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| {{about|cubic equations in one variable|cubic equations in ''two'' variables|cubic plane curve}}
| | To get into it in excel, copy-paste this continued plan down into corpuscle B1. A person's again access an majority of time in abnormal wearing corpuscle A1, the discount in treasures will start in B1.<br><br>If you're purchasing a game your child, appear for an individual that allows several individuals to do together. Gaming might just be a singular activity. Nonetheless, it's important to guide your youngster to be societal, and multiplayer deviate of clans trucos events can do that. They allow siblings as buddies to all park yourself and laugh and play together.<br><br>Gallstones are known as your games primary forex. The Jewels are often purchase resources along via speeding up numerous vitally important tasks. The Other jewels can also be once buy bonus items. Apart from that, additionally, it can let the leader noticeable any undesired debris when you want to obtain a much gems. Players definitely will obtain Gems through rounding out numerous tasks or it could be that using the clash of clans crack available online.<br><br>Up to now, there exists a minimum of social options / abilities with this game i.e. there is not any chat, finding it difficult to team track linked with friends, etc but actually we could expect these to improve soon as Boom Beach continues to remain their Beta Mode.<br><br>Deliver the in-online game songs chance. If, nonetheless, you might end annoyed by using who's soon after one hour or so approximately, don't be worried to mute the telly or personal computer plus play some audio of the very own. You will discover a far more exciting game playing experience performing this and therefore are a whole lot unlikely to get a good frustration from actively actively.<br><br>To defeat higher-level villages, this task aids you to make use of a mixture of troops exactly like Barbarians plus Archers a great bonus those [http://www.britannica.com/search?query=suicide+wall suicide wall] bombers to bust down walls. Goblins can also be a useful [http://Thesaurus.com/browse/feature feature] the combo simply due to the fact attack different buildings. You should understand when you like to begin worrying with reference to higher troops when you have to can''t win battles for Barbarians.<br><br>Why don't we try interpreting the actual abstracts differently. Foresee of it in agreement of bulk with stones to skip 1 moment. Skipping added your time expenses added money, even though you get a enflamed deal. If you have any questions regarding exactly where and how to use clash of clans hack tool; [http://prometeu.net Suggested Online site],, you can speak to us at our own page. Think akin to it as a couple accretion discounts. |
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| [[Image:Polynomialdeg3.svg|thumb|right|210px|Graph of a cubic function with 3 [[real number|real]] [[root of a function|roots]] (where the curve crosses the horizontal axis—where ''y'' = 0). It has 2 [[critical point (mathematics)|critical points]]. Here the function is ''ƒ''(''x'') = (''x''<sup>3</sup> + 3''x''<sup>2</sup> − 6''x'' − 8) / 4.]]
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| In [[mathematics]], a '''cubic function''' is a [[function (mathematics)|function]] of the form
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| :<math>f(x)=ax^3+bx^2+cx+d,\,</math>
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| where ''a'' is nonzero; or in other words, a function defined by a [[polynomial]] of [[Degree of a polynomial|degree]] three. The [[derivative]] of a cubic function is a [[quadratic function]]. The [[integral]] of a cubic function is a [[quartic function]].
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| Setting ''ƒ''(''x'') = 0 produces a cubic [[equation]] of the form:
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| :<math>ax^3+bx^2+cx+d=0.\,</math>
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| Usually, the [[coefficients]] ''a'', ''b'',''c'', ''d'' are real numbers. However, most of the theory is also valid if they belong to any [[field (mathematics)|field]] of [[characteristic (field)|characteristic]] other than 2 or 3.
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| To solve a cubic equation is to find the roots (zeros) of a cubic function.
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| There are various ways to solve a cubic equation. The roots of a cubic function, like those of a quadratic or quartic (fourth degree) function but no higher degree function (by the [[Abel–Ruffini theorem]]), can always be found [[Algebraic function|algebraically]] (as a formula involving simple functions like the [[square root]] and [[cube root]] functions). The roots can also be found [[Trigonometry|trigonometrically]]. Alternatively, one can find a [[numerical approximation]] of the roots in the [[Field (mathematics)|field]] of the [[real number|real]] or [[complex numbers]]. This may be obtained by any [[root-finding algorithm]], like [[Newton's method]].
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| Solving cubic equations is a necessary part of solving the general quartic equation, since solving the latter requires solving its [[resolvent cubic]] equation.
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| ==History==
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| Cubic equations were known to [[Greek mathematics|ancient Greek]] mathematician [[Diophantus]];<ref>Van de Waerden, Geometry and Algebra of Ancient Civilizations, chapter 4, Zurich 1983 ISBN 0-387-12159-5</ref> even earlier to [[Babylonian mathematics|ancient Babylonians]] who were able to solve certain cubic equations;<ref>British Museum BM 85200</ref> and also to the [[Egyptian mathematics|ancient Egyptians]]. [[Doubling the cube]] is the simplest and oldest studied cubic equation, and one which the ancient Egyptians considered to be impossible.<ref>{{Harvtxt|Guilbeau|1930|p=8}} states, "The Egyptians considered the solution impossible, but the Greeks came nearer to a solution."</ref> [[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]],<ref name=Guilbeau>{{Harvtxt|Guilbeau|1930|pp=8–9}}</ref> 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]],<ref name=Guilbeau/> 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|Archimedes']] works, disagree, putting forward evidence that Archimedes really solved cubic equations using intersections of two [[Cone (geometry)|cones]], but also discussed the conditions where the [[root]]s are 0, 1 or 2.<ref>The works of Archimedes, translation by T. L. Heath</ref>
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| [[Image:Graph of cubic polynomial.svg|255px|left|thumb|[[Two-dimensional graph]] of a cubic, the [[polynomial]] ''ƒ''(''x'') = 2''x''<sup>3</sup> − 3''x''<sup>2</sup> − 3''x'' + 2.]]
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| 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 <math>x^3+px^2+qx=N</math>, 23 of them with <math>p,q \ne 0</math>, and two of them with <math>q = 0</math>.<ref>{{Citation
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| |first= Yoshio
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| |last= Mikami
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| |author-link= Yoshio Mikami
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| |title= The Development of Mathematics in China and Japan
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| |chapter= Chapter 8 Wang Hsiao-Tung and Cubic Equations
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| |pages= 53–56
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| |publisher= Chelsea Publishing Co.
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| |location= New York
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| |year= 1974
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| |edition= 2nd
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| |origyear= 1913
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| |isbn= 978-0-8284-0149-4
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| |doi=}}</ref>
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| In the 11th century, the [[Persian literature|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 [[#Omar Khayyám's solution|geometric solution]].<ref>A paper of Omar Khayyam, Scripta Math. 26 (1963), pages 323–337</ref><ref>In {{MacTutor|id=Khayyam|title=Omar Khayyam}} one may read ''This problem in turn led Khayyam to solve the cubic equation'' ''x''<sup>3</sup> + 200''x'' = 20''x''<sup>2</sup> + 2000 ''and he found a positive root of this cubic by considering the intersection of a rectangular hyperbola and a circle. An approximate numerical solution was then found by interpolation in trigonometric tables''. The ''then'' in the last assertion is erroneous and should, at least, be replaced by ''also''. The geometric construction was perfectly suitable for Omar Khayyam, as it occurs for solving a problem of geometric construction. At the end of his article he says only that, for this geometrical problem, if approximations are sufficient, then a simpler solution may be obtained by consulting [[Generating trigonometric tables|trigonometric tables]]. Textually: ''If the seeker is satisfied with an estimate, it is up to him to look into the table of chords of Almagest, or the table of sines and versed sines of Mothmed Observatory.'' This is followed by a short description of this alternate method (seven lines).</ref> 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 section]]s.<ref>J. J. O'Connor and E. F. Robertson (1999), [http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Khayyam.html Omar Khayyam], [[MacTutor History of Mathematics archive]], states, "Khayyam himself seems to have been the first to conceive a general theory of cubic equations."<!-- quotation is in MacTutor--></ref><ref>{{Harvtxt|Guilbeau|1930|p=9}} states, "Omar Al Hay of Chorassan, about 1079 AD did most to elevate to a method the solution of the algebraic equations by intersecting conics."</ref>
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| In the 12th century, the Indian mathematician Bhaskara II attempted the solution of cubic equations without general success. However, he gave one example of a cubic equation:<ref>Datta and Singh, History of Hindu Mathematics, p. 76,Equation of Higher Degree; Bharattya Kala Prakashan, Delhi, India 2004 ISBN 81-86050-86-8</ref>
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| <math>x^3+12x=6x^2+35</math>
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| In the 12th century, another [[Mathematics in medieval Islam|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's rule|Ruffini]]-[[Horner scheme|Horner]] method" to [[Numerical analysis|numerically]] approximate the [[root of a function|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.<ref>{{MacTutor|id=Al-Tusi_Sharaf|title=Sharaf al-Din al-Muzaffar al-Tusi}}</ref> He understood the importance of the [[discriminant]] of the cubic equation to find algebraic solutions to certain types of cubic equations.<ref>{{Citation |first=J. L. |last=Berggren |year=1990 |title=Innovation and Tradition in Sharaf al-Din al-Tusi's Muadalat |journal=Journal of the American Oriental Society |volume=110 |issue=2 |pages=304–309 |doi= 10.2307/604533}}</ref>
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| Leonardo de Pisa, also known as [[Fibonacci]] (1170–1250), was able to find the positive solution to the cubic equation ''x''<sup>3</sup> + 2''x''<sup>2</sup> + 10''x'' = 20, using the [[Babylonian numerals]]. He gave the result as 1,22,7,42,33,4,40 (equivalent to 1 + 22/60 + 7/60<sup>2</sup> + 42/60<sup>3</sup> + 33/60<sup>4</sup> + 4/60<sup>5</sup> + 40/60<sup>6</sup>),<ref>"The life and numbers of Fibonacci" [http://pass.maths.org.uk/issue3/fibonacci/index.html], ''Plus Magazine''</ref> which differs from the correct value by only about three trillionths.
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| 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''<sup>3</sup> + ''mx'' = ''n''. In fact, all cubic equations can be reduced to this form if we allow ''m'' and ''n'' to be negative, but [[negative number]]s 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.
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| [[Image:Niccolò Tartaglia.jpg|thumb|160px|Niccolò Fontana Tartaglia]]
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| 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''<sup>3</sup> + ''mx'' = ''n'', for which he had worked out a general method. Fiore received questions in the form ''x''<sup>3</sup> + ''mx''<sup>2</sup> = ''n'', which proved to be too difficult for him to solve, and Tartaglia won the contest.
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| 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 (Gerolamo Cardano)|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.<ref>{{Citation |last=Katz |first=Victor |title=A History of Mathematics |page=220 |location=Boston |publisher=Addison Wesley |year=2004 |isbn= }}</ref>
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| 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 number]]s 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.
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| [[François Viète]] (1540–1603) independently derived the trigonometric solution for the cubic with three real roots, and [[René Descartes]] (1596–1650) extended the work of Viète.<ref name=Nickalls/>
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| ==Derivative==
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| {{Cubic_graph_special_points.svg}}
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| Through the [[quadratic formula]] the [[root of a function|roots]] of the [[derivative]] ''f'' ′(''x'') = 3''ax''<sup>2</sup> + 2''bx'' + ''c'' are given by
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| :<math style="vertical-align:-70%;">x=\frac{-b \pm \sqrt {b^2-3ac}}{3a}</math> <!-- please do not change this into x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}. The expression is correct. -->
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| and provide the [[critical point (mathematics)|critical points]] where the slope of the cubic function is zero. If ''b''<sup>2</sup> − 3''ac'' > 0, then the cubic function has a [[Maxima and minima|local maximum]] and a [[Maxima and minima|local minimum]]. If ''b''<sup>2</sup> − 3''ac'' = 0, then the cubic's [[inflection point]] is the only critical point. If ''b''<sup>2</sup> − 3''ac'' < 0, then there are no critical points. In the cases where ''b''<sup>2</sup> − 3''ac'' ≤ 0, the cubic function is strictly [[monotonic]].
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| ==Roots of a cubic function==
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| The general cubic equation has the form
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| :<math>ax^3+bx^2+cx+d=0 \qquad(1)</math>
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| with <math>a\neq 0\,.</math>
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| 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 number]]s, but most of the results apply when they belong to any [[Field (mathematics)|field]] of [[Characteristic (algebra)|characteristic]] not 2 or 3.
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| ===The nature of the roots===
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| Every cubic equation (1) with [[real number|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]],
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| :: <math> \Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2. \,</math>
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| The following cases need to be considered:
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| <ref>{{citation
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| |title=Integers, polynomials, and rings
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| |first1=Ronald S.
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| |last1=Irving
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| |publisher=Springer-Verlag New York, Inc.
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| |year=2004
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| |isbn=0-387-40397-3
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| |url=http://books.google.com/?id=B4k6ltaxm5YC}}, [http://books.google.com/books?id=B4k6ltaxm5YC&pg=PA154 Chapter 10 ex 10.14.4 and 10.17.4, pp. 154–156]</ref>
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| * If Δ > 0, then the equation has three distinct real roots.
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| * If Δ = 0, then the equation has a [[multiple root]] and all its roots are real.
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| * If Δ < 0, then the equation has one real root and two nonreal complex conjugate roots.
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| ''For information about the location in the [[complex plane]] of the roots of a polynomial of any degree, including degree three, see [[Properties of polynomial roots]] and [[Routh–Hurwitz stability criterion]]''
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| ===General formula for roots===
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| For the general cubic equation
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| :<math>a x^3 + b x^2 + c x + d = 0</math>
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| the general formula for the roots, in terms of the coefficients, is as follows:<ref>{{cite book
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| |title=Numerical Recipes in Fortran 77: The Art of Scientific Computing
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| |first1=William H.
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| |last1=Press
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| |first2=William T.
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| |last2=Vetterling
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| |publisher=Cambridge University Press
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| |year=1992
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| |isbn=0-521-43064-X
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| |page=179
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| |url=http://books.google.com/books?id=gn_4mpdN9WkC}}, [http://books.google.com/books?id=gn_4mpdN9WkC&pg=PA179 Extract of page 179]
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| </ref><ref>Output of [[Maple (software)|Maple]]'s function "solve".</ref>
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| :<math>x_k = - \frac{1}{3a}\left(b\ +\ u_k C\ +\ \frac{\Delta_0}{u_kC}\right)\ , \qquad k \in \{1,2, 3\}</math>
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| where
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| :<math>u_1 = 1\ ,\qquad u_2 = {-1 + i\sqrt{3} \over 2}\ ,\qquad u_3 = {-1 - i\sqrt{3} \over 2}</math>
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| are the three [[root of unity|cube roots of unity]], and where
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| :<math>C = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}} \qquad \qquad {\color{white}.}</math> (see [[#Special cases | below]] for special cases)
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| with
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| :<math>\Delta_0 = b^2-3 a c</math>
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| :<math>\Delta_1 = 2 b^3-9 a b c+27 a^2 d</math>
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| and
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| :<math>\Delta_1^2 - 4 \Delta_0^3 = -27\,a^2\,\Delta\ ,</math> where <math>\Delta</math> is the [[discriminant]] discussed above.
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| In these formulae, <math>\sqrt{~~}</math> and <math>\sqrt[3]{~~}</math> denote any choice for the square or cube roots. Changing of choice for the square root amounts to exchanging <math>x_2</math> and <math>x_3</math>. Changing of choice for the cube root amounts to [[cyclic permutation|circularly permute]] the roots. Thus the freeness of choosing a determination of the square or cube roots corresponds exactly to the freeness for numbering the roots of the equation.
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| Four centuries ago, [[Gerolamo Cardano]] proposed a similar formula (see [[#Cardano's_method | below]]), which still appears in many textbooks:
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| :<math>x_k = - \frac{1}{3a}(b\ +\ u_k C\ +\ \bar u_k \bar C)</math>
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| where
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| :<math>\bar C = \sqrt[3]{\frac{\Delta_1 - \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}</math>
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| and <math>\bar u_k</math> is the [[complex conjugate]] of <math>u_k</math> (note that <math>C\bar C=\Delta_0</math>).
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| However, this formula is applicable without further explanation ''only'' when ''a'', ''b'', ''c'', ''d'' are [[real number]]s and the operand of the square root <math>\Delta_1^2 - 4 \Delta_0^3 \ge 0</math> is non-negative. 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 everywhere in the solution). 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).
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| The next sections describe how these formulas may be obtained.
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| ====Special cases====
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| If <math>\Delta \neq 0</math> and <math>\Delta_0 = 0,</math> the sign of <math>\sqrt{\Delta_1^2 - 4 \Delta_0^3}=\sqrt{\Delta_1^2} </math> has to be chosen to have <math>C \neq 0,</math> that is one should define <math>\sqrt{\Delta_1^2} =\Delta_1,</math> whichever is the sign of <math>\Delta_1.</math>
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| If <math>\Delta = 0</math> and <math>\ \Delta_0 = 0,</math> the three roots are equal:
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| :<math>x_1=x_2=x_3=-\frac{b}{3a}.</math>
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| If <math>\Delta=0</math> and <math> \Delta_0 \neq 0,</math> 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,
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| :<math> x_1=x_2=\frac{9ad-bc}{2\Delta_0},</math>
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| and a simple root
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| :<math> x_3=\frac{4abc-9a^2d-b^3}{a\Delta_0}.</math>
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| ===Reduction to a depressed cubic===
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| Dividing Equation (1) by <math>a</math> and substituting <math>x</math> by <math> t-\frac{b}{3a}</math> (the [[Tschirnhaus transformation]]) we get the equation
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| :<math>t^3+pt+q=0 \qquad(2)</math>
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| where
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| :<math> \begin{align}
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| p=&\frac{3ac-b^2}{3a^2}\\
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| q=&\frac{2b^3-9abc+27a^2d}{27a^3}.
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| \end{align}
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| </math>
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| The left hand side of equation (2) is a [[monic polynomial|monic]] [[trinomial]] called a '''depressed cubic'''.
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| Any formula for the roots of a depressed cubic may be transformed into a formula for the roots of Equation (1) by substituting the above values for <math>p</math> and <math>q</math> and using the relation <math> x=t-\frac{b}{3a}</math>.
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| Therefore, only Equation (2) is considered in the following.
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| ===Cardano's method===
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| The solutions can be found with the following method due to [[Scipione del Ferro]] and [[Niccolò Fontana Tartaglia|Tartaglia]], published by [[Gerolamo Cardano]] in 1545.<ref>{{Harvnb|Jacobson|2009|p=210}}</ref>
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| This method applies to the depressed cubic
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| :<math> t^3 + pt + q = 0\,. \qquad (2)</math>
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| We introduce two variables ''u'' and ''v'' linked by the condition
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| :<math>u+v=t\,</math>
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| and substitute this in the depressed cubic (2), giving
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| :<math> u^3+v^3+(3uv+p)(u+v)+q=0 \qquad (3)\,</math>.
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| At this point Cardano imposed a second condition for the variables ''u'' and ''v'':
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| :<math> 3uv+p=0\,</math>.
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| As the first parenthesis vanishes in (3), we get <math> u^3+v^3=-q</math> and <math> u^3v^3=-p^3/27</math>. Thus <math> u^3</math> and <math> v^3</math> are the two roots of the equation
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| :<math> z^2 + qz - {p^3\over 27} = 0\,.</math>
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| At this point, Cardano, who did not know [[complex numbers]], supposed that the roots of this equation were real, that is that <math> \frac{q^2}{4}+\frac{p^3}{27} >0\,. </math>
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| Solving this equation and using the fact that <math> u</math> and <math> v</math> may be exchanged, we find
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| :<math> u^{3}=-{q\over 2} + \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}</math> and <math>v^{3}=-{q\over 2} - \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}</math>.
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| As these expressions are real, their cube roots are well defined and, like Cardano, we get
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| :<math> t_1=u+v=\sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} +\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}</math>
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| The two complex roots are obtained by considering the complex cube roots; the fact <math> uv</math> is real implies that they are obtained by multiplying one of the above cube roots by <math>\,\tfrac{-1}{2} + i\tfrac{\sqrt{3}}{2}\,</math> and the other by <math>\,\tfrac{-1}{2} - i\tfrac{\sqrt{3}}{2}\,</math>.
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| If <math> \frac{q^2}{4}+\frac{p^3}{27}\, </math> is not necessarily positive, we have to choose a cube root of <math>u^3</math>. As there is no direct way to choose the corresponding cube root of <math>v^3</math>, one has to use the relation <math>v=-\frac{p}{3u}</math>, which gives
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| :<math> u=\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} \qquad (4) </math>
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| and
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| :<math>t=u-\frac{p}{3u}\,.</math>
| |
| | |
| Note that the sign of the square root does not affect the resulting <math>t</math>, because changing it amounts to exchanging <math>u</math> and <math>v</math>. We have chosen the minus sign to have <math>u\ne 0</math> when <math>p = 0</math> and <math>q\ne 0</math>, in order to avoid a division by zero. With this choice, the above expression for <math>t</math> always works, except when <math>p = q=0</math>, where the second term becomes 0/0. In this case there is a triple root <math> t=0</math>.
| |
| | |
| Note also that in several cases the solutions are expressed with fewer square or cube roots
| |
| | |
| :If <math>p=q=0</math> then we have the triple real root
| |
| ::<math>t=0.\,</math>
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| :If <math>p=0</math> and <math>q\ne 0</math> then
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| ::<math>u=-\sqrt[3]{q} \text{ and } v = 0</math>
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| :and the three roots are the three cube roots of <math>-q</math>.
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| :If <math>p\ne 0</math> and <math>q=0</math> then
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| ::<math>u=\sqrt{{p\over 3}} \qquad \text{and} \qquad v=-\sqrt{{p\over 3}},</math>
| |
| :in which case the three roots are
| |
| ::<math>t=u+v=0 , \qquad t=\omega_1u-{p\over 3\omega_1u}=\sqrt{-p} , \qquad t={u\over \omega_1}-{\omega_1p\over 3u}=-\sqrt{-p} ,</math>
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| :where
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| ::<math>\omega_1=e^{i\frac{2\pi}{3}}=-\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i.</math>
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| :Finally if <math>4p^3+27q^2=0 \text{ and } p\ne 0</math>, there is a double root and a simple root which may be expressed rationally in term of <math>p \text{ and } q </math>, but this expression may not be immediately deduced from the general expression of the roots:
| |
| ::<math> t_1=t_2= -\frac{3q}{2p}\quad \text{and} \quad t_3=\frac{3q}{p}\,.</math>
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| | |
| To pass from these roots of <math>t</math> in Equation (2) to the general formulas for roots of <math>x</math> in Equation (1), subtract <math>\frac{b}{3a}</math> and replace <math> p</math> and <math> q</math> by their expressions in terms of <math>a,b,c,d</math>.
| |
| | |
| === Vieta's substitution ===
| |
| Starting from the depressed cubic
| |
| :<math>t^3 + pt + q = 0,</math>
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| we make the following substitution, known as Vieta's substitution:
| |
| :<math>t = w - \frac{p}{3w}</math>
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| | |
| This results in the equation
| |
| :<math>w^3 + q - \frac{p^3}{27w^3} = 0.</math>
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| | |
| Multiplying by ''w''<sup>3</sup>, it becomes a [[sextic equation]] in ''w'', which is in fact a quadratic equation in ''w''<sup>3</sup>:
| |
| :<math>w^6 + qw^3 - \frac{p^3}{27} = 0</math>
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| | |
| The [[quadratic formula]] allows to solve it in ''w''<sup>3</sup>. If ''w''<sub>1</sub>, ''w''<sub>2</sub> and ''w''<sub>3</sub> are the three [[cube root]]s of one of the solutions in ''w''<sup>3</sup>, then the roots of the original depressed cubic are
| |
| :<math>t_1 = w_1 - \frac{p}{3w_1}, \quad t_2 = w_2 - \frac{p}{3w_2}\quad\text{and} \quad t_3 = w_3 - \frac{p}{3w_3}.</math>
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| | |
| ===Lagrange's method===<!--This section is linked from [[Lagrange resolvents]] and [[Lagrange resolvent]] (singular) -->
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| | |
| 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.<ref name="efei">{{citation
| |
| |title=Elliptic functions and elliptic integrals
| |
| |first1=Viktor
| |
| |last1=Prasolov
| |
| |first2=Yuri
| |
| |last2=Solovyev
| |
| |publisher=AMS Bookstore
| |
| |year=1997
| |
| |isbn=978-0-8218-0587-9
| |
| |url=http://books.google.com/?id=fcp9IiZd3tQC
| |
| }}, [http://books.google.com/books?id=fcp9IiZd3tQC&pg=PA134#PPA134,M1 §6.2, p. 134]</ref><ref>{{citation
| |
| |first=Morris
| |
| |last=Kline
| |
| |title=Mathematical Thought from Ancient to Modern Times
| |
| |publisher=Oxford University Press US
| |
| |year=1990
| |
| |isbn=978-0-19-506136-9
| |
| |url=http://books.google.com/?id=aO-v3gvY-I8C
| |
| }}, [http://books.google.com/books?id=aO-v3gvY-I8C&printsec=frontcover#PPA597,M1 Algebra in the Eighteenth Century: The Theory of Equations]</ref><ref name="laz">Daniel Lazard, "Solving quintics in radicals", in [[Olav Arnfinn Laudal]], [[Ragni Piene]], ''The Legacy of Niels Henrik Abel'', pp. 207–225, Berlin, 2004,. ISBN 3-540-43826-2</ref> 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.<ref name="laz" />
| |
| | |
| In the case of cubic equations, Lagrange's method gives the same solution as Cardano's, where the latter may seem almost magical to the modern reader. But Cardano explains in his book ''Ars Magna'' how he arrived at the idea of considering the unknown of the cubic equation as a sum of two other quantities, by drawing attention to a geometrical problem that involves two cubes of different size. Lagrange's method may also be applied directly to the general cubic equation (1) without using the reduction to the depressed cubic equation (2). Nevertheless the computation is much easier with this reduced equation.
| |
| | |
| Suppose that ''x''<sub>0</sub>, ''x''<sub>1</sub> and ''x''<sub>2</sub> are the roots of equation (1) or (2), and define <math>\zeta = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i</math> (a complex cube root of 1, i.e. a [[root of unity|primitive third root of unity]]) which satisfies the relation <math>\zeta^2+\zeta+1=0</math>. We now set
| |
| :<math>s_0 = x_0 + x_1 + x_2,\,</math>
| |
| :<math>s_1 = x_0 + \zeta x_1 + \zeta^2 x_2,\,</math>
| |
| :<math>s_2 = x_0 + \zeta^2 x_1 + \zeta x_2.\,</math>
| |
| 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 ''s''<sub>''i''</sub> by inverting the above linear transformation via the inverse discrete Fourier transform, giving
| |
| :<math>x_0 = \tfrac13(s_0 + s_1 + s_2),\,</math>
| |
| :<math>x_1 = \tfrac13(s_0 + \zeta^2 s_1 + \zeta s_2),\,</math>
| |
| :<math>x_2 = \tfrac13(s_0 + \zeta s_1 + \zeta^2 s_2).\,</math>
| |
| | |
| The polynomial <math>s_0</math> is an [[elementary symmetric polynomial]] and is thus equal to <math>-b/a</math> 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 <math>s_1</math> and <math>s_2</math> are not [[symmetric polynomial|symmetric functions]] of the roots: <math>s_0</math> is invariant, while the two non-trivial [[cyclic permutation]]s of the roots send <math>s_1</math> to <math>\zeta s_1</math> and <math>s_2</math> to <math>\zeta^2 s_2</math>, or <math>s_1</math> to <math>\zeta^2 s_1</math> and <math>s_2</math> to <math>\zeta s_2</math> (depending on which permutation), while transposing <math>x_1</math> and <math>x_2</math> switches <math>s_1</math> and <math>s_2</math>; other transpositions switch these roots and multiply them by a power of <math>\zeta.</math>
| |
| | |
| Thus, <math>s_1^3</math>, <math>s_2^3</math> and <math>s_1 s_2</math> are left invariant by the cyclic permutations of the roots, which multiply them by <math>\zeta^3=1</math>. Also <math>s_1 s_2</math> and <math>s_1^3+s_2^3</math> are left invariant by the transposition of <math>x_1</math> and <math>x_2</math> which exchanges <math>s_1</math> and <math>s_2</math>. As the [[permutation group]] <math>S_3</math> of the roots is generated by these permutations, it follows that <math>s_1^3+s_2^3</math> and <math>s_1 s_2</math> are [[symmetric polynomial|symmetric functions]] of the roots and may thus be written as polynomials in the [[elementary symmetric polynomial]]s and thus as [[rational function]]s of the coefficients of the equation. Let <math>s_1^3+s_2^3=A</math> and <math>s_1 s_2=B</math> in these expressions, which will be explicitly computed below.
| |
| | |
| We have that <math>s_1^3</math> and <math>s_2^3</math> are the two roots of the quadratic equation
| |
| :<math>z^2-Az+B^3 = 0 \,.</math>
| |
| Thus the resolution of the equation may be finished exactly as described for Cardano's method, with <math>s_1</math> and <math>s_2</math> in place of <math>u</math> and <math>v</math>.
| |
| | |
| ====Computation of ''A'' and ''B''====
| |
| | |
| Setting <math>E_1=x_0+x_1+x_2</math>, <math>E_2=x_0x_1+x_1x_2+x_2x_0</math> and <math>E_3=x_0x_1x_2</math>, the elementary symmetric polynomials, we have, using that <math>\zeta^3=1</math>:
| |
| :<math>s_1^3=x_0^3+x_1^3+x_2^3+3\zeta (x_0^2x_1+x_1^2x_2+x_2^2x_0) +3\zeta^2 (x_0x_1^2+x_1x_2^2+x_2x_0^2) +6x_0x_1x_2\,. </math>
| |
| The expression for <math>s_2^3</math> is the same with <math>\zeta</math> and <math>\zeta^2</math> exchanged. Thus, using <math>\zeta^2+\zeta=-1</math> we get
| |
| :<math>
| |
| A=s_1^3+s_2^3=2(x_0^3+x_1^3+x_2^3)-3(x_0^2x_1+x_1^2x_2+x_2^2x_0+x_0x_1^2+x_1x_2^2+x_2x_0^2)+12x_0x_1x_2\,,
| |
| </math>
| |
| and a straightforward computation gives
| |
| :<math>
| |
| A=s_1^3+s_2^3=2E_1^3-9E_1E_2+27E_3\,.
| |
| </math>
| |
| | |
| Similarly we have
| |
| :<math>
| |
| B=s_1s_2=x_0^2+x_1^2+x_2^2+(\zeta+\zeta^2)(x_0x_1+x_1x_2+x_2x_0)=E_1^2-3E_2\,.
| |
| </math>
| |
| | |
| When solving Equation (1) we have
| |
| :<math>E_1=-b/a</math>, <math>E_2=c/a</math> and <math>E_3=-d/a</math>
| |
| With Equation (2), we have <math>E_1=0</math>, <math>E_2=p</math> and <math>E_3=-q</math> and thus:
| |
| :<math>A=-27q</math> and <math>B=-3p</math>.
| |
| | |
| Note that with Equation (2), we have <math>x_0 = \tfrac13(s_1 + s_2)</math> and <math>s_1s_2=-3p</math>, while in Cardano's method we have set <math>x_0 = u+v</math> and <math>uv=-\frac13p\,.</math>
| |
| Thus we have, up to the exchange of <math>u</math> and <math>v</math>:
| |
| :<math>s_1=3u</math> and <math> s_2=3v</math>.
| |
| 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===<!-- linked from redirect [[Chebyshev cube root]] -->
| |
| | |
| 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 function]]s,<ref>Zucker, I. J., "The cubic equation — a new look at the irreducible case", ''Mathematical Gazette'' 92, July 2008, 264–268.</ref> or more elementarily in terms of [[trigonometry|trigonometric functions]], specifically in terms of the [[cosine]] and [[arccosine]] functions.
| |
| | |
| The formulas which follow, due to [[François Viète]],<ref name=Nickalls>{{Citation |last=Nickalls |first=R. W. D. |title=Viète, Descartes and the cubic equation |url=http://www.nickalls.org/dick/papers/maths/descartes2006.pdf |journal=[[Mathematical Gazette]] |volume=90 |date=July 2006 |pages=203–208 |doi= }}</ref> 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), <math>t^3+pt+q=0</math>, let us set <math>t=u\cos\theta\,.</math> The idea is to choose <math>u</math> to make Equation (2) coincide with the identity
| |
| :<math>4\cos^3\theta-3\cos\theta-\cos(3\theta)=0\,.</math>
| |
| In fact, choosing <math>u=2\sqrt{-\frac{p}{3}}</math> and dividing Equation (2) by <math>\frac{u^3}{4}</math> we get
| |
| :<math>4\cos^3\theta-3\cos\theta-\frac{3q}{2p}\sqrt{\frac{-3}{p}}=0\,.</math>
| |
| Combining with the above identity, we get
| |
| :<math>\cos(3\theta)=\frac{3q}{2p}\sqrt{\frac{-3}{p}}</math>
| |
| and thus the roots are<ref name="crc">{{citation
| |
| |title=CRC Standard Mathematical Tables
| |
| |first=Samuel
| |
| |last=Shelbey
| |
| |publisher=CRC Press
| |
| |year=1975
| |
| |isbn=0-87819-622-6
| |
| }}</ref>
| |
| :<math>t_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right) \quad \text{for} \quad k=0,1,2 \,.</math>
| |
| | |
| This formula involves only real terms if <math>p<0</math> and the argument of the arccosine is between −1 and 1. The last condition is equivalent to <math> 4p^3+27q^2\leq 0\,,</math> which implies also <math>p<0</math>. Thus the above formula for the roots involves only real terms if and only if the three roots are real.
| |
| | |
| Denoting by <math>C(p,q)</math> the above value of ''t''<sub>0</sub>, and using the inequality <math>-\pi\le \arccos(u) \le \pi</math> for a real number ''u'' such that <math>-1\le u\le 1\,,</math> the three roots may also be expressed as
| |
| :<math>t_0=C(p,q),\qquad t_2=-C(p,-q), \qquad t_1=-t_0-t_2\,.</math>
| |
| If the three roots are real, we have
| |
| :<math>t_0\ge t_1\ge t_2\,.</math>
| |
| | |
| 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 depressed cubic|Reduction to a depressed cubic]].
| |
| | |
| When there is only one real root (and ''p'' ≠ 0), it may be similarly represented using [[hyperbolic function]]s, as<ref>These are Formulas (80) and (83) of Weisstein, Eric W. 'Cubic Formula'. From MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/CubicFormula.html, rewritten for having a coherent notation.</ref><ref>Holmes, G. C., "The use of hyperbolic cosines in solving cubic polynomials", ''[[Mathematical Gazette]]'' 86. November 2002, 473–477.</ref>
| |
| :<math>t_0=-2\frac{|q|}{q}\sqrt{-\frac{p}{3}}\cosh\left(\frac{1}{3}\operatorname{arcosh}\left(\frac{-3|q|}{2p}\sqrt{\frac{-3}{p}}\right)\right) \quad \text{if } \quad 4p^3+27q^2>0 \text{ and } p<0\,,</math>
| |
| :<math>t_0=-2\sqrt{\frac{p}{3}}\sinh\left(\frac{1}{3}\operatorname{arsinh}\left(\frac{3q}{2p}\sqrt{\frac{3}{p}}\right)\right) \quad \text{if } \quad p>0\,.</math>
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| If ''p'' ≠ 0 and the inequalities on the right are not satisfied the formulas remain valid but involve complex quantities.
| |
| | |
| When <math> p=\pm 3</math>, the above values of <math>t_0</math> are sometimes called the '''Chebyshev cube root.'''<ref>Abramowitz, Milton; Stegun, Irene A., eds. ''Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables'', Dover (1965), chap. 22 p. 773</ref> More precisely, the values involving cosines and hyperbolic cosines define, when <math> p=-3</math>, the same [[analytic function]] denoted <math>C_{\frac13}(q)</math>, which is the proper Chebyshev cube root. The value involving hyperbolic sines is similarly denoted <math>S_{\frac13}(q),</math> when <math> p=3</math>.
| |
| | |
| ===Factorization===
| |
| | |
| If the cubic equation <math>ax^3 + bx^2 + cx +d=0</math> 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.
| |
| | |
| The rational root test may also be used for a cubic equation with rational coefficients: by multiplication by the [[lowest common denominator]]) of the coefficients, one gets an equation with integer coefficients which has exactly the same roots.
| |
| | |
| The rational root test is particularly useful when there are three real roots because the algebraic solution unhelpfully expresses the real roots in terms of complex entities. The rational root test is also helpful in the presence of one real and two complex roots because it allows all of the roots to be written without the use of cube roots.
| |
| | |
| If ''r'' is any root of the cubic, then we may factor out (''x''–''r'' ) using [[polynomial long division]] to obtain
| |
| :<math>\left (x-r\right )\left (ax^2+(b+ar)x+c+br+ar^2 \right ) = ax^3+bx^2+cx+d\,.</math>
| |
| Hence if we know one root we can find the other two by using the [[quadratic formula]] to solve the quadratic <math>ax^2+(b+ar)x+c+br+ar^2 </math>, giving
| |
| :<math> \frac{-b-ra \pm \sqrt{b^2-4ac-2abr-3a^2r^2}}{2a} </math>
| |
| for the other two roots.
| |
| | |
| ===Geometric interpretation of the roots===
| |
| | |
| ====Three real roots====
| |
| [[File:Trigonometric interpretation of a cubic equation with three real roots.svg|thumb|right|150px|For the cubic <math>\scriptstyle x^3+bx^2+cx+d=0</math> with three real roots, the roots form an equilateral triangle with vertices A, B, and C in the circle.]]
| |
| | |
| Viète's trigonometric expression of the roots in the three-real-roots case lends itself to a geometric interpretation in terms of a circle.<ref name=Nickalls/><ref>{{Citation
| |
| |first= R. W. D.
| |
| |last= Nickalls
| |
| |date= November 1993
| |
| |title= A new approach to solving the cubic: Cardan's solution revealed
| |
| |url= http://www.nickalls.org/dick/papers/maths/cubic1993.pdf
| |
| |journal= The Mathematical Gazette
| |
| |volume= 77
| |
| |issue= 480
| |
| |pages= 354–359
| |
| |issn= 0025-5572
| |
| |doi=10.2307/3619777
| |
| |jstor= 3619777 }} See esp. Fig. 2.</ref> When the cubic is written in depressed form as above as <math>t^3+pt+q=0</math>, as shown above the solution can be expressed as
| |
| | |
| :<math>t_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right) \quad \text{for} \quad k=0,1,2 \,.</math>
| |
| | |
| Here <math>\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)</math> is an angle in the unit circle; taking <math>\tfrac{1}{3}</math> of that angle corresponds to taking a cube root of a complex number; adding <math>-k\frac{2\pi}{3}</math> for ''k'' = 1, 2 finds the other cube roots; and multiplying the cosines of these resulting angles by <math>2\sqrt{-\frac{p}{3}}</math> corrects for scale.
| |
| | |
| For the non-depressed case <math>x^3+bx^2+cx+d=0</math> (shown in the accompanying graph), the depressed case as indicated previously is obtained by defining ''t'' such that <math>x=t-\tfrac{b}{3}</math> so <math>t=x+\tfrac{b}{3}</math>. Graphically this corresponds to simply shifting the graph horizontally when changing between the variables ''t'' and ''x'', without changing the angle relationships.
| |
| | |
| ====One real and two complex roots====
| |
| | |
| =====In the Cartesian plane=====
| |
| [[File:Graphical interpretation of the complex roots of cubic equation.svg|thumb|right|300px|The slope of line RA is twice that of RH. Denoting the complex roots of the cubic as ''g''±''hi'', ''g'' = <math>\scriptstyle\overline{OM}</math> (negative here) and ''h'' = <math>\scriptstyle\sqrt{\tan ORH}</math> = <math>\scriptstyle\sqrt{\text{slope of line RH}}</math> = <math>\scriptstyle\overline{BE}</math> = <math>\scriptstyle\overline{DA}</math>.]]
| |
| | |
| If a cubic is plotted in the Cartesian plane, the real root can be seen graphically as the horizontal intercept of the curve. But further,<ref>{{Citation |last=Henriquez |first=Garcia |title=The graphical interpretation of the complex roots of cubic equations |journal=[[American Mathematical Monthly]] |volume=42 |issue=6 |pages=383–384 |date=June–July 1935 |doi=10.2307/2301359 }}</ref><ref>{{Citation |last=Barr |first=C. F. |title= |journal=[[American Mathematical Monthly]] |volume=25 |issue= |page=268 |year=1918 |doi= }}</ref><ref>{{Citation |last=Barr |first=C. F. |journal=Annals of Mathematics |volume=19 |issue= |page=157 |year=1917 |doi= }}</ref> if the complex conjugate roots are written as ''g''+''hi'', then ''g'' is the abscissa (the positive or negative horizontal distance from the origin) of the tangency point of a line that is tangent to the cubic curve and intersects the horizontal axis at the same place as does the cubic curve; and |''h''| is the square root of the tangent of the angle between this line and the horizontal axis.
| |
| | |
| =====In the complex plane=====
| |
| | |
| With one real and two complex roots, the three roots can be represented as points in the complex plane, as can the two roots of the cubic's derivative. There is an interesting geometrical relationship among all these roots.
| |
| | |
| The points in the complex plane representing the three roots serve as the vertices of an isosceles triangle. (The triangle is isosceles because one root is on the horizontal (real) axis and the other two roots, being complex conjugates, appear symmetrically above and below the real axis.) [[Marden's Theorem]] says that the points representing the roots of the derivative of the cubic are the foci of the [[Steiner inellipse]] of the triangle—the unique ellipse that is tangent to the triangle at the midpoints of its sides. If the angle at the vertex on the real axis is less than <math>\tfrac{\pi}{3}</math> then the major axis of the ellipse lies on the real axis, as do its foci and hence the roots of the derivative. If that angle is greater than <math>\tfrac{\pi}{3}</math>, the major axis is vertical and its foci, the roots of the derivative, are complex. And if that angle is <math>\tfrac{\pi}{3}</math>, the triangle is equilateral, the Steiner inellipse is simply the triangle's incircle, its foci coincide with each other at the incenter, which lies on the real axis, and hence the derivative has duplicate real roots.
| |
| | |
| {{clear}}
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| ==== Omar Khayyám's solution ====
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| [[File:Omar Kayyám - Geometric solution to cubic equation.svg|thumb|right|200px|Omar Khayyám's geometric solution of a cubic equation.]] As shown in this graph, to solve the third-degree equation <math>x^3 + a^2x = b</math> where <math>b>0,</math> [[Omar Khayyám]] constructed the parabola <math>y=x^2/a,</math> the circle with diameter <math>b/a^2</math> having its center on the positive x-axis and intersecting the origin, and a vertical line through the point above the x-axis where the circle and parabola intersect. The solution is given by the length of the horizontal line segment from the origin to the intersection of the vertical line and the x-axis.
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| ==See also==
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| *[[Algebraic equation]]
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| *[[Linear equation]]
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| *[[Newton's method]]
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| *[[Polynomial]]
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| *[[Quadratic equation]]
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| *[[Quartic equation]]
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| *[[Quintic equation]]
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| *[[Spline (mathematics)]]
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| ==Notes==
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| {{Reflist|30em}}
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| ==References==
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| * {{Citation
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| |first= W. S.
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| |last= Anglin
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| |first2= Joachim
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| |last2= Lambek
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| |title= The Heritage of Thales
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| |year= 1995
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| |chapter= Mathematics in the Renaissance
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| |pages= 125–131
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| |publisher= Springers
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| |url=http://books.google.com/?id=mZfXHRgJpmQC&pg=PA125&lpg=PA125&dq=%22mathematics+in+the+renaissance%22+heritage+thales&q
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| |isbn= 978-0-387-94544-6
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| }} Ch. 24.
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| * {{Citation |last=Dence |first=T. |title=Cubics, chaos and Newton's method |journal=Mathematical Gazette |volume=81 |issue= |date=November 1997 |pages=403–408 |publisher=[[Mathematical Association]] |issn=0025-5572 |doi= }}
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| * {{Citation |last=Dunnett |first=R. |title=Newton–Raphson and the cubic |journal=Mathematical Gazette |volume=78 |issue= |date=November 1994 |pages=347–348 |publisher=[[Mathematical Association]] |issn=0025-5572 |doi=}}
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| * {{Citation
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| |first= Lucye
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| |last= Guilbeau
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| |year= 1930
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| |title= The History of the Solution of the Cubic Equation
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| |journal= Mathematics News Letter
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| |volume= 5
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| |issue= 4
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| |pages= 8–12
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| |doi=10.2307/3027812
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| |jstor= 3027812}}
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| * {{Citation| last=Jacobson| first=Nathan| author-link=Nathan Jacobson| year=2009| title=Basic algebra| edition=2nd| volume = 1 | series= | publisher=Dover| isbn = 978-0-486-47189-1}}
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| * {{Citation |last=Mitchell |first=D. W. |title=Solving cubics by solving triangles |journal=Mathematical Gazette |volume=91 |issue= |date=November 2007 |pages=514–516 |publisher=[[Mathematical Association]] |issn=0025-5572 |doi= }}
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| * {{Citation |last=Mitchell |first=D. W. |title=Powers of φ as roots of cubics |journal=Mathematical Gazette |volume=93 |issue= |date=November 2009 |pages=??? |publisher=[[Mathematical Association]] |issn=0025-5572 |doi= }}
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| * {{Citation | last1=Press | first1=WH | last2=Teukolsky | first2=SA | last3=Vetterling | first3=WT | last4=Flannery | first4=BP | year=2007 | title=Numerical Recipes: The Art of Scientific Computing | edition=3rd | publisher=Cambridge University Press | publication-place=New York | isbn=978-0-521-88068-8 | chapter=Section 5.6 Quadratic and Cubic Equations | chapter-url=http://apps.nrbook.com/empanel/index.html?pg=227}}
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| * {{Citation |last=Rechtschaffen |first=Edgar |title=Real roots of cubics: Explicit formula for quasi-solutions |journal=Mathematical Gazette |volume=92 |issue= |date=July 2008 |pages=268–276 |publisher=[[Mathematical Association]] |issn=0025-5572 |doi= }}
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| * {{Citation |last=Zucker |first=I. J. |title=The cubic equation – a new look at the irreducible case |journal=Mathematical Gazette |volume=92 |issue= |date=July 2008 |pages=264–268 |publisher=[[Mathematical Association]] |issn=0025-5572 |doi= }}
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| ==External links==
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| {{commons category|Cubic polynomials}}
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| *{{springer|title=Cardano formula|id=p/c020350}}
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| *[http://home.pipeline.com/~hbaker1/sigplannotices/sigcol07.pdf Solving a Cubic by means of Moebius transforms]
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| *[http://home.pipeline.com/~hbaker1/cubic3realroots.htm Interesting derivation of trigonometric cubic solution with 3 real roots]
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| *[http://www.freewebs.com/brianjs/ultimateequationsolver.htm Calculator for solving Cubics (also solves Quartics and Quadratics)]
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| *[http://mathdl.maa.org/convergence/1/?pa=content&sa=viewDocument&nodeId=1345&bodyId=1491 Tartaglia's work (and poetry) on the solution of the Cubic Equation] at [http://mathdl.maa.org/convergence/1/ Convergence]
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| *[http://www.akiti.ca/Quad3Deg.html Cubic Equation Solver].
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| *[http://www-history.mcs.st-and.ac.uk/history/HistTopics/Quadratic_etc_equations.html Quadratic, cubic and quartic equations] on [[MacTutor archive]].
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| *{{planetmath reference|id=1407|title=Cubic Formula}}
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| *[http://www25.brinkster.com/denshade/cardano.html Cardano solution calculator as java applet] at some local site. Only takes natural coefficients.
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| *[http://www.mathopenref.com/cubicexplorer.html Graphic explorer for cubic functions] With interactive animation, slider controls for coefficients
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| *[http://numericalmethods.eng.usf.edu/mws/gen/03nle/mws_gen_nle_bck_exactcubic.pdf On Solution of Cubic Equations] at Holistic Numerical Methods Institute
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| *[http://arxiv.org/abs/math.HO/0310449 Dave Auckly, Solving the quartic with a pencil] American Math Monthly 114:1 (2007) 29—39
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| * [http://demonstrations.wolfram.com/CubicEquation/ "Cubic Equation"] by [[Eric W. Weisstein]], [[The Wolfram Demonstrations Project]], 2007.
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| {{Polynomials}}
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| {{DEFAULTSORT:Cubic Function}}
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| [[Category:Elementary algebra]]
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| [[Category:Equations]]
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| [[Category:Polynomials]]
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