Quadratic False Position

Quadratic False
Position and the
quadratic formula
Babylonians
And
Diophantus
Quadratic False
Position and the
quadratic formula
But first let’s recall what
quadratic equations look
like “nowadays” and how we solve them:
This type of problem appears
on a Babylonian tablet
~1700BC
x
y
xy = 16
x + y = 10
I am thinking of a rectangle which has …
(Finish this sentence.)
€
This type of problem appears
on a Babylonian tablet
~1700BC
x
y
xy = 16
x + y = 10
Write this down in your notes.
Let’s think how we might guess:
€
As in false position, let’s make
a guess about this rectangle:
  5 x 5 = 25 sq. units,
but we wanted
xy = 16
  How far off are
we?
  25 - 16 = 9 units of
area.
5
5
  Error = 9 sq. units
5
5
3
5
3
5
3
3
5
5
5
2
3
3
2
3
3
2
x
y
xy = 16
x + y = 10
Our solutions for the sides of the rectangle
are the lengths 8 and 2.
€Let’s try another:
€
Try this one at your table:
x
y
xy = 45
x + y = 18
I am looking for two numbers whose
product is 45 and whose sum is 18.
(Hint: x=y=9 works in the bottom equation.)
€
Try this one on our own:
x
y
xy = 45
x + y = 18
Draw the pictures for this one:
€
What did you get?
x
y
xy = 45
x + y = 18
The two numbers whose product is 45
and whose sum is 18 are … 3 and 15.
€
Diophantus (200 - 284)
“There are, however, many other
types of problems considered
by Diophantus.” (MACTUTOR Biography)
Use the Babylonian system on this
problem of Diophantus:
mn = 9
m + n = 10
Solve this pair of equations.
Try 5. (What is my system for choosing the first guess?)
Did you get 9 and 1 for m and n ?
==============
Diophantus has a different idea:
He wants us to plan ahead a bit.
He introduces x to be the difference that we
would soon be adding and subtracting from 5
if we were to do it the Babylonian’s way.
So he replaces m by (5 + x) and n by (5 - x):
mn = 9 becomes (5 + x)(5 - x) = 9
mn = 9 becomes
(5 + x)(5 - x) = 9
25 - x2 = 9
x2 = 16
So, X = 4 **
So we get m = 9 and n = 1
just as Diophantus tells us to do.
What’s been the point of the
two talks on False Position?
  To see and understand how early
mathematicians solved equations;
  To experience a style of doing
algebra that is different from the
way we have been taught; and
  To wonder at how much early
scribes and others really
understood.
Thanks for your
attention and
hard work !
How does this system
relate to “our”quadratic
formula?
Let’s consider:
mn = c
m + n = −b
* Then our first guess is -1/2 b.
€
* Our error will
€ be (-1/2 b)2 - c.
* Take the square root of that:
€
So that square root is the amount that we must add and
subtract from -1/2b. This is what we get for solutions:
m=
−1
1 2
b+
b −c
2
4
2
€
−b + b − 4c
2
2
=
−1
b − 4c
b+
2
4
2
=
€
−b + b − 4ac
2a
The value for n looks nearly the same.
€
=
(when a = 1)
Remember we started with these two equations:
mn = c
m + n = −b
So:
n = −b − m
2
mn = m(−b
−
m)
=
−bm
−
m
=
c
€
€ the last equality in x instead:
If we were to write
€
€
€
2
x + bx + c = 0
OK,
now we’re done!