- MESOPOTAMIA 401
2. Mesopotamia
If we interpret Mesopotamian algebra in our own terms, we can credit the math-
ematicians of that culture with knowing how to solve some systems of two linear
equations in two unknowns, any quadratic equation having at least one real positive
root, some systems of two equations where one of the equations is linear and the
other quadratic, and a potentially complete set of cubic equations. Of course, it
must be remembered that these people were solving problems, not equations. They
did not have any classification of equations in which some forms were solvable and
others not. What they knew was that they could find certain numbers from certain
data.2.1. Linear and quadratic problems. As mentioned in Section 4 of Chapter 6,
the Mesopotamian approach to algebraic problems was to associate with every
pair of numbers another pair: their average and their semidifference. These linear
problems arise frequently as a subroutine in the solution of more complex problems
involving squares and products of unknowns. In Mesopotamia, quadratic equations
occur most often as problems in two unknown quantities, usually the length and
width of a rectangle. The Mesopotamian mathematicians were able to reduce a large
number of problems to the form in which the sum and product or the difference
and product of two unknown numbers are given. We shall consider an example that
has been written about by many authors. It occurs on a tablet from the Louvre in
Paris, known as AO 8862.^2
A loose translation of the text of this tablet, made from Neugebauer's German
translation, reads as follows:I have multiplied the length and width so as to make the area. Then
I added to the area the amount by which the length exceeds the
width, obtaining 3,3. Then I added the length and width together,
obtaining 27. What are the length, width, and area?
27 3,3 the sums
15 length
3,0 area
12 width
You proceed as follows:
Add the sum (27) of the length and width to 3,3. You thereby ob-
tain 3,30. Next add 2 to 27, getting 29. You then divide 29 in half,
getting 14;30. The square of 14;30 is 3,30,15. You subtract 3,30
from 3,30;15, leaving the difference of 0;15. The square root of 0;15
is 0;30. Adding 0;30 to the original 14;30 gives 15, which is the
length. Subtracting 0;30 from 14;30 gives 14 as width. You then
subtract 2, which was added to the 27, from 14, giving 12 as the
final width.
The author continues, verifying that these numbers do indeed solve the problem.
This text requires some commentary, since it is baffling at first. Knowing the
general approach of the Mesopotamian mathematicians to problems of this sort,
one can understand the reason for dividing 29 in half (so as to get the average of
two numbers) and the reason for subtracting 3,30 from the square of 14;30 (the
(^2) AO stands for Antiquites Orientales.