}$ 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 ${\displaystyle a_{n}}$ is the coefficient of ${\displaystyle x^{n}}$ (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.)

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
${\displaystyle t={\frac {(2a_{1}^{3}-9a_{0}a_{1}a_{2}+27a_{0}^{2}a_{3})^{2}}{(a_{1}^{2}-3a_{0}a_{2})^{3}}}}$
${\displaystyle t={\frac {(2a_{2}^{3}-9a_{1}a_{2}a_{3}+27a_{0}^{2}a_{3})^{2}}{(a_{2}^{2}-3a_{1}a_{3})^{3}}}.}$
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 ${\displaystyle a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}}$ 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 ${\displaystyle a_{n}}$ 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 ${\displaystyle C_{\frac {1}{3}}}$ is confusing: we have ${\displaystyle C_{\frac {1}{3}}(0)=2\operatorname {arccosh} {\frac {1}{3}}}$ which is not a real number, but both the power series and the statement that this function solves ${\displaystyle x^{3}-3x=t}$ 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 ${\displaystyle {\sqrt {3}}}$ for ${\displaystyle C_{\frac {1}{3}}(0)}$. 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 ${\displaystyle {\sqrt {3}}}$, ${\displaystyle -{\sqrt {3}}}$, 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:
Here's a discussion of Omar Khayyam's geometric method, which is closely related to this business:
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:

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

which has roots 0, -1, and 1. According to the revised page's notation, this has: ${\displaystyle \alpha _{3}=1}$, ${\displaystyle \alpha _{1}=-1}$, and ${\displaystyle \alpha _{2}=\alpha _{0}=0}$. Then, according to the page, ${\displaystyle \Delta =-4<0}$, 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.

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

For the form:

ax3 + bx2 + cx + d = 0.

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

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)
${\displaystyle E={1 \over 2}{\begin{vmatrix}0&3A&2B\\3A&B&C\\B&C&3D\end{vmatrix}}={9ABC-27A^{2}D-2B^{3} \over 2}{\mbox{ and its sub-determinant }}}$
${\displaystyle F=-{\begin{vmatrix}3A&B\\B&C\end{vmatrix}}={\begin{vmatrix}B&3A\\C&B\end{vmatrix}}=B^{2}-3AC{\mbox{ form Depressed Cubic for }}3AX+B:}$
${\displaystyle (3AX+B)^{3}-3F(3AX+B)-2E=0={\Big (}AX^{3}+BX^{2}+CX+D{\Big )}*3^{3}A^{2}\quad (1)}$
${\displaystyle (t_{1}+t_{2})^{3}-3t_{1}*t_{2}(t_{1}+t_{2})-{\Big (}t_{1}^{3}+t_{2}^{3}{\Big )}=0;{\mbox{ if }}3AX+B=t_{1}+t_{2}\Rightarrow F=t_{1}*t_{2}{\mbox{ and }}}$
${\displaystyle 2E=t_{1}^{3}+t_{2}^{3}{\mbox{ along with }}F^{3}=t_{1}^{3}*t_{2}^{3}{\mbox{ form Quadratic for }}t^{3}:}$
${\displaystyle {\Big (}t^{3}{\Big )}^{2}-2Et^{3}+F^{3}=0\Rightarrow (t_{1,2})_{k}={\sqrt[{3}]{E\pm \ {\sqrt {E^{2}-F^{3}}}}}*e^{\pm \ i{2k\pi \over 3}}}$
${\displaystyle {\mbox{ where k = 0; 1; 2 and }}e^{\pm \ i{2k\pi \over 3}}=\operatorname {cos} {2k\pi \over 3}\pm \ i\operatorname {sin} {2k\pi \over 3}={\sqrt[{3}]{1}}{\mbox{ (Euler formula) }}}$
${\displaystyle X_{k}={\frac {-B+{\sqrt[{3}]{E+{\sqrt {E^{2}-F^{3}}}}}*e^{+i{2k\pi \over 3}}+{\sqrt[{3}]{E-{\sqrt {E^{2}-F^{3}}}}}*e^{-i{2k\pi \over 3}}}{3A}}\quad (2)}$
${\displaystyle S={B \over 3A}\Rightarrow X_{0}=-S+{\frac {{\sqrt[{3}]{E+{\sqrt {E^{2}-F^{3}}}}}+{\sqrt[{3}]{E-{\sqrt {E^{2}-F^{3}}}}}}{3A}}{\mbox{ real root and }}}$
${\displaystyle X_{1,2}=-S-{t_{1}+t_{2}\pm {\sqrt {-3}}(t_{1}-t_{2}) \over 6A}=-S-{X_{0}+S \over 2}\pm \ {\sqrt {{\frac {F}{3A^{2}}}-3{\Bigg (}{X_{0}+S \over 2}{\Bigg )}^{2}}}}$
${\displaystyle 0\leq E^{2}\gtreqless F^{3}\gtreqless 0}$ discriminates 8 cases; X1 & X2 nature in 5 Marginal ones is:
${\displaystyle {\mbox{I. }}F=0{\mbox{ and }}E=0\Rightarrow X_{0}=X_{1}=X_{2}=-S{\mbox{ (all real and equal) }}}$
${\displaystyle {\mbox{ II. }}F=0\Rightarrow X_{0}=-S+{{\sqrt[{3}]{2E}} \over 3A}{\mbox{ and }}X_{1,2}=-S-{{\sqrt[{3}]{2E}} \over 6A}{\Big (}1\pm \ i{\sqrt {3}}{\Big )}{\mbox{ (conjugate) }}}$
${\displaystyle {\mbox{ III. }}F^{3}=E^{2}\Rightarrow X_{0}=-S+{2{\sqrt[{3}]{E}} \over 3A}{\mbox{ and }}X_{1}=X_{2}=-S-{{\sqrt[{3}]{E}} \over 3A}{\mbox{ (real and equal) }}}$
${\displaystyle {\mbox{ IV. }}E=0{\mbox{ and }}F<0\Rightarrow X_{0}=-S{\mbox{ and }}X_{1,2}=-S\pm \ i{{\sqrt {-3F}} \over 3A}{\mbox{ (conjugate) }}}$
${\displaystyle {\mbox{ V. }}E=0{\mbox{ and }}F>0\Rightarrow X_{0}=-S{\mbox{ and }}X_{1,2}=-S\pm \ {{\sqrt {3F}} \over 3A}{\mbox{ (real) }}}$
As shown above Xk-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).
${\displaystyle R={2E \over 3|E|A}{\sqrt {f*F}}\gtrless 0{\mbox{ if }}E\gtrless 0{\mbox{ where }}f={\frac {F}{|F|}}=\pm \ 1\Rightarrow f*F>0}$
${\displaystyle x={X+S \over R}\Rightarrow 3AX+B=3ARx={2Ex \over |E|}{\sqrt {f*F}}{\mbox{ inserted into (1) gives }}}$
${\displaystyle {\mbox{ Primeval Cubic: }}4x^{3}\mp \ 3x={\cfrac {|E|}{\sqrt {f*F^{3}}}}=g={\frac {e^{3z}\pm \ e^{-3z}}{2}}\geq 0\quad (3)\Rightarrow }$
${\displaystyle x_{k}={\frac {{\sqrt[{3}]{g+{\sqrt {g^{2}\mp \ 1}}}}*e^{i{2k\pi \over 3}}+{\sqrt[{3}]{g-{\sqrt {g^{2}\mp \ 1}}}}*e^{-i{2k\pi \over 3}}}{2}}={\frac {e^{z+i{2k\pi \over 3}}\pm \ e^{-z-i{2k\pi \over 3}}}{2}}}$
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, X1,2= – R(sinhu ± i 3½coshu)/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, X1,2 = – R(coshv ± i 3½sinhv)/2 – S; g = 1 → III.
VIII. F > 0 i.e. f = + 1 ≥ g = cos3w = 4cos3w – 3cosw → z(g) = i w = i arccos(⅓g),
xk = cos(w + k×120°), Xk = 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)
${\displaystyle Q={\begin{vmatrix}b&4a&4a&3b\\0&0&b&4a\\4e&b&4d&4c\\b&4a&0&0\end{vmatrix}},\quad R={\begin{vmatrix}0&2a&b\\2a&b&c\\b&c&2e\end{vmatrix}},\quad S={\begin{vmatrix}2a&3b\\b&4c\end{vmatrix}}\,}$ ${\displaystyle (4ax+b)^{4}+2S(4ax+b)^{2}+8R(4ax+b)+Q={\Big (}ax^{4}+bx^{3}+cx^{2}+dx+e{\Big )}*4^{4}a^{3}=}$
${\displaystyle ={\Big [}(4ax+b)^{2}-P(4ax+b)+Q_{1}{\Big ]}*{\Big [}(4ax+b)^{2}+P(4ax+b)+Q_{2}{\Big ]}=0\Rightarrow }$ ${\displaystyle P^{2}+2S=Q_{1}+Q_{2},\,{8R \over P}=Q_{1}-Q_{2}\Rightarrow (P^{2}+2S)^{2}-{(8R)^{2} \over P^{2}}=4Q_{1}Q_{2}=4Q}$ ${\displaystyle {\Big [}P^{6}+4SP^{4}+4(S^{2}-Q)P^{2}-(8R)^{2}{\Big ]}*3^{3}=0=(3P^{2}+4S)^{3}-3F(3P^{2}+4S)-2E}$ ${\displaystyle E={1 \over 2}{\begin{vmatrix}0&3&8S\\3&4S&4(S^{2}-Q)\\4S&4(S^{2}-Q)&-3(8R)^{2}\end{vmatrix}}=8(S^{3}-9QS+108R^{2})\Rightarrow F=4(S^{2}+3Q)}$
${\displaystyle P_{k}^{2}={\frac {{\sqrt[{3}]{E+{\sqrt {E^{2}-F^{3}}}}}*e^{+i{2k\pi \over 3}}+{\sqrt[{3}]{E-{\sqrt {E^{2}-F^{3}}}}}*e^{-i{2k\pi \over 3}}-4S}{3}}>0}$
${\displaystyle x={P_{k}-2b \over 8a}\pm \ {1 \over 8a}{\sqrt {{\frac {16R}{P_{k}}}-4S-P_{k}^{2}}}\quad (4)}$
Positive & negative value of Pk (k satisfying Pk2 > 0 should be chosen) results into
2 (two) symmetric pairs of either real or conjugate (if 16R/Pk < Pk2 + 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 [itex]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 ${\displaystyle x^{3}-2x^{2}}$ 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
${\displaystyle \Delta _{2}=\mathrm {res} (f,f'')=-216a^{3}d+72a^{2}bc-16ab^{3}.}$
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 ${\displaystyle d=0}$ 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:
${\displaystyle {\mbox{ Normalized form of Primeval Cubic: }}{\cfrac {x^{3}}{{\tfrac {1}{4}}g}}\mp \ {\cfrac {x}{{\tfrac {1}{3}}g}}=1}$
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)

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.