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The Cubic Equation Formula WARNING: Do not assign this in MAT 150 for a grade.

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The Cubic Equation Let: We wish to solve the reduced equation:

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Next let: Then: Clearing the fractions gives us:

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By the Quadratic Formula: Let Then the two values of z 3 are:

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Then: If we let: then:

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Since y = z + ( p/3z), we get the solution: we get: Since:

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the real (since p, q are real) solution to the reduced equation y 3 + py + q = 0 is: From and

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Relabel: So if y 3 + py + q = 0, we can rewrite the real solution as: But there should be two others!

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By DeMoivre’s Theorem, if A is the real cube root of: then are the other two (complex) roots, where: Similarly, since B is the real cube root of then the other two (complex) roots are:.

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. The result y = A + B as a solution of: y 3 + 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

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We get that the three solutions of: y 3 + py + q = 0 are: where: Add b/3 to each solution to get the solutions of: x 3 +bx 2 + cx + d = 0.

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Example: x 3 3x + 2 = 0 The equation is in reduced form with p = 3 and q = 2. So x 1 = 2 is our first solution.

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Using our previous notation, x 1 = A + B, where A = B = 1. The second solution is x 2 = A + 2 B, where Thus: Similarly the third solution x 3 = 2 A + B = 1. Thus the three solutions of x 3 3x + 2 = 0 are: x = 2, 1, 1.

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Example x 3 +12x x + 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: y 3 + 6y 20 = 0. (Cardano’s Example) Here p = 6 and q = 20.

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Can be made to make sense?? YES!!

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Let us start by assuming that: Cubing both sides yields: Equating coefficients and factoring yields: This works for a = b = 1. Thus:

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Similarly: So: Letting: we see that y 1 = A + B, as before.

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Now, for the other solutions: and:

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Thus the three solutions of y 3 + 6y 20 = 0 are: y = 2, 1 3i. Remember that x = y 4. Thus the solutions to the original equation: x 3 +12x x + 68 = 0 are:

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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 ax 3 + bx 2 + cx + d = 0 is unsolvable. c Scipione del Ferro solves x 3 + px = q, x 3 + q = px and x 3 = px + q, where p, q > 0. He does not publish it. c Del Ferro reveal his result to his colleague Antonio Maria Fior 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.

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Chronology of the Cubic 1535 On the night of February 12 13, Tartaglia solves all thirty problems. He generously passes on the dinner 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 x 3 + px = q to Cardano. He swears Cardano to an oath of secrecy. Cardano extends the result to general cubic equations.

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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 A furious Tartaglia sues. Ferrari issues a public challenge. The case is argued in a Milan court on August 10 th. Tartaglia leaves before the case is settled. He loses A group of scholars at a KY teaching conference learns about Cardano’s (and Tartaglia’s, and del Ferro’s) formula.

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Bibilography Dobbs, David and Hanks, Robert. A Modern Course on the Theory of Equations. Polygonal Publishing House, Dunham, William. Journey Through Genius: The Great Theorems of Mathematics. Wiley, Irving, Ron. Beyond the Quadratic Formula. MAA Inc., 2013

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