Some constructions were known in Ancient Greece, many involving conic sections (parabolas and hyperbolas).
Since Arabic mathematicians did not use negative numbers and did not allow zero coefficients, many of Khayyam’s constructions required side conditions to guarantee the existence of positive solutions. While his work was impressive for the time, it did not result in a means for arriving at the numerical solution for the cubic equation.
Nctm2012 2 26-12
12 Annua l Meeting20 rvice & In -ServicePrese Gallery W orkshop Merging History and Technology: Immersive Strategies for Solving Cubic Polynomials Frederick W. Sakon, Angelina Kuleshova, & Christopher G. Thompson
Workshop Overview• Build historical timeline regarding the resolution of cubic polynomials.• Explore multiple representations and technological advances that help students conceptualize and visualize cubic equations.• Utilize historical methods and technology to solve cubic polynomials.• Discuss relationship between historical development of mathematics content and pedagogical content knowledge.
Workshop ApplicationsMathematics Teacher Educators Preservice Teacher Training In-service Teacher Professional DevelopmentMathematics Teachers Classroom Instruction
The Problem Begins x + ax + bx + c = 0 3 2400 B.C. Interest in solving cubic equations arose from geometric problems considered by Greek mathematicians. Particularly, the question of how to β trisect a given angle amounted to solving β a cubic equation. β
Some Progression Omar Khayyam 1048 - 1131Khayyam still approached the problem from the geometricperspective, using intersecting conic sections to construct aline segment with a length satisfying the equation.Arabic algebra was expressed entirely in words, from whichKhayyam constructed and worked with 14 different types ofcubic equations (e.g. “a cube and roots are equal to anumber” equivalent to the form x3 + bx = c).
The Italian Renaissance Scipione del Ferro 1465 - 1526 Niccolò Fontana 1500 - 1557• Del Ferro and Fontana, known as Tartaglia, both discovered methods to solve particular cubics but kept them secret.• Italian scholars of the time had to prove their academic competence in public competition against another scholar.• Del Ferro passed on his method to solve cubics of the form x3 + bx = c to his student, Antonio Maria Fiore.
The Italian Renaissance 1535 1539In 1535, Tartaglia boasts about his ability to solve cubicequations, prompting a challenge from Fiore.News of Tartaglia’s victory against Fiore spread andinspired Girolamo Cardano (1501 – 1576), to contactTartaglia in 1539.With promises of secrecy, Cardano requested that Tartagliashare his methods, to which Tartaglia eventually agreed.
The Italian Renaissance 1545 1539Cardano had obtained methods for solving cases of thecubic in the form x3 + ax2 = c and x3 + bx = c.After six years of work, Cardano and his assistant, LodovicoFerrari (1522 – 1565) managed to solve the generalequation completely.Cardano published the Ars Magna (The Great Art),documenting a complete account of how to solve any typeof cubic equation.
Cardano’s Explanation: Solving a cubic equationA cube and things equal to a number Cube the third part of the number of things, to which you add the square of half the number and take the square root of the whole which you will use, in the one case adding the half of the number which you just multiplied by itself, in the other case subtracting the same half, and you will have a “binomial” and “apotome” respectively; then subtract the cube root of the apotome from the cube root of the binomial. This is the value of the thing.
Translating Cardano’s Explanation: “A cube and things x3+cx = d equal to a number” 3 c “cube the third part of the number of things” 3 2 d “add the square of half + the number” 2
Translating Cardano’s Explanation: “take the square root of 3 2 c d the whole” + 3 2 “in the one case adding the half of the number 3 c d 2 d which you just + + 3 2 2 multiplied by itself”
Translating Cardano’s Explanation: “in the other case subtracting 3 2 the same half” c d d + − 3 2 2 “then subtract the cube root of the apotome from the cube root of the binomial” 3 3 c d d 3 2 c d 2 d 3 + + − + − 3 2 2 3 2 2
Activity 1 Derive solutions to each of the given cubic equations by applying Cardano’s method: 3 3 c d 2 d 3 c d d 23 + + − + − 3 2 2 3 2 2
ReflectionThere are some limitations or problems we might runinto given only Cardano’s method to solve cubicequations. Let’s consider 3 cases
Case 1: x3 + 6x = 20 3 3 6 20 2 20 6 20 20 2x= 3 + + − 3 3 + 2 − 2 3 2 2 x= 3 108 + 10 − 3 108 − 10 When we apply the method, we cannot simplify to the solution, x = 2, without utilizing a calculator or computer to resolve the non-perfect square and cube roots.
Case 2: x3 = 30x + 36 3 −30 36 36 3 2 −30 36 36 2x= 3 + + − 3 3 + 2 − 2 3 2 2 x= 3 −676 + 18 − 3 −676 − 18 Here, we cannot simplify to the solution, x = 6, without complex number theory. After Cardano’s publication of Ars Magna, Rafael Bombelli (1526 – 1572) developed methods for resolving such cases.
Case 3: x3 + 6x2 + 11x + 6 = 0• A key element to Cardano’s discovery is that he actually proved that his method worked.• However, Cardano sought the solution of a particular type of cubic: x3 + cx = d. Obviously, not all cubic equations resemble this particular form.• For cases such as this, he proposed a means of reducing, or as he called it depressing, the original cubic to the form which fit his method.• How might cubic equations be reduced so that they fit Cardano’s proven method?
Reducing Cubic EquationsGiven a cubic equation, x + ax + bx + n = 0, 3 2 awe can apply the substitution, x = y − , 3to obtain an equivalent equation, x 3 + cx + d = 0.
A Reduction ExampleGiven a cubic equation, x + 6x + 11x + 6 = 0, 3 2 6we can apply the substitution, x = y − , or x = y − 2. 3(y − 2)3 + 6(y − 2)2 + 11(y − 2) + 6 = 0y 3 − 4y 2 + 4y − 2y 2 + 8y − 8 + 6y 2 − 24y + 24 + 11y − 22 + 6 = 0y3 − y = 0Now, Cardano’s method could be applied, where c = – 1 and d = 0.Note, the solution to the original cubic equation requiressubstitution of the intermediary solution for y into x = y – 2.
Activity 2Convert each cubic equation to the form x3 + cx = d, for which Cardano’s method may be applied. a x = y− 3
Reflection Solving cubic equations was a mathematical process that took many years to develop. Each additional development granted us greater flexibility in our problem solving capacity.Let’s consider some of the tools we have developed and how they can expand our perspective on solving cubic equations
A “simple” cubic example, continuedThe equation, x3 + 60x2 + 1200x = 4000, can bethought of as being very close to a perfect cube.With this in mind, can you “complete the cube”?Try this now, thinking of what you would need todo to complete the cube on x3 + 60x2 + 1200x.(Recall that: (a+b)3=a3+3a2b+3ab2+b3)
Example, continuedIf they considered the x + bx + cx = d“general” cubic as: 3 2Then the “cubic formula” 3became: c c x=3 +d −However, not all cubics were of b bthis form!!!(And similar to otherequations solved, we only findone solution with this“formula.”)
Various forms of cubic equations…and ways to solve them Example 1 (from the Italian Renaissance): Solve x3 + 60x2 + 1200x = 4000 The solution is: x = 3 (1200 / 60) 3 + 4000 − (1200 ) = 3 12000 − 20 60