Impossible Equations: Algebraic Equations in the Renaissance
Ray Streater
Mathematics Department, King's College London
The present account of the history of cubic and quartic
equations comes from the following sources.
References
 [1]
 Katz, V. J; A History of MathematicsAn Introduction,
AddisonWesley, 1998.
 [2]
 Calinger, C., A Contextual History of Mathematics,
PrenticeHall, 1999.
1 Completing the Square
The Greeks knew how to solve some quadratic equations by completing
the square, for example
The lefthand side of the equation is the shaded area of the diagram. The
missing square has area 25, so
(x+5)^{2} = x^{2}+10x+25 = 39+25 = 64. 
 (2) 
Hence
so x=3 or 13. Negative numbers or even zero were not considered to
be solutions, as numbers were lengths, so: x=3. In modern notation,
x=unknown, and we solve the general equation
by writing
x^{2}+bx = (x+b/2)^{2}b^{4}/4 
 (5) 
so if (4) is true, then
This is the modern formula with a = 1, which is no loss of generality.
Naturally, if b^{2} < 4c we would say that there are complex solutions
but to the Greeks there were no such numbers.
In 1494 Luca Pacioli concluded that (in his opinion) the general cubic
(quartic, quintic and higher) could not be solved by a formula. The historical development of the study of these equations is as follows:
 Cubic.

solved by Scipione in 1515:
 Quartic.

solved by Ferrari in 1545:
x^{4}+bx^{3}+cx^{2}+d = 0 
 (9) 
 Quintic.

showed to be unsolvable by Abel in 1824:
x^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f = 0 
 (10) 
Giralamo Cardano (15011576) invented, or at least popularised, the method of
``linear substitution''. Write any equation as an equivalent equation for
the unknown y, where y = x+p, with p a parameter to be chosen
conveniently. Then x = yp and the quadratic equation becomes
which can be expanded to give a quadratic equation for y:
y^{2}+(b2p)y+p^{2}bp+c = 0. 
 (12) 
Seems more complicated than the original equation, but we can choose
p = b/2, so second term is zero, and we see that y^{2} = a known quantity.
When this method is applied to the cubic equation, we can choose p = b/3 to
remove the coefficient of y^{2}, or by solving a quadratic equation for p,
we can remove the coefficient of y; but unless we are lucky, we cannot
remove both the y^{2} and the y terms, to ``complete the cube''. This
happens only when a special relation exists between b and c.
2 Solving by factorisation
If we can write
then eq. (4) is true if x = a or if x = b. Conversely,
if eq. (4) is true, then either xa = 0 or xb = 0.
The righthand side of eq. (13) is
x^{2}(a+b)x+ab, and is true for all x,
so we can identify
``When the coefficient of x^{2} is 1, then the sum of the roots
of a quadratic is b and their product is c''.
This helps us solve the quadratic for the second solution, if we are
given the first.
Exercise. Show that if ac ¹ 0, then the two solutions to
ax^{2}+bx+c = 0 can be written
Similar things occur for cubics, quartics etc. Write



x^{3}+bx^{2}+cx+d = (xa)(xb)(xg) 
 

x^{3}(a+b+g)x^{2}+(ab+bg+ga)xabg. 

 

Hence



 (16)  

sum of products of roots, two at a time 
 (17)  

product of the three roots. 
 (18) 
 

Similarly, if a,b,g,d are the roots of a quartic,



x^{4}+bx^{3}+cx^{2}+dx+e = (xa)(xb)(xg)(xd) 
 

 

 


 

Therefore, for the quartic with unit coefficient of x^{4}:



 (19)  

sum of products of roots two at a time 
 (20)  

 sum of products of roots 3 at a time 
 (21)  

product of the four roots. 
 (22) 
 

These relations were known for positive roots to Viète 1591, and for complex roots
to Albert Girard (1629).
3 Occurence of cubics etc in the theory of finance
From about 1344, Dardi could solve special cubics arising in finance.
Question At what % compound interest will an investor gain half
his capital after 3 years?
Answer. Interest is compound, so we get the equation

æ ç
è

1+ 
x
100


ö ÷
ø

3

= 1+ 
1
2

. 
 (23) 
Hence we find y: = 1+x/100
y: = 1+x/100 = (3/2)^{1/3} 
 (24) 
and hence x. When we expand eq. (23) we get a cubic equation,
but of the special form that can be solved by ``completing the cube''. The
general cubic was not really needed for finance, but became a challenge
after 1494 and Pacioli's remark. By posing problems as challenges, the
scolars competed for students and the few paid positions.
4 The solution of the cubic
Around 1500, cubics were classified into 13 different types, depending on
whether a coefficient was positive or not, rather than on whether there were
real or complex solutions. The cubic of the form
with any positive c,d,
was solved within 25 years of Pacioli's remarks by Scipione
dal Ferro, probably before 1515, in Bologna. He kept it secret, except
from his student Antonio Fiore, who challenged Niccolo
Tartaglia (14991557) of Venice to a contest, because Tartaglia was
claiming he could solve x^{3}+bx^{2} = d, with b and d as usual, positive.
During the contest, Tartaglia worked out how to solve Fiore's problems,
of the form (25), and won the challenge, as Fiore could not
solve equations with a quadratic term.
At school, I read a book by Sir J. H. Jeans on astronomy, written in about
1920. In it, Sir James remarks
in a threeline footnote that the solution to the cubic was hardly a great
discovery; one just needs to write x = p+q, where p and q are unknowns,
and compare the identity
(p+q)^{3}3pq(p+q)(p^{3}+q^{3}) = 0 
 (26) 
with the equation x^{3}+cx+d = 0 to get
Then, knowing p^{3}+q^{3} and p^{3}q^{3} (sum and product of unknowns), we see
that p^{3} and q^{3} are the solutions to a known quadratic equation, and
hence can be solved. This is Viète's later explanation. It is
probable that Scipione hit on a version of this identity.
5 Arab and Medieval notation
Jeans's comments are unfair. Notation in algebra used words and some
shorthand, and was very cumbersome.
+ was written piu or p
 was written meno or m
x was written cosa, or co, meaning ``thing''
x^{2} was written censo or ce
x^{3} was written co ce
x^{4} was written cece, or æ
= was written  
Ö was written radix or R
(...)^{1/3} was written cub R
and a typical equation was
co ce p 3ce p 3 co  7
Notation changed slowly. Bombelli (15261572) introduced powers:
x^{3} was written
2x^{2} was written
3x was written
Bombelli introduced brackets, but continued to use Rad, cub Rad,
while Ö had been introduced in Germany in 1525 by Rudolff, with
ÖÖ as cuberoot and ÖÖÖ as fourth root
(only the first survived). Also in Germany, Stifel used letters A,B,C,D,F
and also x as unknowns, and in France, Viète writes A,E,I,O,U,Y
as unknowns, and B,C,D,...as given numbers. He also introduced
for division of A by B. Even 50 years after
Pacioli, it must have been hard to manipulate equations. Descartes later
introduced a,b,c¼ for knowns and x,y,z for unknowns.
6 Tartaglia's quarrel with Cardano
Tartaglia had won the contest with Scipioni's student Fiore,
but had kept the method secret. Cardano persuaded Tartaglia to tell him,
on oath not to put it in his, Cardano's, forthcoming book on arithmetic.
He kept his oath, but then Cardano and his student Ferrari (15221565)
visited Bologna where Scipione's soninlaw, Annibale della Nave,
was in charge of Scipione's papers, which they were allowed to see.
Cardano confirmed that Scipione had the solution 20 years before Tartaglia,
and that Scipione's special case could be used to solve Tartaglia's cases,
by linear substitution. So he felt under no further obligation to keep the
secret, and he published the solutions in Ars Magna (The Great Work)
in 1545, which included a solution to the quartic, by then solved by Ferrari.
Tartaglia was furious, though Cardano had given him credit for independently
solving the cubic) and challenged Ferrari to a contest, of which Ferrari
was deemed the winner.
To this day, the formula for the cubic is known as Cardano's formula.
He never claimed that he found it himself, but he wrote a good book
about it.
7 Cardano's Formula for the Cubic
Suppose
Write x = p+q. Then
p^{3}q^{3} = C^{3}/27; p^{3}+q^{3} = D. 

So, p^{3},q^{3} are the two solutions to
y^{2}+DyC^{3}/27 = 0, so 

y = {D±Ö(D^{2}+4C^{3}/27)}/2 

One solution is p^{3}, the other is q^{3}. Therefore
x = 
ì í
î


D+Ö(D^{2}+4C^{3}/27)
2


ü ý
þ

1/3

+ 
ì í
î


DÖ(D^{2}+4C^{3}/27)
2


ü ý
þ

1/3

. 
 (29) 
8 Ferrari's solution to the quartic
First reduce the equation by removing the x^{3} term by linear substitution:
put y=x+b/4 in eq. (9). Then rearrange, to get
Now add 2px^{2}+p^{2} to each side, where p is a parameter to be conveniently
chosen later. We get
x^{4}+2px^{2}+p^{2} = (2pc)x^{2}dx+(p^{2}e) 
 (31) 
The lefthand side is a perfect square, whatever we choose for p: it is
(x^{2}+p)^{2}. Choose p so that the righthand side is a quadratic
expression with two equal roots, say a. This means ``b^{2} = 4ac'', that is
d^{2} = 4(2pc)(p^{2}e). 

This is a cubic in p with known coefficients, so can be
solved by Cardano's formula. Also we then know a = d/(2(2pc)), from
the quadratic with equal roots. Then we may take the square root of
each side of eq. (31), to give
a quadratic equation to find x.
9 My method of solution
Following Jeans's note, I tried for a year to solve quartics, and succeeded
in November 1953. We write x = p+q+r and compare the identity below
with the equation:



2(p^{2}+q^{2}+r^{2})x^{2}8pqr x+(p^{2}+q^{2}+r^{2})^{2}4(p^{2}q^{2}+p^{2}r^{2}+q^{2}r^{2}) = 0 
 


 

This gives the symmetric polynomials of p^{2},q^{2},r^{2}:



 


4(p^{2}q^{2}+p^{2}r^{2}+q^{2}r^{2})+c^{2}/4 



 

This gives a cubic, whose three solutions are p^{2},q^{2},r^{2}. I spent most of
the year with another, less useful identity, which led nowhere. Then I
found this one, and got a cubic for p^{4},q^{4},r^{4} instead of the squares.
I had it stuck in my mind that the analogue of Viète's method should require
fourth roots in the answer.
10 Bombelli's (15261572) Complex Numbers
Cardano often listed negative solutions, but had doubts as to their meaning.
He also met complex solutions to quadratics, but concluded that they were
useless and that algebra had run away to a point beyond its usefulness.
He also noticed that the solution to x^{3}+6x = 20 is given (by the famous
formula) as
x = {(Ö(108)+10)}^{1/3}{(Ö(108)10)}^{1/3} 

This is the solution 2, but how can we see it? Worse, for the equation
x^{3} = 15x+4, the ``Cardano'' formula gives
x = {(2+Ö(121))}^{1/3}+{(2Ö(121))}^{1/3} 

though the solution is 4, a real number. Bombelli realised that he
had to go via complex numbers; he argued that (if the solution
is an integer) it must be 4, by introducing complex numbers and the rules
for their manipulation. He had the first notation for complex numbers:
23i was written 2m di m3 

This was done at a time when even negative numbers were suspect.
11 Viète
We have mentioned Viète's better notation. He also explained the Cardano
formula using the algebra in Jeans's footnote. He showed how to use
trigonometry to solve the cuberoot problem of complex numbers. Suppose that
we can use linear substitution y = x+p and scaling z = qy to transform
the cubic to
Then complex numbers will be needed in the solution. But we can compare with
cos^{3}q 
3
4

cosq = 
1
4

cos(3q). 

Thus put z = cosq, and we find q from cos(3q) = 4A, from
tables of cosines, which were then becoming available. This paved the way
for de Moivre's formula.
Another 280 years were to pass without any
progress on the quintic. Abel in 1824 proved that there exists NO
formula involving a finite number of surds giving the solution to the
general quintic. Galois (aged 20) gave a general theory of when a quintic
has a solution using surds, and invented group theory and classfield
theory in his thesis. He wrote it up the day before dying in a duel.
File translated from T_{E}X by T_{T}H, version 2.01.
On 3 Jul 2001, 21:00.