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The quadratic formula is in every school book. A formula for cubic equations exists too, and it was first found in 16th-century Italy amid public contests, oaths and a feud that lasted years. At the centre was a poor, self-taught teacher with a scarred face. Our post on Gerolamo Cardano tells the story from Cardano's side; this one is about Tartaglia, his contest, his rule and his other work.
Who was Tartaglia?
The MacTutor biography says Niccolò Fontana was born in Brescia in 1499 or 1500, the son of a mail rider who was murdered when Niccolò was six. In 1512, when French troops sacked Brescia, the 12-year-old was struck in the face by a soldier's sabre, which cut his jaw and palate. He survived through his mother's care but spoke with difficulty for the rest of his life, hence the nickname Tartaglia, "the stammerer".
He was self-taught in mathematics. He taught in Verona and moved to Venice in 1534, where he built a reputation by winning public debates. He died there in poverty on 13 December 1557.
How did he win the 1535 contest?
Scipione del Ferro had solved one type of cubic, "unknowns and cubes equal to numbers", x³ + px = q, and told only his student Antonio Fior. Fior challenged Tartaglia. Each set the other thirty problems. MacTutor's collection of the contest documents, Tartaglia versus Cardan, lists some of Fior's. Problem 3 asks for a number which, added to its own cube, makes five: x³ + x = 5.
Every one of Fior's thirty problems was of this single type. Tartaglia said afterwards that he set Fior thirty problems "each of a different kind" to show his range. MacTutor reports that the method came to Tartaglia in the early hours of 13 February 1535, after which he solved all thirty in under two hours. Fior made little headway on Tartaglia's questions.
What did the poem say?
In March 1539 Cardano persuaded Tartaglia to reveal the method, swearing never to publish it. Tartaglia gave it as a poem. MacTutor's English translation begins:
"When the cube and things together / Are equal to some discreet number, / Find two other numbers differing in this one. / Then you will keep this as a habit / That their product should always be equal / Exactly to the cube of a third of the things."
The poem continues: subtract the cube roots of the two numbers, and the result is the unknown. Cardano published the method in *Ars Magna* in 1545, crediting del Ferro and Tartaglia, after learning that del Ferro had found it first. Tartaglia felt betrayed, and the dispute ended with a public contest against Cardano's student Ferrari in 1548, from which Tartaglia left after the first day.
Worked example 1: x³ + 6x = 20
Here p = 6 and q = 20, so we need u − v = 20 and uv = (6/3)³ = 8.
From u = v + 20: v(v + 20) = 8, so v² + 20v − 8 = 0.
v = −10 + √108 = −10 + 10.392305 = 0.392305, and u = 20.392305.
∛u = 2.732051 and ∛v = 0.732051.
x = 2.732051 − 0.732051 = 2. Check: 8 + 12 = 20.
The cube roots are 1 + √3 and √3 − 1, which is why the decimals cancel so neatly.
Worked example 2: Fior's problem 3, x³ + x = 5
Now p = 1 and q = 5, so u − v = 5 and uv = 1/27.
v² + 5v − 1/27 = 0 gives v = −2.5 + √(6.25 + 0.037037) = −2.5 + 2.507396 = 0.007396, and u = 5.007396.
∛u = 1.710819 and ∛v = 0.194838.
x = 1.710819 − 0.194838 = 1.515981, or 1.51598 to five decimals. Check: 1.51598³ = 3.48402, and 3.48402 + 1.51598 = 5.00000.
What mistakes do students make with cubics?
Using the rule on the wrong form. It needs no x² term and the form x³ + px = q. Other cubics must be rearranged or shifted first.
Forgetting that a cubic can have three roots. The rule gives one real root. Divide by (x − root) to find the others: x³ + 6x − 20 = (x − 2)(x² + 2x + 10), and the quadratic has no real roots.
Assuming the square root is always real. For some cubics, such as x³ = 15x + 4, it is not, even though the roots are real. Our post on Rafael Bombelli and complex numbers shows how that puzzle was resolved.
If the algebra in step 1 loses you, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
What else did Tartaglia do?
MacTutor credits him with *Nova Scientia* (1537) on mathematics applied to artillery, which included the first firing tables; a popular arithmetic text; the first Italian translation of Euclid's *Elements* (1543); and Latin editions of Archimedes. He was a working mathematician who applied his subject to real problems.
What can a student take from Tartaglia?
He prepared for range, not for one trick, and that is why he won in 1535. He also shows the cost of secrecy: by keeping his method for future contests, he lost the chance to publish it first. Share what you work out, and practise many kinds of problem rather than one.
Try it now: use the rule to solve x³ + 3x = 14, then check your answer by substitution.
