The solution of cubic and quartic equations - 1

In the 16th century in Italy, there occurred the first progress on polynomial equations beyond the quadratic case. The person credited with the solution of a cubic equation is Scipione del Ferro (1465-1526), who lectured in arithmetic and geometry at the University of Bologna from 1496 until 1526. He wrote down a solution of the cubic equation in a manuscript, which passed to his student Annibale dalla Nave. The manuscript is now lost, and no work of del Ferro has survived. News of the solution spread by word of mouth and reached the mathematician Niccol`o Tartaglia (whose surname means ‘The Stammerer’: he had difficulty in speaking after he received a sword cut to the tongue during a French siege of Brescia, where he lived). Girolamo Cardano heard that Tartaglia was in possession of the solution and he implored him to share it. Tartaglia eventually told Cardano the solution but swore him to secrecy. Later, Nave told Cardano of the existence of del Ferro’s manuscript on the solution of the cubic and thereafter Cardano felt no longer bound by the terms of his oath to Tartaglia, as Tartaglia was not the originator of the method.

Cardano (1501-76) was an important figure in the development of early modern science, and was eager to hear of new developments, such as the solution of the cubic equation. He was also a famous physician, whose skills were sought throughout Europe. He wrote several books, but for our purposes, the most important is Artis magnae, sive de regulis algebraicis (Of the great art, or concerning the algebraic rules), better known as Ars magna, which was published in Nuremberg in 1545. Cardano is credited as the originator of the modern theory of algebraic equations, for he presented here for the first time to the public the solution of the cubic equation, as he had learnt it from Tartaglia. It is important to realize that work on algebra was hindered by lack of good notation, which only began to appear in the 17th century in France. It is not possible for the novice to read Cardano’s work in its original form, as the notation is so unfamiliar, but a translation into modern English with standard notation is now available.

It was in his solution of the cubic equation that Cardano first encountered what we have come to know as complex numbers, and he may to some extent be credited with their discovery. He was, however, apparently unconvinced about the validity of the use of these imaginary or sophistic numbers. It might be added that there was a general reserve about using negative numbers at this time, and Cardano often looked at different forms of equations to avoid introducing negative numbers (writing them down in what we would consider to be repetitive but apparently different ways, so that all coefficients were positive). Nonetheless, the need for negative numbers must have been appreciated, for otherwise the complex numbers would never have appeared. It might be added that anxiety about negative numbers (not to mention imaginary numbers) was felt well into the 19th century.

Let us see what was involved in Cardano’s solution. We will depart from his notation, but his basic idea remains the same and has not been improved to this day. We take a monic cubic polynomial

f(x) = x3 + ax2 + bx + c,

where a, b and c are the coefficients, which we can take to be real numbers, and we look for a root of this polynomial: this is a number α (which might turn out not to be real!) satisfying f(α) = 0. The first thing that Cardano did was to change variables so as to form a cubic in a new variable y with the property that the coefficient of y2 is 0. We do this as follows: Let y = x + r, where r is some number to be chosen, and evaluate f(x) = f(y - r) in terms of y. We get

f(x) = f(y - r) = y3 - 3ry2 + 3r2y - r3 + a(y2 - 2ry + r2) + b(y - r) + c = y3 + (a - 3r)y2 + dy + e,

where d and e are two new coefficients (which we will not write down explicitly). We can see that if we choose r so that a - 3r = 0, the coefficient of y2 is 0. This trick is used quite frequently in polynomial theory (trying to make several coefficients become zero by transformation of variables). It was extended by Tschirnhaus in the 17th century. For the purposes of finding the roots of a cubic polynomial, this transformation shows that it will suffice to be able to find the roots of one of the form f(x) = x3 + ax + b.

We now introduce two new unknowns u and v and write x = u + v. Then on substituting for x, we get

f(x) = f(u + v) = (u + v)3 + a(u + v) + b

= u3 + 3u2v + 3uv2 + v3 + au + av + b = u3 + v3 + (3uv + a)(u + v) + b.

We now choose u and v so that 3uv +a = 0, which removes the u + v term. This gives us $v = - \frac{a}{3u}$ So if f(x) = 0, we get, in terms of u and v,

If we multiply through by u3, we get

$u^6 + b u^3 - \frac{a^3}{27}$

which we recognize as a quadratic equation in u3. We solve for u3 using the quadratic formula:

$u^3 + v^3 + b - 0 - u^3 - \frac{a^3}{27 u^3} + b, \ \ u^3 = \frac{- b \pm \sqrt{b^2 + (\frac{4a^3}{27})}}{2}$

We get two solutions for u3 and on taking cube roots, we obtain six solutions for u. Now the relation$v = - \frac{a}{3u}$gives

This shows that in general u3 and v3 are what are called algebraic conjugate expressions. We can find a root u+v of the original cubic by extracting cube roots in these expressions.

Let’s try an example that Cardano gave in the Ars magna, namely, find the roots of x3 +6x-20 = 0, which Cardano would have expressed as x3 +6x = 20, to avoid negative quantities. Here, a = 6, b = -20. Then

$b^2 = 400, \ \frac{4a^3}{27} = 32, \ b^2 + \frac{4a^3}{27} = 432 = (12)^2 \times 3.$

Therefore,$u^3 = 10 + 6 \sqrt{3}$ and, in principle we can extract cube roots to find three possible

values for u (two of which involve complex cube roots of unity). From what we have said above, $v^3 = 10 - 6 \sqrt{3}$ and thusarootofthe cubic is

This is a rather complicated expression. It is not so hard to show that it in fact equals 2, something that Cardano noted, but was not able to prove.

An interesting problem arises in the application of Cardano’s method, which is quite difficult to appreciate. Cardano’s formula involves the extraction of a square root and then of cube roots. This seems to be quite satisfactory, for it must be realized that when we talk of algebraic solutions of polynomial equations, we understand that the solutions should be given in terms of so called radicals–that is, expressions that involve extracting square, cube, and higher roots. Other types of solution may be given, in terms say of trigonometric functions, but these are taken to lie outside the domain of algebra. Now we will inevitably be drawn into the realm of complex numbers if the expression b2 +(4a3/27) is negative, since we have to extract its square root. There are examples known where we have cubic polynomials whose three roots are all real and yet where Cardano’s method leads inevitably to the use of complex number methods. This is not simply a defect of his method: it can be proved that in certain cases any expression for the roots in terms of radicals, even if the roots are real, must involve non-real expressions under the radical signs.

$\begin{eqnarray} v^3 &=& - \frac{a^3}{27 u^3} \\ \text{ If, say, we take} \ \ \ u^3 &=& \frac{- b + \sqrt{b^2 + (\frac{4a^3}{27})}}{2}, \\ \text{then this leads to} \ \ \ v^3 &=& \frac{- b - \sqrt{b^2 + (\frac{4a^3}{27})}}{2} \end{eqnarray}$

This became known as the casus irreducibilis or irreducible case of Cardano’s method, and it has demonstrated the relevance and indeed inevitably of the use of complex numbers. Cardano’s was aware of the problem but he was unable to develop the algebraic tools to clarify what is happening.

Solution of cubic and quartic equations - 2

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