- EUROPE 431
6.5. Solution of cubic and quartic equations. In Europe algebra was confined
to linear and quadratic equations for many centuries, whereas the Chinese and
Japanese had not hesitated to attack equations of any degree. The difference in the
two approaches is a result of different ideas of what constitutes a solution. This
distinction is easy to make nowadays: The European mathematicians were seeking
an exact solution using only arithmetic operations and root extractions, what is
called solution by radicals. However, it will not do to press the distinction too far:
It is impossible to do good numerical work without a sound theoretical basis. As
we saw in the work of Sharaf al-Tusi, the coefficients that appear in the course of
his numerical solution have theoretical significance.
The Italian algebraists of the early sixteenth century made advances in the
search for a general algorithm for solving higher-degree equations. We discussed
the interesting personal aspects of the solution of cubic equations in Section 4 of
Chapter 3. Here we concentrate on the technical aspects of the solution.
The verses Tartaglia had memorized say, in modern language, that to solve the
problem x^3 +px = q, one should look for two numbers u and í satisfying u-v = q,
uv = (p/3)^3. The problem of finding u and í is that of finding two numbers given
their difference and their product, and of course, that is merely a matter of solving
a quadratic equation, a problem that had already been completely solved. Once
this quadratic has been solved, the solution of the original cubic is ÷ = yfu — ffi.
The solution of the cubic has thus been reduced to solving a quadratic equation,
taking the cube roots of its two roots, and subtracting. Cardano illustrated with
the case of "a cube and six times the side equal to 20." Using his complicated rule
(complicated because he stated it in words), he gave the solution as
He did not add that this number equals 2.
Ludovico Ferrari. Cardano's student Ludovico Ferrari worked with him in the so-
lution of the cubic, and between them they had soon found a way of solving certain
fourth-degree equations. Ferrari's solution of the quartic was included near the end
of Cardano's Ars magna. Counting cases as for the cubic, one finds a total of 20
possibilities. The principle in most cases is the same, however. The idea is to make
a perfect square in x^2 equal to a perfect square in ÷ by adding the same expression
to both sides. Cardano gives the example
It is necessary to add to both sides an expression rx^2 + s to make them squares,
that is, so that both sides of
are perfect squares. Now the condition for this to happen is well known: ax^2 +bx+c
is a perfect square if and only if b^2 — Aac = 0. Hence we need to have simultaneously
Solving the second of these equations for s in terms of r and substituting in the
first leads to the equation
60x = ÷^4 + 6x^2 + 36.rx^2 + 60x -)- s = x^4 + (6 + r)x^2 + (36 + s)3600 - 4sr = 0, (6 + r)^2 - 4(36 + s) = 0.r^3 + 12r^2 = 108r + 3600.