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In about 300 BC Euclid developed a geometrical approach which, although later mathematicians used it to solve quadratic equations, amounted to finding a length which in our notation was the root of a quadratic equation. Euclid had no notion of equation, coefficients etc. but worked with purely geometrical quantities.
Hindu mathematicians took the Babylonian methods further so that Brahmagupta (598-665 AD) gives an, almost modern, method which admits negative quantities. He also used abbreviations for the unknown, usually the initial letter of a colour was used, and sometimes several different unknowns occur in a single problem.
The Arabs did not know about the advances of the Hindus so they had
neither negative quantities nor abbreviations for their unknowns. However al-Khwarizmi
(c 800) gave a classification of different types of quadratics (although only
numerical examples of each). The different types arise since al-Khwarizmi
had no zero or negatives. He has six chapters each devoted to a different type
of equation, the equations being made up of three types of quantities namely:
roots, squares of roots and numbers i.e. x, x2 and
numbers.
Abraham bar Hiyya Ha-Nasi, often known by the Latin name Savasorda, is famed for his book Liber embadorum published in 1145 which is the first book published in Europe to give the complete solution of the quadratic equation.
A new phase of mathematics began in Italy around 1500. In 1494 the
first edition of Summa de arithmetica, geometrica, proportioni et
proportionalita, now known as the Suma, appeared. It was written by
Luca Pacioli
although it is quite hard to find the author's name on the book, Fra Luca
appearing in small print but not on the title page. In many ways the book is
more a summary of knowledge at the time and makes no major advances. The
notation and setting out of calculations is almost modern in style: Pacioli
does not discuss cubic equations but does discuss quartics. He says that, in our
notation, x4 = a + bx2 can be solved
by quadratic methods but x4 + ax2 = b
and x4 + a = bx2 are impossible at
the present state of science.
Scipione dal Ferro
(1465-1526) held the Chair of Arithmetic and Geometry at the University of
Bologna and certainly must have met Pacioli
who lectured at Bologna in 1501-2. dal Ferro
is credited with solving cubic equations algebraically but the picture is
somewhat more complicated. The problem was to find the roots by adding,
subtracting, multiplying, dividing and taking roots of expressions in the
coefficients. We believe that dal Ferro
could only solve cubic equation of the form x3 + mx =
n. In fact this is all that is required.
x3 + mx = n where m = c -
b2/3, n = d - bc/3 +
2b3/27. Fior was a mediocre mathematician and far less good at keeping
secrets than dal Ferro.
Soon rumours started to circulate in Bologna that the cubic equation had been
solved. Nicolo of Brescia, known as Tartaglia
meaning 'the stammerer', prompted by the rumours managed to solve equations of
the form x3 + mx2 = n and made no
secret of his discovery.
Fior challenged Tartaglia
to a public contest: the rules being that each gave the other 30 problems with
40 or 50 days in which to solve them, the winner being the one to solve most but
a small prize was also offered for each problem. Tartaglia
solved all Fior's problems in the space of 2 hours, for all the problems Fior
had set were of the form x3 + mx = n as he
believed Tartaglia
would be unable to solve this type. However only 8 days before the problems were
to be collected, Tartaglia
had found the general method for all types of cubics.
News of Tartaglia's
victory reached Girolamo Cardan
in Milan where he was preparing to publish Practica Arithmeticae (1539).
Cardan
invited Tartaglia
to visit him and, after much persuasion, made him divulge the secret of his
solution of the cubic equation. This Tartaglia
did, having made Cardan
promise to keep it secret until Tartaglia
had published it himself. Cardan
did not keep his promise. In 1545 he published Ars Magna the first Latin
treatise on algebra.
Notice that (a - b)3 + 3ab(a -
b) = a3 - b3 After Tartaglia
had shown Cardan
how to solve cubics, Cardan
encouraged his own student, Lodovico Ferrari,
to examine quartic equations. Ferrari
managed to solve the quartic with perhaps the most elegant of all the methods
that were found to solve this type of problem. Cardan
published all 20 cases of quartic equations in Ars Magna. Here, again in
modern notation, is Ferrari's
solution of the case: x4 + px2 + qx +
r = 0. First complete the square to obtain
Now we know how to solve cubics, so solve for y. With this value of
y the right hand side of (*) is a perfect square so, taking the square
root of both sides, we obtain a quadratic in x. Solve this quadratic and
we have the required solution to the quartic equation.
The irreducible case of the cubic, namely the case where Cardan's
formula leads to the square root of negative numbers, was studied in detail by
Rafael Bombelli
in 1572 in his work Algebra.
In the years after Cardan's
Ars Magna many mathematicians contributed to the solution of cubic and
quartic equations. Vičte,
Harriot,
Tschirnhaus,
Euler,
Bezout
and Descartes
all devised methods. Tschirnhaus's
methods were extended by the Swedish mathematician E S Bring
near the end of the 18th Century.
Thomas Harriot
made several contributions. One of the most elementary to us, yet showing a
marked improvement in understanding, was the observation that if x =
b, x = c, x = d are solutions of a cubic then
the cubic is
Leibniz
wrote a letter to Huygens
in March 1673. In it he made many contributions to the understanding of cubic
equations. Perhaps the most striking is a direct verification of the Cardan-Tartaglia
formula. This Leibniz
did by reconstructing the cubic from its three roots (as given by the formula)
as Harriot
claimed in general. Nobody before Leibniz
seems to have thought of this direct method of verification. It was the first
true algebraic proof of the formula, all previous proofs being geometrical in
nature.
Article by: J J O'Connor and E F
Robertson Other Web sites:
6.p.R.10
18.m.R.90
____________________________
108.m.R.3240.p.R.3240.m.R.90
hoc est 78.
In our notation
(6 +
The last term in the answer 90 is an early misprint and should be
900 but the margin was too narrow so the printer missed out the final 0!
10)
(18 -
90) =
(108-
3240 +
3240 -
900)
which is 78.
For, given the general cubic y3 -
by2 + cy - d = 0, put y = x +
b/3 to get
However, without the
Hindu's knowledge of negative numbers, dal Ferro
would not have been able to use his solution of the one case to solve all cubic
equations. Remarkably, dal Ferro
solved this cubic equation around 1515 but kept his work a complete secret until
just before his death, in 1526, when he revealed his method to his student
Antonio Fior.
Here, in modern notation, is Cardan's
solution of x3 + mx = n.
Cardan
noticed something strange when he applied his formula to certain cubics. When
solving x3 = 15x + 4 he obtained an expression
involving
so if a and
b satisfy 3ab = m and a3 -
b3 = n then a - b is a solution of
x3 + mx = n.
But now b =
m/3a so a3 -
m3/27a3 = n,
i.e.
a6 - na3 - m3/27 =
0.
This is a quadratic equation in a3, so solve for
a3 using the usual formula for a quadratic.
Now a
is found by taking cube roots and b can be found in a similar way (or
using b=m/3a).
Then x = a - b is
the solution to the cubic.
-121. Cardan
knew that you could not take the square root of a negative number yet he also
knew that x = 4 was a solution to the equation. He wrote to Tartaglia
on 4 August 1539 in an attempt to clear up the difficulty. Tartaglia
certainly did not understand. In Ars Magna Cardan
gives a calculation with 'complex numbers' to solve a similar problem but he
really did not understand his own calculation which he says is as subtle as
it is useless.
x4 + 2px2 +
p2 = px2 - qx - r +
p2
Now the clever bit. For any y we have
i.e.
(x2 + p)2
= px2 - qx - r + p2
(x2 + p + y)2 =
px2 - qx - r + p2 +
2y(x2 + p) + y2
Now the right
hand side is a quadratic in x and we can choose y so that it is a
perfect square. This is done by making the discriminant zero, in this case
=
(p + 2y)x2 - qx + (p2
- r + 2py + y2) (*) (-q)2 -4(p +
2y)(p2 - r + 2py +
y2) = 0.
Rewrite this last equation as
(q2 - 4p3 + 4 pr) +
(-16p2 + 8r)y - 20 py2 -
8y3 = 0
to see that it is a cubic in y.
(x - b)(x - c)(x - d) =
0.
Harriot
also had a nice method for solving cubics. Consider the cubic
x3 + 3b2x =
2c3
Put x = (e2 -
b2)/e. Then
e6 - 2c3e3 =
b6
which is a quadratic in e3,
and so can be solved for e3 to get
e3 = c3 +
However
(b6 +
c6). e3(e3 - 2c3)
= b6 so that b6/e3 =
-c3 +
Now x = e - b2/e and both
e and b2/e are cube roots of expressions given
above.
(b6 +
c6).