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The Cubic Equation Formula WARNING: Do not assign this in MAT 150 for a grade. The Cubic Equation f ( x ) x bx cx d 0 3 2 Let: b g( y) f y 3 2b b bc 3 3 y c y d y py q 0 3 3 27 2 3 We wish to solve the reduced equation: h ( y ) y py q 0 . 3 Next let: y z p . 3z Then: 3 3 p p p 3 h(z) z q0 pz q z 3 3z 3z 27 z Clearing the fractions gives us: z qz 6 3 p 3 27 0 By the Quadratic Formula: q q 2 4p 3 27 z 3 2 Let p 3 q 27 q 2 . 4 Then the two values of z3 are: 2 , q 2 3 27 2 q p q 2 4 Then: q 2 3 q q 2 3 2 q p 3 2 3 If we let: z 3 q 2 then: z 3 q 2 p p 3 3 Since: z 3 q 2 p 3 we get: 3 q 2 p . 3z Since y = z + (p/3z), we get the solution: y 3 q 2 3 q 2 . From p 3 27 q 2 y and 3 q 3 2 4 q , 2 the real (since p, q are real) solution to the reduced equation y3 + py + q = 0 is: y 3 q 2 p 3 27 q 2 4 3 q 2 p 3 27 q 2 4 . Relabel: z 3 q A; 3 2 q B 2 So if y3 + py + q = 0, we can rewrite the real solution as: y 3 q 2 p 3 27 q 2 3 4 But there should be two others! q 2 p 3 27 q 2 4 A B . By DeMoivre’s Theorem, if A is the real cube root of: q 2 p 3 27 q 2 , 4 then A and 2 A are the other two (complex) roots, where: 1 3 i 2 2 i e 3 . 2 Similarly, since B is the real cube root of q 2 p 3 27 q 2 , 4 then the other two (complex) roots are: B and 2 B . . The result y=A+B as a solution of: y3 + py + q = 0 will also hold if A and B are replaced by the other respective cube roots, so long as the product of the terms is AB = p/3. Since 3 = 1, this will also hold true for the pairs A, B 2 and A, B . 2 We get that the three solutions of: y3 + py + q = 0 are: y A B, A B, 2 A B 2 where: A 3 q 2 p 3 27 q 2 and B 4 3 q 2 Add b/3 to each solution to get the solutions of: x3 +bx2 + cx + d = 0. p 3 27 q 2 4 Example: x3 3x + 2 = 0 The equation is in reduced form with p = 3 and q = 2. x 3 3 q p 3 2 27 2 ( 3) q 3 27 1 1 3 1 3 1 3 3 q p 2 2 3 3 1 1 1 1 1 1 2 So x1 = 2 is our first solution. 2 2 4 3 27 2 4 2 2 q 2 4 ( 3) 27 3 2 2 4 Using our previous notation, x1 = A + B, where A = B = 1. The second solution is x2 = A + 2B, where 2 i e Thus: 3 1 2 1 3 x2 i ( 1) 2 2 i 3 . 2 1 3 i ( 1) 1. 2 2 Similarly the third solution x3 = 2A + B = 1. Thus the three solutions of x3 3x + 2 = 0 are: x = 2, 1, 1. Example x3 +12x2 + 54x + 68 = 0 Here we have b = 12, c = 54, and c = 68. We transform to reduced form by letting x = y b/3 = y 4. Reduced Form: y3 + 6y 20 = 0. (Cardano’s Example) Here p = 6 and q = 20. x1 x1 3 q 2 3 p 3 27 10 6 3 q 2 3 4 3 10 6 3 q 2 p 3 27 q 2 4 Can 3 10 6 3 3 10 6 3 be made to make sense?? YES!! Let us start by assuming that: 3 10 6 3 a b 3 , Cubing both sides yields: 10 6 3 ( a 9 ab ) (3 a b 3 b ) 3 . 3 2 2 Equating coefficients and factoring yields: a (a 9b ) 10 2 2 b(a b ) 2 2 2 This works for a = b = 1. Thus: 3 10 6 3 1 3. 3 Similarly: 3 10 6 3 1 3. So: y1 3 10 6 3 3 10 6 3 1 3 1 Letting: A 1 3, we see that y1 = A + B, as before. B 1 3, 3 2 Now, for the other solutions: y2 A B 2 1 3 i 1 2 2 1 3 3 i 1 2 2 3 1 3i 1 3 3 i 1 2 2 3 1 3i and: y3 A B 2 1 3 i 1 2 2 Thus the three solutions of y3 + 6y 20 = 0 are: y = 2, 1 3i. Remember that x = y 4. Thus the solutions to the original equation: x3 +12x2 + 54x + 68 = 0 are: x 2, 5 3i Chronology of the Cubic 1494 In his book Summa de Arithmetica, the Italian Luca Pacioli states that, given the state of mathematics, the general cubic ax3 + bx2 + cx + d = 0 is unsolvable. c. 1500 Scipione del Ferro solves x3 + px = q, x3 + q = px and x3 = px + q, where p, q > 0. He does not publish it. c. 1535 Del Ferro reveal his result to his colleague Antonio Maria Fior. 1535 Fior challenges the Venetian mathematician Niccolò Fontana (aka “Tartaglia”) to a contest where each poses thirty math problems to the other. The loser is to pay for a dinner for thirty. All of Fior’s problems involve the above cubics. Chronology of the Cubic 1535 On the night of February 1213, Tartaglia solves all thirty problems. He generously passes on the dinner. 1539 The scholar Girolamo Cardano asks Tartaglia (through a third party) what his solution to the cubic is. Tartaglia refuses. Cardano then invites Tartaglia to Milan as his guest. He offers Tartaglia the opportunity to show off his military inventions to the commander of Milan. (He also gives Tartaglia a dinner.) 1539 On March 29, Tartaglia reveals his solution to x3 + px = q to Cardano. He swears Cardano to an oath of secrecy. Cardano extends the result to general cubic equations. Chronology of the Cubic 1545 Cardano publishes Tartaglia ’s result, along with that of his protégé, Ludovico Ferrari. Ferrari shows how to use the reduced cubic to solve quartic equations. Cardano cites del Ferro and Tartaglia in his treatise. 1548 A furious Tartaglia sues. Ferrari issues a public challenge. The case is argued in a Milan court on August 10th. Tartaglia leaves before the case is settled. He loses. 2014 A group of scholars at a KY teaching conference learns about Cardano’s (and Tartaglia’s, and del Ferro’s) formula. Bibilography Dobbs, David and Hanks, Robert. A Modern Course on the Theory of Equations. Polygonal Publishing House, 1980. Dunham, William. Journey Through Genius: The Great Theorems of Mathematics. Wiley, 1990. Irving, Ron. Beyond the Quadratic Formula. MAA Inc., 2013