The Solution of Cubic and Quartic Equations – Part 2

In Part 1 we saw how del Ferro, Tartaglia and Cardano found a formula for the cubic equation, and how that formula can lead to square roots of negative numbers even when every root is real. This page tells what happened next: how Rafael Bombelli learned to calculate with these "impossible" numbers, how Lodovico Ferrari solved the equation of the fourth degree, and how the quarrel over the cubic ended.

On this page: Bombelli's Algebra · Bombelli's notation · Bombelli's cubic ${x^3=15x+4}$ · Ferrari and the quartic · Ferrari's method on an example · Dividing 10 into three parts · The general quartic · The quarrel with Tartaglia · After the quartic

Bombelli's Algebra

Rafael Bombelli (1526–1572) was born in Bologna. He was an engineer by profession and spent years on the draining of marshland in the Val di Chiana in Tuscany. During breaks in this work he wrote a textbook of algebra, meant to explain the new methods of the Ars Magna more clearly than Cardano had done.

While he was writing, Bombelli saw a Greek manuscript of the Arithmetica of Diophantus in the Vatican Library and began translating it with Antonio Maria Pazzi, a professor in Rome. The translation was never finished, but the discovery changed the book: Bombelli replaced many of his own exercises with problems taken from Diophantus. It is a good example of how Renaissance scholars recovered the mathematics of ancient Greece and joined it to the new algebra.

Three books of Bombelli's L'Algebra were printed in 1572, the year of his death. Two more books, on geometry, survived only in manuscript and were found in a library in Bologna in 1923. The book starts from elementary arithmetic, moves through equations of the first and second degree and ends with the cubic and the quartic. It was read for a long time; Leibniz still used it a century later when he studied cubic equations.

Bombelli's Notation

Earlier Italian algebra was written almost entirely in words. Bombelli used a handful of abbreviations and signs that made his calculations much shorter:

"p" for più (plus) and "m" for meno (minus), as in older Italian books; the signs + and − already appeared in German books and later replaced them everywhere;
"R.q." (radice quadrata) for a square root and "R.c." (radice cubica) for a cube root;
an L-shaped bracket and its mirror image to enclose a long expression, a forerunner of our parentheses;
for powers of the unknown, a small number written over a little arc.

With the last sign the polynomial ${x^3+6x^2-3x}$ was written as

${1\overset{3}{\smile}\ \text{p}\ 6\overset{2}{\smile}\ \text{m}\ 3\overset{1}{\smile}}$.

Bombelli's boldest step was to treat square roots of negative numbers as numbers in their own right. He called ${+\sqrt{-1}}$ più di meno ("plus of minus") and ${-\sqrt{-1}}$ meno di meno ("minus of minus"). In this language ${2+11\sqrt{-1}}$ was "2 p di m 11" and ${2-11\sqrt{-1}}$ was "2 m di m 11". He then stated rules for multiplying these new quantities, just as one states the rules of signs for negative numbers. Writing $i$ for $\sqrt{-1}$, his rules say:

Bombelli's words	Meaning
più via più di meno fa più di meno	${(+1)\cdot(+i)=+i}$
meno via più di meno fa meno di meno	${(-1)\cdot(+i)=-i}$
più di meno via più di meno fa meno	${(+i)\cdot(+i)=-1}$
più di meno via meno di meno fa più	${(+i)\cdot(-i)=+1}$
meno di meno via meno di meno fa meno	${(-i)\cdot(-i)=-1}$

With these rules he could add, subtract, multiply and divide expressions of the form ${a+b\sqrt{-1}}$, which we now call complex numbers. He admitted that the whole idea seemed to rest on sophistry rather than on truth, but he showed that it gave correct answers.

Bombelli's Cubic: ${x^3=15x+4}$

The equation ${x^3=15x+4}$ was the example of the irreducible case from Part 1. It has the root 4, since ${4^3=64=15\cdot 4+4}$, and dividing by ${x-4}$ gives

${x^3-15x-4}={(x-4)\left(x^2+4x+1\right)}$,

so the other two roots are ${-2\pm\sqrt{3}}$. All three roots are real. Yet Cardano's formula, with ${p=-15}$ and ${q=-4}$, gives

${\frac{q^2}{4}+\frac{p^3}{27}}={4-125}={-121}$, ${\sqrt{-121}}={11\sqrt{-1}}$,

${x}={\sqrt[3]{2+11\sqrt{-1}}}+{\sqrt[3]{2-11\sqrt{-1}}}$.

Bombelli noticed that the two numbers under the cube roots differ only in the sign in front of $\sqrt{-1}$. He guessed that their cube roots must have the same form, so he looked for them as

${\sqrt[3]{2+11\sqrt{-1}}}={c+d\sqrt{-1}}$, ${\sqrt[3]{2-11\sqrt{-1}}}={c-d\sqrt{-1}}$.

Their sum is $2c$. If this sum is to be the root 4, then ${2c=4}$ and ${c=2}$. To find $d$, cube ${2+d\sqrt{-1}}$ with his rule ${\sqrt{-1}\cdot\sqrt{-1}=-1}$:

${\left(2+d\sqrt{-1}\right)^3}={8+3\cdot 4\cdot d\sqrt{-1}}$

${+}{3\cdot 2\cdot d^2\cdot(-1)}+{d^3\cdot\left(-\sqrt{-1}\right)}$

${=}{\left(8-6d^2\right)+\left(12d-d^3\right)\sqrt{-1}}$.

This must equal ${2+11\sqrt{-1}}$, so the real parts and the imaginary parts must agree:

${8-6d^2=2}$, ${12d-d^3=11}$.

The first equation gives ${d^2=1}$, and ${d=1}$ also satisfies the second: ${12-1=11}$. Indeed

${\left(2+\sqrt{-1}\right)^3}={8+12\sqrt{-1}-6-\sqrt{-1}}={2+11\sqrt{-1}}$,

and in the same way ${\left(2-\sqrt{-1}\right)^3}={2-11\sqrt{-1}}$. Therefore

${x}={\left(2+\sqrt{-1}\right)+\left(2-\sqrt{-1}\right)}={4}$.

The imaginary parts cancel and the real root appears. Bombelli had to know the answer 4 in advance and had to guess the form of the cube roots, but the calculation itself is completely modern. It showed, for the first time, that computing with square roots of negative numbers could lead to true real results. For most cubics in the irreducible case the cube roots cannot be found by such a guess. The real roots can then be written with cosines instead; this trigonometric solution was found by François Viète at the end of the 16th century.

Ferrari and the Quartic

Lodovico Ferrari (1522–1565) was born in Bologna. At fourteen he came to Milan as a servant in Cardano's house. Cardano soon noticed how gifted the boy was and taught him mathematics, and Ferrari became his assistant and secretary.

In 1540 the mathematician Zuanne de Tonini da Coi proposed a problem that Cardano could not solve:

Divide 10 into three parts in continued proportion so that the product of the first and the second is 6.

The problem leads to an equation of the fourth degree, a quartic. Ferrari, then eighteen, found how to solve it, and his method works for every quartic. It needs the solution of a cubic equation along the way, which is why Cardano could not publish it without publishing the cubic rule as well. Both appeared together in the Ars Magna in 1545.

Ferrari's method uses one simple identity, which he justified with a geometric figure of squares and rectangles:

${\left(A+t\right)^2}={A^2+2tA+t^2}$.

The plan is to rewrite the quartic so that both sides are perfect squares. Let us first see the method on an equation with simple numbers.

Ferrari's Method on an Example

Solve ${x^4-8x^2-4x+3=0}$. The equation has no $x^3$ term; Ferrari always removed it first, as we will see below.

1. Complete the square on the left. Keep the $x^4$ and $x^2$ terms on the left:

${x^4-8x^2=4x-3}$.

Half of the coefficient of $x^2$ is ${-4}$. Adding ${(-4)^2=16}$ to both sides gives a square on the left:

${\left(x^2-4\right)^2=4x+13}$.

2. Add a new unknown $t$ inside the square. By the identity with ${A=x^2-4}$,

${\left(x^2-4+t\right)^2}={\left(x^2-4\right)^2+2t\left(x^2-4\right)+t^2}$

${=}{4x+13+2tx^2-8t+t^2}$

${=}{2t\,x^2+4x+\left(t^2-8t+13\right)}$.

The left side is a square for every $t$. We now choose $t$ so that the right side is a square too.

3. The resolvent cubic. A quadratic ${Ax^2+Bx+C}$ with ${A\gt 0}$ is the square of a linear expression exactly when its discriminant is zero, ${B^2=4AC}$. Here ${A=2t}$, ${B=4}$ and ${C=t^2-8t+13}$:

${4^2}={4\cdot 2t\left(t^2-8t+13\right)}$

${2}={t\left(t^2-8t+13\right)}$

${t^3-8t^2+13t-2=0}$.

This cubic equation is called the resolvent cubic. In general it is solved by Cardano's formula; here we can see the root ${t=2}$ directly: ${8-32+26-2=0}$.

4. Take square roots. For ${t=2}$ the right side is ${4x^2+4x+1=(2x+1)^2}$ and the left side is ${\left(x^2-2\right)^2}$, so

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

Two squares are equal when the numbers being squared are equal or opposite:

${x^2-2=2x+1}$ or ${x^2-2=-(2x+1)}$,

${x^2-2x-3=0}$ or ${x^2+2x-1=0}$.

The quartic has split into two quadratics. They give

${x_1=3}$, ${x_2=-1}$, ${x_{3,4}=-1\pm\sqrt{2}}$.

Check one root: ${3^4-8\cdot 3^2-4\cdot 3+3}={81-72-12+3}={0}$.

Dividing 10 into Three Parts

Now the problem of da Coi. Call the three parts $a$, $b$ and $c$. "Continued proportion" means ${a:b=b:c}$, that is ${b^2=ac}$. The conditions are

${a+b+c=10}$, ${ab=6}$, ${b^2=ac}$.

From the second condition ${a=\frac{6}{b}}$, and from the third ${c=\frac{b^2}{a}=\frac{b^3}{6}}$. Substituting into the first:

${\frac{6}{b}+b+\frac{b^3}{6}=10}$.

Multiplying by $6b$ gives ${36+6b^2+b^4=60b}$, which is the quartic

${b^4+6b^2-60b+36=0}$.

Ferrari's steps work exactly as before. Write ${b^4+6b^2=60b-36}$ and add 9 to both sides:

${\left(b^2+3\right)^2=60b-27}$.

Add $t$ inside the square:

${\left(b^2+3+t\right)^2}={60b-27+2t\left(b^2+3\right)+t^2}={2t\,b^2+60b+\left(t^2+6t-27\right)}$.

The right side is a square when ${60^2=4\cdot 2t\left(t^2+6t-27\right)}$, that is ${3600=8t^3+48t^2-216t}$, or

${t^3+6t^2-27t-450=0}$.

This resolvent cubic has no rational root. It has one real root, which Cardano's formula gives as an expression with cube roots; its value is ${t\approx 7.0098}$. Taking square roots then leaves two quadratic equations in $b$, and the real solutions are

${b\approx 3.0999}$ and ${b\approx 0.6444}$.

With ${b\approx 3.0999}$ the three parts are approximately

${a\approx 1.9356}$, ${b\approx 3.0999}$, ${c\approx 4.9646}$.

Check: the sum is 10, ${ab\approx 6.000}$, and the ratios ${\frac{b}{a}\approx\frac{c}{b}\approx 1.6015}$ are equal. (The second value of $b$ gives the same kind of division with the first and third parts much more unequal: about ${9.3110}$, ${0.6444}$ and ${0.0446}$.) So da Coi's problem has an answer, but not a pretty one; that is exactly why it needed a new method.

The General Quartic

Ferrari's method works for every quartic equation. Start from

${x^4+ax^3+bx^2+cx+d=0}$.

Just as for the cubic, the substitution ${x=y-\frac{a}{4}}$ removes the cubic term and leaves an equation of the form

${y^4+py^2+qy+r=0}$.

Complete the square on the left:

${y^4+py^2=-qy-r}$,

${\left(y^2+\frac{p}{2}\right)^2=\frac{p^2}{4}-qy-r}$.

Add $t$ inside the square:

${\left(y^2+\frac{p}{2}+t\right)^2}={\left(y^2+\frac{p}{2}\right)^2+2t\left(y^2+\frac{p}{2}\right)+t^2}$

${=}{2t\,y^2-qy+\left(t^2+pt+\frac{p^2}{4}-r\right)}$.

The right side is a perfect square in $y$ when its discriminant is zero:

${q^2=8t\left(t^2+pt+\frac{p^2}{4}-r\right)}$.

This is a cubic equation for $t$, the resolvent cubic, and it can be solved by Cardano's formula. With a root ${t\ne 0}$ both sides are squares, and taking square roots splits the quartic into two quadratic equations:

${y^2+\frac{p}{2}+t}={\pm\sqrt{2t}\left(y-\frac{q}{4t}\right)}$.

In the example above, ${p=-8}$, ${q=-4}$, ${r=3}$, and the condition becomes ${16=8t\left(t^2-8t+13\right)}$, the same resolvent cubic we found there.

The Ars Magna presented Ferrari's method for the many separate types of quartic that Cardano distinguished (again to keep all coefficients positive), and Bombelli later gave a clear and systematic treatment of all the cases in his Algebra.

The Quarrel with Tartaglia

When the Ars Magna appeared in 1545, Tartaglia felt robbed of his greatest discovery. In 1546 he published his own book, Quesiti et inventioni diverse, where he printed the poem with the rule, told his version of the meeting of 1539 and attacked Cardano for breaking his oath.

Cardano did not answer himself; Ferrari defended his master. In February 1547 Ferrari sent Tartaglia a printed challenge, a cartello, accusing him of misrepresenting the facts and inviting him to a public debate on any subject in mathematics. Over the next year and a half each side published six such pamphlets, full of insults as well as problems for the opponent. The pamphlets survive and are still read by historians.

The two men finally met in a public disputation in Milan on 10 August 1548, in front of a large crowd. Tartaglia left the city after the first day, and Ferrari was considered the winner. Ferrari's reputation was made: he was soon offered well-paid positions and became a wealthy man. Tartaglia, on the other hand, lost the teaching post in Brescia that he had just obtained and returned to Venice. Few episodes in the history of mathematics produced so much publicity and so much bad feeling.

The formula is still usually called "Cardano's formula", although the honour of the discovery belongs to del Ferro and Tartaglia.

After the Quartic

By the middle of the 16th century, the Italian algebraists could express the roots of every cubic and every quartic equation through the coefficients by arithmetic operations, square roots and cube roots, if necessary of negative numbers. This was the greatest advance in algebra since Diophantus in the third century. It was also natural that such formulas need roots: even the simplest equations ${x^3=c}$ and ${x^4=d}$ are solved only by extracting a cube root and two square roots.

The obvious next goal was the equation of the fifth degree, the quintic. Mathematicians expected a formula with square, cube and fifth roots, but for more than two centuries every attempt failed except for special equations. The reason was finally found in the 19th century. Paolo Ruffini (1799) and, with a complete proof, Niels Henrik Abel (1824) showed that the general quintic cannot be solved in radicals at all, and Évariste Galois explained which equations can and which cannot.