The Babylonian Solution of The Cubic
We have already seen that the Babylonians were adept at solving the quadratic equation.
We now turn to a study of the cubic equation.
The Babylonians were not able to solve the general cubic equation.
This is not surprising when one considers the complicated form of the genreal cubic solution.
It is difficult enough tostate or understand in symbolic notation.
It would be very difficult to write in prose.
There are some special cubic equations which have fairly simple solutions and we will begin with these.
A Babylonian tablet has been discovered which has the values of
n3 + n2
for values of n from 1 to 30. We shall construct such a table for n= 1 to 10 and show that it can
be used to solve cubic equations. Let
f(n) = n3 + n2
then
- f(1) =2
- f(2) =12
- f(3) =36
- f(4) =80
- f(5) =150
- f(6) =252
- f(7) =392
- f(8) =576
- f(9) =810
- f(10) =1100
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- f(11) =1452
- f(12) =1872
- f(13) =2366
- f(14) =2940
- f(15) =3600
- f(16) =4352
- f(17) =5202
- f(18) =6156
- f(19) =7220
- f(20) =8400
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- f(21) =9702
- f(22) =11132
- f(23) =12696
- f(24) =14400
- f(25) =16250
- f(26) =18252
- f(27) =20412
- f(28) =22736
- f(29) =25230
- f(30) =27900
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Verify that this table is correct. Now consider the following equation
x3 + 2x2 = 3136
We wish to use the table to solve this equation.
The first observationis that if the solution is an integer, then it can not be larger than 14. Now recall that
143 + 142= 2940
But we wish to compute
143 + 2(142) = 2940 + 142 = 2940 + 196 = 3136.
This is essentially a method of trial and error. But consider another approach . Let y = 2x.
Then we have the equation
( 2y)3 + 2(2y) 2 =8 y3 + 8y 2 = 3136
Divide 8 into 3136. This may be done by halving repeatedly
(this is likely what the Babylonians would have done). That is
8 divided into 3136
4 into 1568
2 into 784
1 into 392
Then the equation is
y3 + y 2 = 392
which has solution y=7 and x = 14.
A more common method of division is to to double the divisor until we obtaiin the largest
number less than the dividend. This number is subtracted and the process is continued. This is essentially a method of writing the quotient as a sum of powers of two. Consider the division problem we solved above. We start with the divisor 8 and list a column of doubles as follows
- 8 is 8 x 1
- 16 is 8 x 2
- 32 is 8 x 4
- 64 is 8 x 8
- 128 is 8 x 16
- 256 is 8 x 32
- 512 is 8 x 64
- 1024 is 8 x 128
- 2048 is 8 x 256
We see that
- 3136 -2048 leaves 1088 , or 3136 - (8 x 256) = 1088,
- 1088 less 1024 leaves 64, or 1088 - (8 x 128) = 64,
- 64 less 64 leaves 0, or 64 - (8 x 8) = 0
then 8 x ( 256 + 128 + 8) = 3136, or 8 x 392 = 3136.
Consider another example. We wish to divide 12 into 492. We list the doubles of 12.
- 12 is 12 x 1
- 24 is 12 x 2
- 48 is 12 x 4
- 96 is 12x 8
- 192 is 12 x 16
- 384 is 12 x 32
- 492 -384 leaves 108 , or 492 - (12 x 32 ) = 108,
- 108 less 96 leaves 12, or 108- (12 x 8) = 12,
- 12 less 12 leaves 0, or 12 - (12 x 1) = 0
then 12 x (32 + 8 + 1) = 492, or 12 x 41 = 492.
This method was developed for grouping numeral systems. The method is just as easy using Roman Numerals. Let's apply the method using Roman Numerals. Divide CCCCLXXXXII by XXII
- XII is XII x I
- XXIII is XII x II
- XXXXVIII is XII x IIII
- LXXXVI is XII x VIII
- CLXXXXII is 1XII x XVI
- CCCLXXXIII is XII x XXXII
- CCCCLXXXXII - CCCLXXXIIII - leaves CVIII,
- CVIII less LXXXXVI leaves XII,
- XII less XII leaves 0,
then XII x( XXXII + XVI + I) is XII x XXXXI which is CCCCLXXXXII.
Homework Assignment 2.
- Multiply using the method of doubling in the given system.
- 23 x 30
- 15 x 18
- XXI x XVII
- MCX x CXXV
- Divide using the method of halving.
- 7 into 21
- 3 into 12
- XXV / V
- XXXII / IIII
- Solve the cubic equations using the methods outlined above.
- x3 + 5 x2 = 1936
- x3 + x2 = 810
- x3 + 10 x2 = 2000
- x3 + 3x2 = 21870
To continue see The Renassaince solution of the cube
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