- Cubic function
Setting ƒ(x) = 0 produces a cubic equation of the form:
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 2 or 3. To solve a cubic equation is to find the roots (zeros) of a cubic function. There are various ways to solve a cubic equation. The roots of a cubic, like those of a quadratic or quartic (fourth degree) function but no higher degree function, can always be found algebraically (as a formula involving simple functions like the square root and cube root functions). The roots can also be found trigonometrically. Alternatively, one can find a numerical approximation of the roots in the field of the real or complex numbers. This may be obtained by any root-finding algorithm, like Newton's method.
Solving cubic equations is a necessary part of solving the general quartic equation, since solving the latter requires solving its resolvent cubic equation.
- 1 History
- 2 Derivative
- 3 Roots of a cubic function
- 4 See also
- 5 Notes
- 6 References
- 7 External links
Cubic equations were known to ancient Greek mathematician Diophantus; even earlier to ancient Babylonians who were able to solve certain cubic equations; and also to the ancient Egyptians. Doubling the cube is the simplest and oldest studied cubic equation, and one which the ancient Egyptians considered to be impossible. Hippocrates reduced this problem to that of finding two mean proportionals between one line and another of twice its length, but could not solve this with a compass and straightedge construction, a task which is now known to be impossible. Hippocrates, Menaechmus and Archimedes are believed to have come close to solving the problem of doubling the cube using intersecting conic sections, though historians such as Reviel Netz dispute whether the Greeks were thinking about cubic equations or just problems that can lead to cubic equations. Some others like T. L. Heath, who translated all Archimedes' works, disagree, putting forward evidence that Archimedes really solved cubic equations using intersections of two cones, but also discussed the conditions where the roots are 0, 1 or 2.
In the 7th century, the Tang dynasty astronomer mathematician Wang Xiaotong in his mathematical treatise titled Jigu Suanjing systematically established and solved 25 cubic equations of the form x3 + px2 + qx = N, 23 of them with , and two of them with q = 0.
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 and stated that it cannot be solved using compass and straightedge constructions. He also found a geometric solution. 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, the Indian mathematician Bhaskara II attempted cubic equations without general success. However, he gave one example of a cubic equation:
x3 + 12x = 6x2 + 35
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 to find algebraic solutions to certain types of cubic equations.
Leonardo de Pisa, also known as Fibonacci (1170–1250), was able to find the positive solution to the cubic equation x3+2x2+10x = 20, using the Babylonian numerals. He gave the result as 1,22,7,42,33,4,40 which is equivalent to: 1+22/60+7/602+42/603+33/604+4/605+40/606.
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 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 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 x3 + mx = n, for which he had worked out a general method. Fiore received questions in the form x3 + mx2 = 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.
Through the quadratic formula the roots of the derivative f′(x) = 3ax2 + 2bx + c are given by
and provide the critical points where the slope of the cubic function is zero. If b2 − 3ac > 0, then the cubic function has a local maximum and a local minimum. If b2 − 3ac = 0, then the cubic's inflection point is the only critical point. If b2 − 3ac < 0, then there are no critical points. In the cases where b2 − 3ac ≤ 0, the cubic function is strictly monotonic.
Roots of a cubic function
The general cubic equation has the form
This section describes how the roots of such an equation may be computed. The coefficients a, b, c, d are generally assumed to be real numbers, but most of the results apply when they belong to any field of characteristic not 2 or 3.
The nature of the roots
Every cubic equation (1) with real coefficients has at least one solution x among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminant,
The following cases need to be considered: 
- If Δ > 0, then the equation has three distinct real roots.
- If Δ = 0, then the equation has a multiple root and all its roots are real.
- If Δ < 0, then the equation has one real root and two nonreal complex conjugate roots.
General formula of roots
For the general cubic equation (1) with real coefficients, the general formula for the roots, in terms of the coefficients, is as follows. Note that the expression under the square root sign in what follows is , where Δ is the above-mentioned discriminant.
However, this formula is applicable without further explanation only when the operand of the square root is non-negative and a,b,c,d are real coefficients. When this operand is real and non-negative, the square root refers to the principal (positive) square root and the cube roots in the formula are to be interpreted as the real ones. Otherwise, there is no real square root and one can arbitrarily choose one of the imaginary square roots (the same one in both parts of the solution for each xi). For extracting the complex cube roots of the resulting complex expression, we have also to choose among three cube roots in each part of each solution, giving nine possible combinations of one of three cube roots for the first part of the expression and one of three for the second. The correct combination is such that the two cube roots chosen for the two terms in a given solution expression are complex conjugates of each other (whereby the two imaginary terms in each solution cancel out).
Another way of writing the solution may be obtained by noting that the proof of above formula shows that the product of the two cube roots is rational. This gives the following formula in which or stands for any choice of the square or cube root, if
If and , the sign of has to be chosen to have .
If and , the three roots are equal:
If Q = 0 and , the above expression for the roots is correct but misleading, hiding the fact that no radical is needed to represent the roots. In fact, in this case, there is a double root,
and a simple root
The next sections describe how these formulas may be obtained.
Reduction to a monic trinomial
Dividing Equation (1) by a and substituting x by (the Tschirnhaus transformation) we get the equation
Any formula for the roots of Equation (2) may be transformed into a formula for the roots of Equation (1) by substituting the above values for p and q and using the relation .
Therefore, only Equation (2) is considered in the following.
We first apply preceding reduction, giving the so-called depressed cubic
We introduce two variables u and v linked by the condition
and substitute this in the depressed cubic (2), giving
At this point Cardano imposed a second condition for the variables u and v:
As the first parenthesis vanishes in (3), we get u3 + v3 = − q and u3v3 = − p3 / 27. Thus u3 and v3 are the two roots of the equation
At this point, Cardano, who did not know complex numbers, supposed that the roots of this equation were real, that is that
Solving this equation and using the fact that u and v may be exchanged, we find
- and .
As these expressions are real, their cube roots are well defined and, like Cardano, we get
The two complex roots are obtained by considering the complex cubic roots; the fact uv is real implies that they are obtained by multiplying one of the above cubic roots by and the other by .
If is not necessarily positive, we have to choose a cube root of u3. As there is no direct way to choose the corresponding cube root of v3, one has to use the relation , which gives
Note that the sign of the square root does not affect the resulting t, because changing it amounts to exchanging u and v. We have chosen the minus sign to have when p = 0 and , in order to avoid a division by zero. With this choice, the above expression for t always works, except when p = q = 0, where the second term becomes 0/0. In this case there is a triple root t = 0.
Note also that in several cases the solutions are expressed with fewer square or cube roots
- If p = q = 0 then we have the triple real root
- If p = 0 and then
- and the three roots are the three cube roots of − q.
- If and q = 0 then
- in which case the three roots are
- Finally if , there is a double root and a simple root which may be expressed rationally in term of p and q, but this expression may not be immediately deduced from the general expression of the roots:
To pass from these roots of t in Equation (2) to the general formulas for roots of x in Equation (1), subtract and replace p and q by their expressions in terms of a,b,c,d.
In his paper Réflexions sur la résolution algébrique des équations ("Thoughts on the algebraic solving of equations"), Joseph Louis Lagrange introduced a new method to solve equations of low degree.
This method works well for cubic and quartic equations, but Lagrange did not succeed in applying it to a quintic equation, because it requires solving a resolvent polynomial of degree at least six. This is explained by the Abel–Ruffini theorem, which proves that such polynomials cannot be solved by radicals. Nevertheless the modern methods for solving solvable quintic equations are mainly based on Lagrange's method.
In the case of cubic equations, Lagrange's method gives the same solution as Cardano's, but avoids its seemingly magical aspect (Why did Cardano choose these auxiliary variables?). Moreover, it may also be applied directly to the general cubic equation (1) without using the reduction to the trinomial equation (2). Nevertheless the computation is much easier with this reduced equation.
This is the discrete Fourier transform of the roots: observe that while the coefficients of the polynomial are symmetric in the roots, in this formula an order has been chosen on the roots, so these are not symmetric in the roots. The roots may then be recovered from the three si by inverting the above linear transformation via the inverse discrete Fourier transform, giving
The polynomial s0 is an elementary symmetric polynomial and is thus equal to − b / a in case of Equation (1) and to zero in case of Equation (2), so we only need to seek values for the other two.
The polynomials s1 and s2 are not symmetric functions of the roots: s0 is invariant, while the two non-trivial cyclic permutations of the roots send s1 to ζs1 and s2 to ζ2s2, or s1 to ζ2s1 and s2 to ζs2 (depending on which permutation), while transposing x1 and x2 switches s1 and s2; other transpositions switch these roots and multiply them by a power of ζ.
Thus, , and s1s2 are left invariant by the cyclic permutations of the roots, which multiply them by ζ3 = 1. Also s1s2 and are left invariant by the transposition of x1 and x2 which exchanges s1 and s2. As the permutation group S3 of the roots is generated by these permutations, it follows that and s1s2 are symmetric functions of the roots and may thus be written as polynomials in the elementary symmetric polynomials and thus as rational functions of the coefficients of the equation. Let and s1s2 = B in these expressions, which will be explicitly computed below.
We have that and are the two roots of the quadratic equation
Thus the resolution of the equation may be finished exactly as described for Cardano's method, with s1 and s2 in place of u and v.
Computation of A and B
Setting E1 = x0 + x1 + x2, E2 = x0x1 + x1x2 + x2x0 and E3 = x0x1x2, the elementary symmetric polynomials, we have, using that ζ3 = 1:
The expression for is the same with ζ and ζ2 exchanged. Thus, using ζ2 + ζ = − 1 we get
and a straightforward computation gives
Similarly we have
When solving Equation (1) we have
- E1 = − b / a, E2 = c / a and E3 = − d / a
With Equation (2), we have E1 = 0, E2 = p and E3 = − q and thus:
- A = − 27q and B = − 3p.
Note that with Equation (2), we have and s1s2 = − 3p, while in Cardano's method we have set x0 = u + v and Thus we have, up to the exchange of u and v:
- s1 = 3u and s2 = 3v.
In other words, in this case, Cardano's and Lagrange's method compute exactly the same things, up to a factor of three in the auxiliary variables, the main difference being that Lagrange's method explains why these auxiliary variables appear in the problem.
Trigonometric (and hyperbolic) method
When a cubic equation has three real roots, the formulas expressing these roots in terms of radicals involve complex numbers. It has been proved that when none of the three real roots is rational—the casus irreducibilis— one cannot express the roots in terms of real radicals. Nevertheless, purely real expressions of the solutions may be obtained using hypergeometric functions, or more elementarily in terms of trigonometric functions, specifically in terms of the cosine and arccosine functions.
The formulas which follow, due to François Viète, are true in general (except when p=0), are purely real when the equation has three real roots, but involve complex cosines and arccosines when there is only one real root.
Starting from Equation (2), t3 + pt + q = 0, let us set The idea is to choose u to make Equation (2) coincide with the identity
In fact, choosing and dividing Equation (2) by we get
Combining with the above identity, we get
and thus the roots are
This formula involves only real terms if p < 0 and the argument of the arccosine is between -1 and 1. The last condition is equivalent to which implies also p < 0. Thus the above formula for the roots involves only real terms if and only if the three roots are real.
Denoting by C(p,q) the above value of t0, and using the inequality for a real number u such that the three roots may also be expressed as
If the three roots are real, we have
All these formulas may be straightforwardly transformed into formulas for the roots of the general cubic equation (1), using the back substitution described in Section Reduction to a monic trinomial.
If p≠0 and the inequalities on the right are not satisfied the formulas remain valid but involve complex quantities.
When , the above values of t0 are sometimes called the Chebyshev cube root. More precisely, the values involving cosines and hyperbolic cosines define, when p = − 3, the same analytic function denoted , which is the proper Chebyshev cube root. The value involving hyperbolic sines is similarly denoted when p = 3.
If the cubic equation ax3 + bx2 + cx + d = 0 with integer coefficients has a rational real root, it can be found using the rational root test: If the root is r = m / n fully reduced, then m is a factor of d and n is a factor of a, so all possible combinations of values for m and n can be checked for whether they satisfy the cubic equation. This is particularly useful when there are three real roots, since in this case as indicated previously the algebraic solution unhelpfully expresses the real roots in terms of complex entities.
If r is any root of the cubic, then we may factor using r to obtain
Hence if we know one root we can find the other two by solving the quadratic ax2 + (b + ar)x + c + br + ar2, giving
for the other two roots.
If there are three real roots and none of them is rational, we have the so-called casus irreducibilis in which the cubic cannot be factored into the product of a linear polynomial and a quadratic polynomial each with real coefficients.
- Linear equation
- Quadratic equation
- Quartic equation
- Quintic equation
- Newton's method
- Spline (mathematics)
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- ^ a b Daniel Lazard, "Solving quintics in radicals", in Olav Arnfinn Laudal, Ragni Piene, The Legacy of Niels Henrik Abel, pp. 207–225, Berlin, 2004,. ISBN 3-5404-3826-2
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- Zucker, I. J., "The cubic equation—A new look at the irreducible case," Mathematical Gazette 92, July 2008, 264-268.
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- Solving a Cubic by means of Moebius transforms
- Interesting derivation of trigonometric cubic solution with 3 real roots
- 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.
- Cubic Formula on PlanetMath
- 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.
- The affine equivalence of cubic polynomials at Dynamic Geometry Sketches
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