The Solution of Cubic and Quartic Equations – Part 1

The Babylonians already knew how to solve quadratic equations, but for about three thousand years nobody found a general rule for equations of the third degree. The breakthrough came in Italy in the first half of the 16th century, and it came with a story of secrets, public contests and a broken promise. On this page we follow that story and work through the mathematics: how a cubic is simplified, how Cardano's formula is obtained, and why it sometimes forces us to take square roots of negative numbers.

On this page: A secret formula in Bologna · The contest of 1535 · Cardano and the Ars Magna · Removing the square term · Cardano's formula · Example: ${x^3+6x=20}$ · The irreducible case

A Secret Formula in Bologna

The first person known to have solved a type of cubic equation is Scipione del Ferro (1465–1526), who taught arithmetic and geometry at the University of Bologna from 1496 until his death. Some time around 1515 he found how to solve equations of the form "a cube and some things equal to a number", which we write today as ${x^3+px=q}$ with positive $p$ and $q$.

Del Ferro did not publish his discovery. In his time a university teacher had to renew his position regularly, and a rival could challenge him to a public contest of problems. A method that nobody else knew was a valuable weapon in such a contest, so it made sense to keep it secret. Before he died, del Ferro passed his notes to his son-in-law Annibale della Nave, who later took over his post, and he also told the method to a student, Antonio Maria Fior. None of del Ferro's writings has survived.

The Contest of 1535

Niccolò Fontana of Brescia (about 1500–1557) is known as Tartaglia, "the stammerer". When the French army sacked Brescia in 1512, the boy was badly wounded in the face by a soldier's sword. The cuts to his jaw and palate left him with a lifelong speech defect, and the nickname stayed with him. Tartaglia grew up poor and learned mathematics largely on his own, and he became a well-known teacher in Venice.

Tartaglia let it be known that he could solve some cubic equations, in particular ones with a square term, ${x^3+px^2=q}$. Fior, who knew only del Ferro's single rule, did not believe him and challenged him to a contest early in 1535. Each man gave the other thirty problems. Fior's thirty problems all led to equations of the type ${x^3+px=q}$. A few days before the deadline Tartaglia found the rule for this type himself, solved all of Fior's problems in a short time and won the contest easily, while Fior solved hardly any of Tartaglia's problems.

Like del Ferro, Tartaglia kept his rule to himself. He planned to publish it later in a book of his own.

Cardano and the Ars Magna

Girolamo Cardano (1501–1576) of Milan was one of the most colourful people of the Renaissance. He was a physician famous throughout Europe, an astrologer, an inventor and a passionate gambler who wrote one of the first books on probability, about games of dice. In the 1530s he was working on a large book on arithmetic and algebra, and he wanted to include the solution of the cubic.

After repeated requests, Tartaglia visited Cardano in Milan in March 1539 and gave him the rule in the form of a poem, after Cardano had sworn an oath not to publish it. Cardano and his young assistant Lodovico Ferrari then studied the rule closely. They found proofs for it and extended it to all the different kinds of cubic equations, and Ferrari went on to solve the equation of the fourth degree (see Part 2).

Both results depended on the cubic rule, so the oath stood in the way. In 1543 Cardano and Ferrari went to Bologna and met della Nave, who showed them del Ferro's notes. These contained the same rule, written down years before Tartaglia found it. Cardano decided that he was now free to publish the method, since it was not Tartaglia's original invention.

The book appeared in Nuremberg in 1545 under the title Artis magnae, sive de regulis algebraicis ("The great art, or the rules of algebra"), usually called the Ars Magna. Cardano gave credit to both del Ferro and Tartaglia in it, but Tartaglia was furious, and a bitter public quarrel followed. We return to it in Part 2.

Reading the Ars Magna today is hard work. There were almost no algebraic symbols yet: equations were written mostly in words, and the notation we use now only took shape in the following century, especially in France. Cardano also avoided negative numbers wherever he could. Instead of one cubic equation with arbitrary coefficients, he treated thirteen separate cases, such as "a cube equal to things and a number" or "a cube and things equal to a number", so that every coefficient would be positive. He did write down negative and even imaginary quantities when a problem forced him to, but he called them "fictitious" and did not really trust them. Unease about negative and imaginary numbers lasted well into the 19th century.

In modern notation all these cases can be handled at once, and the method takes two steps.

Step 1: Removing the Square Term

Divide a cubic equation by its leading coefficient. Then it has the form

${x^3+ax^2+bx+c=0}$.

The idea is to shift the unknown by a constant so that the term with $x^2$ disappears. Put ${x=y-\frac{a}{3}}$. The term $y^3$ comes only from ${\left(y-\frac{a}{3}\right)^3}$, and the terms with $y^2$ are

${3y^2\cdot\left(-\frac{a}{3}\right)+a\,y^2}={-ay^2+ay^2}={0}$.

After expanding everything, the equation becomes a depressed cubic, a cubic without a square term:

${y^3+py+q=0}$, where ${p=b-\frac{a^2}{3}}$, ${q=\frac{2a^3}{27}-\frac{ab}{3}+c}$.

Example. Take ${x^3+3x^2+9x-13=0}$. Here ${a=3}$, so we put ${x=y-1}$:

${p}={9-\frac{3^2}{3}}={9-3}={6}$,

${q}={\frac{2\cdot 3^3}{27}-\frac{3\cdot 9}{3}-13}={2-9-13}={-20}$.

So the equation becomes ${y^3+6y-20=0}$, that is ${y^3+6y=20}$. This is the equation we solve below; its root ${y=2}$ gives ${x=y-1=1}$, and indeed ${1+3+9-13=0}$.

The same trick of removing terms by a change of variable was pushed much further by Ehrenfried Walther von Tschirnhaus in 1683, who used more complicated substitutions to remove several terms at once.

Step 2: Cardano's Formula

Now solve ${x^3+px+q=0}$. The key idea is to look for the root as a sum of two numbers, ${x=u+v}$. Cubing a sum gives

${(u+v)^3}={u^3+3u^2v+3uv^2+v^3}={u^3+v^3+3uv(u+v)}$.

Substituting ${x=u+v}$ into the equation therefore gives

${u^3+v^3}+{(3uv+p)(u+v)}+{q}={0}$.

We have two unknowns $u$ and $v$ but only one equation, so we are free to impose one more condition. Choose $u$ and $v$ so that the bracket vanishes:

${3uv+p=0}$, that is ${uv=-\frac{p}{3}}$.

Then the equation reduces to ${u^3+v^3=-q}$, and cubing the condition gives ${u^3v^3=-\frac{p^3}{27}}$. So we know the sum and the product of the two numbers $u^3$ and $v^3$. By Vieta's formulas they are the two roots of the quadratic equation

${t^2+qt-\frac{p^3}{27}=0}$.

The quadratic formula gives

${u^3}={-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$, ${v^3}={-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$.

Taking cube roots and adding, we get the formula published in the Ars Magna:

Cardano's formula. A root of ${x^3+px+q=0}$ is ${x}={\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}}+{\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}}$, where the two cube roots must be chosen so that their product is ${-\frac{p}{3}}$.

The number ${\frac{q^2}{4}+\frac{p^3}{27}}$ under the square root decides what happens. If it is positive, the formula gives one real root directly (the other two roots are complex). If it is negative, we meet the square root of a negative number, which is the subject of the last section.

Cardano's Example: ${x^3+6x=20}$

The first worked example in the chapter of the Ars Magna on "a cube and things equal to a number" is ${x^3+6x=20}$. In our form it is ${x^3+6x-20=0}$, so ${p=6}$ and ${q=-20}$. We apply the formula step by step:

${\frac{q^2}{4}}={\frac{(-20)^2}{4}}={\frac{400}{4}}={100}$,

${\frac{p^3}{27}}={\frac{6^3}{27}}={\frac{216}{27}}={8}$,

${\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}={\sqrt{108}}={6\sqrt{3}}$,

${-\frac{q}{2}}={10}$.

${x}={\sqrt[3]{10+6\sqrt{3}}}+{\sqrt[3]{10-6\sqrt{3}}}$.

Cardano wrote the answer as "the cube root of ${\sqrt{108}+10}$ minus the cube root of ${\sqrt{108}-10}$", which is the same number, because ${10-6\sqrt{3}}$ is negative.

The expression looks complicated, but it is simply 2. To see this, notice that both numbers under the cube roots are perfect cubes:

${\left(1+\sqrt{3}\right)^3}={1+3\sqrt{3}+3\cdot 3+3\sqrt{3}}={10+6\sqrt{3}}$,

${\left(1-\sqrt{3}\right)^3}={1-3\sqrt{3}+3\cdot 3-3\sqrt{3}}={10-6\sqrt{3}}$.

Therefore

${x}={\left(1+\sqrt{3}\right)+\left(1-\sqrt{3}\right)}={2}$.

The condition on the product is also satisfied: ${\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}={1-3}={-2}={-\frac{6}{3}}$. Check: ${2^3+6\cdot 2}={8+12}={20}$.

Dividing by ${x-2}$ gives ${x^3+6x-20}={(x-2)\left(x^2+2x+10\right)}$, and the quadratic factor has the complex roots ${-1\pm 3i}$. So 2 is the only real root, as the positive number under the square root predicted.

Cardano could check that 2 satisfies the equation, but the formula itself gave him no hint that the long expression is equal to 2. Spotting the cubes ${(1\pm\sqrt{3})^3}$ is a matter of luck; in most cases the cube roots cannot be simplified at all.

The Irreducible Case

Cardano's formula first takes a square root and then cube roots. This is exactly what is meant by solving an equation in radicals: the roots are obtained from the coefficients by the four arithmetic operations and by extracting roots. Formulas with other functions, such as sines and cosines, also exist, but they are not counted as algebraic solutions.

The trouble appears when the number ${\frac{q^2}{4}+\frac{p^3}{27}}$ is negative. Then the formula asks for the square root of a negative number, and this happens precisely when the cubic has three different real roots. So the formula breaks down in what seems to be the friendliest case. For example, the equation ${x^3=15x+4}$, that is ${x^3-15x-4=0}$, has ${p=-15}$, ${q=-4}$ and

${\frac{q^2}{4}+\frac{p^3}{27}}={4+\frac{-3375}{27}}={4-125}={-121}$,

so the formula gives

${x}={\sqrt[3]{2+\sqrt{-121}}}+{\sqrt[3]{2-\sqrt{-121}}}$,

although ${x=4}$ is obviously a root: ${64=60+4}$.

This situation was named the casus irreducibilis, the irreducible case. Cardano knew about it but could not resolve it. It is not a defect of his particular method. In the 19th century it was proved that when a cubic with rational coefficients has three real roots and no rational root, any expression of those roots in radicals must pass through square roots of negative numbers. The real roots cannot be reached without leaving the real numbers. This was one of the strongest early reasons for taking complex numbers seriously. How Rafael Bombelli made sense of the expression above and obtained ${x=4}$ from it is told in Part 2.