Polynomial equations (ax² + bx + c = 0)

What this episode is about

This discussion explores polynomial equations, focusing primarily on linear and quadratic types while briefly touching on cubic equations. It covers the definition of polynomials, their degrees, historical methods of solving them, and provides examples from ancient Egyptian mathematics illustrating early problem-solving techniques.

Key takeaways

• Polynomial equations are ancient, but myths like the 'cubic equation' solver remain largely unconfirmed.
• Linear equations? Just degree-1 love letters—simple solutions wrapped in algebraic charm.
• Quadratic equations date back 4,000 years, proving math is a timeless romance with numbers.
• Egyptians used the method of false position, showing that 'guess and check' has historical cred!
• Today’s students grapple with polynomials—yet the ancients tackled complex problems with wit and wisdom.

What is discussed, and when

1. 0:00 – 3:00: Egyptian papyrus
2. 3:00 – 6:00: Egyptian papyrus, Mesopotamian clay tablet
3. 6:00 – 9:00: Mesopotamian clay tablet
4. 9:00 – 12:00: Muhammad Alquarismi
5. 12:00 – 15:00: Tartalia, Antonio Fior, Scipion Delero, Nicolo of Brussia
6. 15:00 – 18:00: Cardano, Tartalia, The Great Art, Ais Galwis

Notable quotes from "Polynomial equations (ax² + bx + c = 0)"

• There are several similar problems on this part of the papyrus which suggest that it may have been used for teaching.

  about Egyptian papyrus · at 3:00

• Remarkably solving quadratic equations can be traced back almost 4,000 years to this Mesopotamian clay tablet which includes several examples of a similar type. One of these is given here.

  about Mesopotamian clay tablet · at 7:49

• Clearly from a geometrical solution presented in Baghdad by the 9th century Persian mathematician Muhammad Alquarismi in a book that he'd written. His title contained the term algebra which gave us the word algebra.

  about Muhammad Alquarismi · at 10:16

• And he revealed his method to his pupil Antonio Fior.

  about Antonio Fior · at 13:02

• But Tartalia was a far better mathematician than Fior, who failed to solve Tartalia's equations.

  about Tartalia · at 13:49

• Cardano gave a solemn oath never to reveal Tartalia's method, but on learning later that it had been discovered by Delero, he went ahead and published it.

  about Cardano · at 16:07

• His work was then extended by the brilliant young French mathematician Ais Galwis who developed powerful algebraic criteria for deciding which equations can be solved.

  about Ais Galwis · at 17:35

Who and what is mentioned in this episode?

People

• Tartalia — Person · Referenced expert · 4 mentions · first at 13:35
  Central historical figure in the episode’s narrative about the first general solutions of certain cubic equations and the famous public contest with Fior.
• Cardano — Person · Referenced expert · 3 mentions · first at 15:32
  Presented as the mathematician who learned Tartaglia’s method, then published it in a major algebra text, catalyzing dissemination of cubic and quartic solutions.
• Ais Galwis — Person · Referenced expert · 1 mention · first at 17:36
  Introduced as extending Abel’s work by creating algebraic criteria (Galois theory) to determine which polynomials are solvable by radicals.
• Muhammad Alquarismi — Person · Referenced expert · 1 mention · first at 10:23
  Cited as a 9th-century mathematician whose geometric method illustrates completing the square and whose book title helped give rise to the term 'algebra'.
• Antonio Fior — Person · Referenced participant · 2 mentions · first at 13:02
  Appears as Del Ferro’s pupil and Tartaglia’s contest opponent, used to illustrate how mathematical contests drove disclosure of methods.
• Scipion Delero — Person · Referenced expert · 1 mention · first at 12:50
  Credited with an early discovery of a method for solving a specific class of cubic equations, later connected to Cardano’s decision to publish Tartaglia’s method.
  Around the year 1500, one of them, Scipion Delero, discovered how to solve cubic equations of the form x cubed plus cx equals d, which we'll call type one.
• Nicolo of Brussia — Person · Alias / referenced expert · 1 mention · first at 13:09
  This appears to be a mis-transcription of Niccolò Fontana (Tartaglia); in the episode it is used to introduce Tartaglia’s identity and backstory.
  Had been slashed by a saber in childhood, found a method for solving cubic equations of the form x cubed + b x²= d, which we'll call type two.

Books & papers

• Egyptian papyrus — Book · Referenced work · 2 mentions · first at 2:23
  Ancient text containing linear equations, used for teaching purposes around 1800 BC.
• Mesopotamian clay tablet — Book · Referenced work · 2 mentions · first at 4:24
  Ancient text containing quadratic equations, used for teaching purposes around 1800 BC.
• The Great Art — Book · Referenced work · 2 mentions · first at 15:40
  Mentioned as a classic algebra text by Cardano that included solutions to cubic and quartic equations.
  Cardano gave a solemn oath never to reveal Tartalia's method, but on learning later that it had been discovered by Delero, he went ahead and published it.

Full transcript with timestamps

0:17 This talk is about polomial equations and how to solve them. But what is a polomial equation? In particular, what are linear and quadratic equations and what methods have been used to solve them? Who first solved cubic equations? And can all polomial equations be solved? A polomial is an algebraic expression like the one shown here. 3 x 5th + 4 x 4 - 8 x^2 - 5 x + 6 which is obtained by adding and subtracting powers of a variable x and their multiples such as 3 * x 5 and - 8 * x^ 2. A polomial equation is then an equation where a polomial is put equal to zero.

1:12 The degree of a polomial is its highest power. Here the degree is five because the highest power appearing is x to ^ 5. Our aim is then to find solutions of the equation. For example, this equation has x= 1 as a solution because when we substitute 1 for x on its powers, we get 3 + 4 - 8 - 5 + 6 which equals zero. It also has four other solutions. In this talk, our concern is mainly with solving linear equations, those of degree 1, such as ax plus b equals 0, where a and b are constants, and quadratic equations, those of degree 2, with a term involving x^2.

2:03 But we'll also look at some cubic equations, those of degree three, with a term involving x cubed. People have been solving equations for thousands of years, although our algebraic notation is only some 400 years old. Here are two problems involving linear equations from an Egyptian papyrus dating from about 1800 BC. The first of these problem 26 shown here asks for a quantity which together with a quarter of it give 15 or in our algebraic notation which they didn't have x + a4x equals 15. The second equation asks for the quantity which together with 2/3 of it 1/2 of it and 17th of it gives 33. And again we can rewrite this algebraically.

3:00 There are several similar problems on this part of the papyrus which suggest that it may have been used for teaching. How did the Egyptians solve such problems? For several of them they used the method of false position where they guessed a solution and then scaled it up or down as necessary. here because of the 1/4th we might try four. Then the quantity in its fourth part is 4 + 1 which is five. But we want 15 not five. And so we scale our guess of four by a factor of three giving us the answer 12. Problem 31 is much more complicated. And using our algebraic notation we can solve it to give the answer 14 and 28 over 97.

3:56 On the papyrus they gave their answer in terms of unit fractions of the form 1 over something as 14 + 1 over 4 + 1 over 56 and so on. Illustrating the amazing facility they had when calculating with these unit fractions. A different type of problem appears with many others of a similar type on this Mesopotamian clay tablet also from around 1800 BC. It asks, "I found a stone but did not weigh it. After I weighed out eight times its weight, added three jin, and added 1/3 of 113th multiplied by 21, I weighed it. One manner. What was the weight of the stone?

4:52 This clearly isn't a practical problem. To find the weight of the stone, we could simply weigh it. And it was probably used for teaching purposes. But using our algebraic notation, let x be the stone's weight in jin. Then 8 * its weight + 3 jin is 8x + 3. We then add 1/3 * 113 * 21 * its amount. And simplifying this, we then get 20 over3 * 8 x + 3. Putting this equal to one mana or 60 jin then gives us x = 4 and a half jin as the weight of the stone. Let's now move on to quadratic equations. Those with an x squar term in them.

5:53 The two most familiar ways of solving these which we learn at school are both shown here. We'll illustrate them with the specific equation x^2 + 10 x - 39 = 0 which we'll meet again later. The first method is to factoriize it. Here the clue is the 39 which is 3 * 13 giving us the factorization x^2 + 10 x - 39 = x - 3 * x + 13. But since their product is zero either x - 3 or x + 13 must be zero. That is x is 3 or -3. The second method uses the familiar quadratic equation formula that is x = - b plus or minus the square roo<unk> of b ^ 2 - 4 a c all divided by 2 a.

7:02 Here a is 1, b is 10 and c is - 39. So x = -10 plus or minus the square root of b ^ 2 that's 100 - 4 * 1 * - 39 which is + 156 all divided by 2 a which is 2. Now 100 + 156 is 256 with square<unk> 16. And so the answers are - 10 plus or -6 all divided by 2 which give us x = 3 or -3 as before. Remarkably solving quadratic equations can be traced back almost 4,000 years to this Mesopotamian clay tablet which includes several examples of a similar type. One of these is given here. I have subtracted the side of my square from the area 1430.

8:11 You write down one the coefficient. You take half of one 030 and 030 you multiply. You add 015 to 1430. Result 1430 15. This is the square of 2930. You add 030 to 2930. Result 30. the side of the square. Whatever this may mean, it's surely a stepbystep procedure or algorithm for solving the problem. Notice first that they use a number system based on 60 rather than our decimal system. So that 1430 means 14 * 60 plus 30 which is 870 in decimals. Also 030 means 30 over 60 which is 12 and so on. Here's the explanation.

9:13 Although we can't literally subtract the side of a square from its area, we can interpret this as x^2 - x and this is equal to 870. How did they solve this? They first took one which is the minus b in the quadratic equation halved it and squared the result to give one quarter. They then added this to 870 to give 870 and a4 and took the square root of this to give 2930 or 29 12. Finally, adding the half they had earlier gave them the correct answer of 30 for x. You can check that these steps described so long ago corresponding to building up the quadratic equation formula.

10:10 This method for solving quadratic equations is sometimes called completing the square. And we can see this more clearly from a geometrical solution presented in Baghdad by the 9th century Persian mathematician Muhammad Alquarismi in a book that he'd written. His title contained the term algebra which gave us the word algebra. He considered the problem one square and 10 roots of the same amount to 39 dirhams which means x^2 + x = 39. Here a darham is a unit of currency. To solve this, he drew the shaded square on the left with side x and area x squared and added to it two rectangles of size 5x.

11:07 These two rectangles have total area 10 x. He then completed the square by adding the shaded 5x5 square on the right. The resulting large square has side x + 5. So its area is x + 5 all squared which multiplies out to give x^2 + 10 x + 25. But x^2 + 10 x is 39. And so we now have 39 + 25 which is 64 as the area of the large square. So the side of the square is the square root of 64 which is 8. That is x + 5 = 8 giving x = 3 as the solution to the problem. Notice that negative numbers were then considered to be meaningless. So the alternative solution x= -13 which we found earlier did not arise.

12:16 What about cubic equations? For many years, these were considered impossible to solve. Indeed, it wasn't until the early 16th century that they were solved in general in Italy. At that time, university teachers had little job security and to show their superiority and to secure their positions, they needed to submit themselves to public problem-solving contests. Around the year 1500, one of them, Scipion Delero, discovered how to solve cubic equations of the form x cubed plus cx equals d, which we'll call type one.

13:02 And he revealed his method to his pupil Antonio Fior. A little later, Nicolo of Brussia, known as Tatalia, the stammer after his face had been slashed by a saber in childhood, found a method for solving cubic equations of the form x cubed + b x²= d, which we'll call type two. But by this time, Del Faroh had died and Fior felt himself free to challenge Tartalia to a public contest, giving him 30 type one cubic equations to solve. In return, Tartalia gave Fior 30 type two equations. But Tartalia was a far better mathematician than Fior, who failed to solve Tartalia's equations.

13:56 Meanwhile, Tatalia after a sleepless night managed to solve all of Fior's type one equations and so won the contest. Here briefly is the method Tartalia used to solve type one cubics. He first taught two numbers U and V with difference D and product C over3 all cubed. Finding these involved solving only quadratic equations. A solution to the equation was then the cube root of u minus the cube root of v. For example, for the equation x cub + 9x = 26, where c is 9 and d is 26, he would seek numbers u and v where u minus v is 26.

14:52 and u * v is 3 cubed or 27. Solving these gives u = 27 and v = 1. And so x is the cube root of 27 minus cube root of 1 giving 3 - 1 = 2. The general solution to equations of this type is the complicated formula given below. But notice that like the quadratic formula, it involves only arithmetical operations such as adding and multiplication and the taking of roots. At this stage, another mathematician named Cardano entered the scene. He was writing what would become a classic algebra text, the great art. And on hearing about the Fior Tartalia contest, he wanted to learn Tartalia's method for type one equations.

15:52 The secretive Tartalia was reluctant to reveal it, but finally agreed to do so when Cardano promised to give him an introduction to the city's Spanish governor who might be approached for funding. Cardano gave a solemn oath never to reveal Tartalia's method, but on learning later that it had been discovered by Delero, he went ahead and published it. Tatalia was outraged and spent his remaining years writing him vitriolic letters. By this time quatic equations, polomial equations of degree four had also been solved and Cardano duly included them in his book.

16:40 Finally, what about polomial equations of degree five or more? Can these be solved with only arithmetic operations and the taking of roots? The search for such methods continued through the 17th and 18th centuries by Joseph Louie Lrange in France who formulated a new approach. This was then taken up by the Italian Paulo Rafini who produced a proof that no such methods or formulas can exist for these higher degrees. But his proof contained a gap. This gap was eventually filled by the Norwegian Neils Henrik Arble in in the 1820s who proved the result shown here.

17:27 No general method or formula exists for solving polomial equations of degree five or more. His work was then extended by the brilliant young French mathematician Ais Galwis who developed powerful algebraic criteria for deciding which equations can be solved. Tragically, both of these young men died very early with Arbal contracting tuberculosis at the age of 26 and Galwis dying in a duel of honor at the age of just 20, having spent his final nights writing up his results for posterity. It would be some years before it was finally realized what a genius the world had lost.

18:17 And with that sad ending, I thank you for listening.

Questions about this episode

What is "Polynomial equations (ax² + bx + c = 0)" about?

This discussion explores polynomial equations, focusing primarily on linear and quadratic types while briefly touching on cubic equations. It covers the definition of polynomials, their degrees, historical methods of solving them, and provides examples from ancient Egyptian mathematics illustrating early problem-solving techniques.

What is mentioned in this Oxford Mathematics episode?

10 people, works and ideas are mentioned, including Tartalia, Cardano, Ais Galwis, Egyptian papyrus, Mesopotamian clay tablet, Muhammad Alquarismi. Each one on this page links to its own page with every episode that mentions it.

How long is "Polynomial equations (ax² + bx + c = 0)"?

The episode runs 19 min; the transcript is about 2,119 words in 20 timestamped sections.

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