Added to Favorites

Popular Searches

Definitions

- This article discusses cubic equations in one variable. For a discussion of cubic equations in two variables, see elliptic curve.

In mathematics, a cubic function is a function of the form

- $f(x)=ax^3+bx^2+cx+d,,$

where a is nonzero; or in other words, a polynomial of degree three. The derivative of a cubic function is a quadratic function. The integral of a cubic function is a quartic function.

If you set $f(x)=0$, you get a cubic equation of the form:

- $ax^3+bx^2+cx+d=0\; ,$

where

- $ane\; 0\; ,$

(if a = 0, then the equation becomes a quadratic equation).

(if both a and b = 0, then the equation becomes a linear equation).

Usually, the coefficients $a$, $b$, $c$, $d$ are real numbers. However, most of the theory is also valid if they belong to a field of characteristic other than two or three.

Solving a cubic equation amounts to finding the roots of a cubic function.

In the 11th century, the Persian poet-mathematician, Omar Khayyám (1048–1131), made significant progress in the theory of cubic equations. In an early paper he wrote regarding cubic equations, he discovered that a cubic equation can have more than one solution, that it cannot be solved using earlier compass and straightedge constructions, and found a geometric solution which could be used to get a numerical answer by consulting trigonometric tables. In his later work, the Treatise on Demonstration of Problems of Algebra, he wrote a complete classification of cubic equations with general geometric solutions found by means of intersecting conic sections.

In the 12th century, another Persian mathematician, Sharaf al-Dīn al-Tūsī (1135-1213), wrote the Al-Mu'adalat (Treatise on Equations), which dealt with eight types of cubic equations with positive solutions and five types of cubic equations which may not have positive solutions. He used what would later be known as the "Ruffini-Horner method" to numerically approximate the root of a cubic equation. He also developed the concepts of a derivative function and the maxima and minima of curves in order to solve cubic equations which may not have positive solutions. He understood the importance of the discriminant of the cubic equation and used an early version of Cardano's formula to find algebraic solutions to certain types of cubic equations.

In the early 16th century, the Italian mathematician Scipione del Ferro (1465-1526) found a method for solving a class of cubic equations, namely those of the form x^{3} + 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 to him at that time. Del Ferro kept his achievement secret until just before his death, when he told his student Antonio Fiore about it.

In 1530, Niccolò Tartaglia (1500-1557) received two problems in cubic equations from Zuanne da Coi and announced that he could solve them. He was soon challenged by Fiore, which led to a famous contest between the two. Each contestant had to put up a certain amount of money and to propose a number of problems for his rival to solve. Whoever solved more problems within 30 days would get all the money. Tartaglia received questions in the form x^{3} + mx = n, for which he had worked out a general method. Fiore received questions in the form x^{3} + mx^{2} = n, which proved to be too difficult for him to solve, and Tartaglia won the contest.

Later, Tartaglia was persuaded by Gerolamo Cardano (1501-1576) to reveal his secret for solving cubic equations. In 1539, Tartaglia did so only on the condition that Cardano would never reveal it and that if he did reveal a book about cubics, that he would give Tartaglia time to publish. Some years later, Cardano learned about Ferro's prior work and published Ferro's method in his book Ars Magna in 1545, meaning Cardano gave Tartaglia 6 years to publish his results (with credit given to Tartaglia for an independent solution). Cardano's promise with Tartaglia stated that he not publish Tartaglia's work, and Cardano felt he was publishing del Ferro's, so as to get around the promise. Nevertheless, this led to a challenge to Cardano by Tartaglia, which Cardano denied. The challenge was eventually accepted by Cardano's student Lodovico Ferrari (1522-1565). Ferrari did better than Tartaglia in the competition, and Tartaglia lost both his prestige and income .

Cardano noticed that Tartaglia's method sometimes required him to extract the square root of a negative number. He even included a calculation with these complex numbers in Ars Magna, but he did not really understand it. Rafael Bombelli studied this issue in detail and is therefore often considered as the discoverer of complex numbers.

- $Delta\; =\; -4b^3d\; +\; b^2c^2\; -\; 4ac^3\; +\; 18abcd\; -\; 27a^2d^2.\; ,$

- If Δ > 0, then the equation has three distinct real roots.
- If Δ < 0, then the equation has one real root and a pair of complex conjugate roots.
- If Δ = 0, then (at least) two roots coincide. It may be that the equation has a double real root and another distinct single real root; alternatively, all three roots coincide yielding a triple real root. A possible way to decide between these subcases is to compute the resultant of the cubic and its second derivative: a triple root exists if and only if this resultant vanishes.

See also: multiplicity of a root of a polynomial

We first divide the standard equation by the leading coefficient to arrive at an equation of the form

- $x^3\; +\; ax^2\; +\; bx\; +c\; =\; 0.\; qquad\; (1)$

- $t^3\; +\; pt\; +\; q\; =\; 0,\; quadmbox\{where\; \}\; p\; =\; b\; -\; frac\{a^2\}3\; quadmbox\{and\}quad\; q\; =\; c\; +\; frac\{2a^3-9ab\}\{27\}.\; qquad\; (2)$

Suppose that we can find numbers u and v such that

- $-u^3-v^3\; =\; q\; quadmbox\{and\}quad\; -uv\; =\; frac\{p\}\{3\}.\; quad\; (3)$

- $t\; =\; v\; +\; u,\; ,$

- $(v+u)^3-3uv(v+u)+(-u^3-v^3)=0\; .$

The system (3) can be solved by solving the second equation for v, which gives

- $v\; =\; -frac\{p\}\{3u\}.$

- $-\; u^3\; +\; frac\{p^3\}\{27u^3\}\; =\; q.$

- $27u^6\; +\; 27qu^3\; -\; p^3\; =\; 0,.$

- $u^\{3\}=-\{qover\; 2\}pm\; sqrt\{\{q^\{2\}over\; 4\}+\{p^\{3\}over\; 27\}\}$

- $u=sqrt[3]\{-\{qover\; 2\}pm\; sqrt\{\{q^\{2\}over\; 4\}+\{p^\{3\}over\; 27\}\}\}.\; quad\; (4)$

- $x=-frac\{p\}\{3u\}+u-\{aover\; 3\}.$

In summary, for the cubic equation

- $x^3\; +\; ax^2\; +\; bx\; +c\; =\; 0$

- $x=-frac\{p\}\{3u\}+u-\{aover\; 3\}$

- $p\; =\; b\; -\; frac\{a^2\}3$

- $q\; =\; c\; +\; frac\{2a^3-9ab\}\{27\}$

- $u=sqrt[3]\{-\{qover\; 2\}pm\; sqrt\{\{q^\{2\}over\; 4\}+\{p^\{3\}over\; 27\}\}\}.$

Although this method is simple and elegant, it fails for the case of three real roots, e.g. when:

- $D\; <\; 0,\; D\; =\; left\; (\{q\; over\; 2\}\; right\; )^2\; +\; left\; (\{p\; over\; 3\}\; right\; )^3$

Actually this method can work for the case where D < 0 and all other cases if we use all three cube roots of u and v above whether real or complex. This is important historically because finding solutions this way helped lead to the acceptance of complex numbers. Unfortunately things then were a bit more complicated.

$text\{We\; know\; that\; \}!!~!!text\{\; \}!!~!!text\{\; \}!!~!!text\{\; \}!!~!!text\{\; \}u^\{3\}=-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}~~~~or~~~~-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}$ .

$text\{But\; since\; u\; and\; v\; must\; satisfy\; \}!!~!!text\{\; \}!!~!!text\{\; \}!!~!!text\{\; \}!!~!!text\{\; \}-u^\{3\}-v^\{3\}=q~~~~and~~~~-uv=frac\{p\}\{3\}$

then one can show that if

$text\{\; \}~~~~~~u^\{3\}=-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}~~~~then~~~~v^\{3\}=-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}$

or vice versa it doesn't matter which so we will chose this way.

Writing out the three cube roots we get

$u=left\{\; begin\{align\}\; \&\; sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; \&\; left(-frac\{1\}\{2\}+ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; \&\; left(-frac\{1\}\{2\}-ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; end\{align\}\; right.~~~and~~~v=left\{\; begin\{align\}\; \&\; sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; \&\; left(-frac\{1\}\{2\}+ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; \&\; left(-frac\{1\}\{2\}-ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; end\{align\}\; right.$

Remembering $~t=u+v~$ we get only three possible values for t because only three combinations of u and v are possible if $-uv=frac\{p\}\{3\}$ is to remain valid as it must - so

$t=left\{\; begin\{align\}\; \&\; sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}+sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; \&\; left(-frac\{1\}\{2\}+ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}+left(-frac\{1\}\{2\}-ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; \&\; left(-frac\{1\}\{2\}-ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}+left(-frac\{1\}\{2\}+ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}\; end\{align\}\; right.$

and because $x=t-frac\{a\}\{3\}$ the three possible values of x are

$x=left\{\; begin\{align\}\; \&\; sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}+sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}-frac\{a\}\{3\}\; \&\; left(-frac\{1\}\{2\}+ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}+left(-frac\{1\}\{2\}-ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}-frac\{a\}\{3\}\; \&\; left(-frac\{1\}\{2\}-ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}+sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}+left(-frac\{1\}\{2\}+ifrac\{sqrt\{3\}\}\{2\}\; right)sqrt[3]\{-frac\{q\}\{2\}-sqrt\{frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}\}\}-frac\{a\}\{3\}\; end\{align\}\; right.$

and the cubic is solved whether $D=frac\{q^\{2\}\}\{4\}+frac\{p^\{3\}\}\{27\}$ is positive, negative or zero.

If D is positive then there is one real and two complex roots.

If D is negative then there are three real roots.

If D = 0 then there is one real root (a triple root) or two real roots (a single and a double root.)

- $s\_0\; =\; r\_0\; +\; r\_1\; +\; r\_2,,$

- $s\_1\; =\; r\_0\; +\; zeta\; r\_1\; +\; zeta^2\; r\_2,,$

- $s\_2\; =\; r\_0\; +\; zeta^2\; r\_1\; +\; zeta\; r\_2.,$

- $r\_0\; =\; (s\_0\; +\; s\_1\; +\; s\_2)/3,,$

- $r\_1\; =\; (s\_0\; +\; zeta^2\; s\_1\; +\; zeta\; s\_2)/3,,$

- $r\_2\; =\; (s\_0\; +\; zeta\; s\_1\; +\; zeta^2\; s\_2)/3.,$

- $(z-s\_1^3)(z-s\_2^3)\; qquad\; (5)$

- $\{z\}^\{2\}+\; left(-9,ba+2,\{a\}^\{3\}+27,c\; right)\; z+\; left(\{a\}^\{2\}-3,bright)^\{3\}.$

- $frac92,ab-\{a\}^\{3\}-\; frac\{27\}\{2\},c\; pm\; frac\{1\}\{2\},sqrt\{D\},$

- $left\; (x-rright\; )left\; (x^2+(a+r)x+b+ar+r^2\; right\; )\; =\; x^3+ax^2+bx+c.$

- $frac12\; left(-a-r\; pm\; sqrt\{-3r^2-2ar+a^2-4b\}right)$

The formula for finding the roots of a cubic function, based on Cardano's method, is fairly complicated. Therefore, it is common to use the rational root test or a numerical solution instead.

If we have

- $f(x)\; =\; ax^3\; +\; bx^2\; +\; cx\; +\; d\; =\; a(x\; -\; x\_1)(x\; -\; x\_2)(x\; -\; x\_3),$

- $q\; =\; frac\{3ac-b^2\}\{9a^2\}$

and

- $r\; =\; frac\{9abc\; -\; 27a^2d\; -\; 2b^3\}\{54a^3\},$

we define the discriminant:

- $Delta\; =\; q^3+r^2.,$

There are two distinct cases:

- $Delta\; geq\; 0,$

- in which case there is one real root and 2 imaginary. We define:

- $s\; =\; sqrt[3]\{r\; +\; sqrt\{q^3+r^2\}\}$

- and

- $t\; =\; sqrt[3]\{r\; -\; sqrt\{q^3+r^2\}\}.$

- $Delta\; <\; 0,$

- in which case we have 3 real roots. We express the complex quantity $r\; +\; isqrt\{-Delta\}$ in polar form:

- $r\; +\; isqrt\{-Delta\}\; =\; (rho,\; theta),$

- $rho\; =\; sqrt\{r^2-Delta\},$

- $theta\; =\; acos(frac\{r\}\{rho\}),$

- and we define:

- $s\; =\; (sqrt[3]\{rho\},\; frac\{theta\}\{3\}),$

- and

- $t\; =\; (sqrt[3]\{rho\},\; -frac\{theta\}\{3\}).,$

In both cases, the solutions are

- $x\_1\; =\; s+t-frac\{b\}\{3a\},$

- $x\_2\; =\; -frac\{1\}\{2\}(s+t)-frac\{b\}\{3a\}+frac\{sqrt\{3\}\}\{2\}(s-t)i,$

- $x\_3\; =\; -frac\{1\}\{2\}(s+t)-frac\{b\}\{3a\}-frac\{sqrt\{3\}\}\{2\}(s-t)i.$

If we have a cubic equation which is already in depressed form, we may write it as $,x^3\; -\; 3px\; -\; q\; =\; 0$. Substituting $x\; =\; sqrt\{p\}\; z$ we obtain $z^3\; -\; 3z\; -\; p^\{-frac\{3\}\{2\}\}q\; =\; 0$ or equivalently

- $z^3\; -\; 3z\; =\; p^\{-frac\{3\}\{2\}\}q\; .$

- $r\_0\; =\; sqrt\{p\},C\_\{1over3\}(p^\{-frac\{3\}\{2\}\}q),,$

- $r\_1\; =\; -sqrt\{p\},C\_\{1over3\}(-p^\{-frac\{3\}\{2\}\}q),,$

- $r\_2\; =\; -r\_0\; -\; r\_1\; .$

- $x^3\; +\; ax^2\; +\; bx\; +c\; =\; 0\; qquad\; (1)$

- $t\_\{a;b;c\}\; =\; p^\{-frac\{3\}\{2\}\}q\; =\; -frac\{2a^3-9ab+27c\}\{(a^2-3b)^\{3/2\}\}.$

This gives us the solutions to (1) as

- $r\_0\; =\; sqrt\{p\},C\_\{1over3\}(t\_\{a;b;c\})-\{aover\; 3\}\; ,,$

- $r\_1\; =\; -sqrt\{p\},C\_\{1over3\}(-t\_\{a;b;c\})-\{aover\; 3\},,$

- $r\_2\; =\; -r\_0\; -\; r\_1\; -\; a\; .$

Suppose the coefficients of (1) are real. If s is the quantity q/r from the section on real roots, then s = t^{2}; hence 0 < s < 4 is equivalent to −2 < t < 2, and in this case we have a polynomial with three distinct real roots, expressed in terms of a real function of a real variable, quite unlike the situation when using cube roots. If s > 4 then either t > 2 and $C\_\{1over3\}(t)$ is the sole real root, or t < −2 and $-C\_\{1over3\}(-t)$ is the sole real root.
If s < 0 then the reduction to Chebyshev polynomial form has given a t which is a pure imaginary number; in this case $iC\_\{1over3\}(-it)-iC\_\{1over3\}(it)$ is the sole real root. We are now evaluating a real root by means of a function of a purely imaginary argument; however we can avoid this by using the function

- $S\_\{1over3\}(t)\; =\; iC\_\{1over3\}(-it)-iC\_\{1over3\}(it)\; =\; 2\; operatorname\{sinh\}left(operatorname\{arcsinh\}left(\{tover2\}right)/3right),,$

which is a real function of a real variable with no singularities along the real axis. If a polynomial can be reduced to the form $x^3\; +\; 3x\; -t$ with real t, this is a convenient way to solve for its roots.

- $y^2\; =\; x(x-a)(x-b),$

where $0\; <\; a\; <\; b$ is called a bipartite cubic. This is from the theory of elliptic curves.

You can graph a bipartite cubic on a graphing device by graphing the function

- $f(x)\; =\; sqrt\{x(x-a)(x-b)\},$

corresponding to the upper half of the bipartite cubic. It is defined on

- $(0,a)\; cup\; (b,+infty).,$

- W. S. Anglin; & J. Lambek (1995). "Mathematics in the Renaissance", in The heritage of Thales, Ch. 24. Springers.
- Lucye Guilbeau (1930). "The History of the Solution of the Cubic Equation", Mathematics News Letter 5 (4), p. 8-12.
- R.W.D. Nickalls (1993). A new approach to solving the cubic: Cardan's solution revealed, The Mathematical Gazette, 77:354–359.

- Calculator for solving Cubics (also solves Quartics and Quadratics)
- Tartaglia's work (and poetry) on the solution of the Cubic Equation at Convergence
- Cubic Equation Solver
- Quadratic, cubic and quartic equations on MacTutor archive.
- Cardano solution calculator as java applet at some local site. Only takes natural coefficients.
- Graphic explorer for cubic functions With interactive animation, slider controls for coefficients
- On Solution of Cubic Equations at Holistic Numerical Methods Institute
- Dave Auckly, Solving the quartic with a pencil American Math Monthly 114:1 (2007) 29--39
- "Cubic Equation" by Eric W. Weisstein, The Wolfram Demonstrations Project, 2007.

Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer)

This article is licensed under the GNU Free Documentation License.

Last updated on Tuesday October 07, 2008 at 22:04:23 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

This article is licensed under the GNU Free Documentation License.

Last updated on Tuesday October 07, 2008 at 22:04:23 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

Copyright © 2015 Dictionary.com, LLC. All rights reserved.