Thursday, 13 November 2014

A 23rd (and the very first known?) way to solve a quadatic

 A very long time ago (2007) in a land far, far away (Lakenheath, England) I completed a somewhat longer than average paper (for me) with the somewhat tongue-in-cheek title of "There must be Twenty Ways to Solve a Quadratic," with reference to an even older song called "Fifty Ways to Leave your Lover."
Among them were several nice geometric approaches, and much later (May 2014) I blogged here about two additional geometric approaches I found quite lovely.
Today while looking at a Kindle sample of Taming the Unknown by Victor Katz and Karen Parshall, I discovered yet another that I had overlooked, and it might well be the oldest quadratic problem ever found, and thus the earliest approach ever printed. The authors refer to it as "the only example extant in the Egyptian Papyri...".  The approach was well known, but only it's application to quadratics had escaped my attention. 
I will describe this 23rd solution to a quadratic with one other very early quadratic problem from the same source to follow.
In the Berlin Papyrus (c 1900 BC) a problem is given to divide a square of 100 sq units into two squares where the ratio of the sides of the squares is 1: 3/4.
This would be written in modern notation as \(x^2 + (\frac{3x}{4})^2 = 100 \)
The scribe's approach was to use the rule of false position. He begins by assuming one side of the larger square has sides of 1 unit, and the other would then be 3/4. From this he obtains a total sum of squares to be \( \frac{25}{16}\) The scribe then chose to take the square root of this ratio since it represents a ratio between areas, not the side lengths he was comparing, to obtain 5/4. Knowing that the sides of the original square were ten units, he computed the number of multiples of 5/4 to achieve ten by dividing, and realized that his final sides must be 8 times as large as his false trial, or 8 units for one square and six for the other, producing squares with areas of 64 sq units and 36 sq units to make the desired total of 100 sq units. 
Not that the problem solved in this example is of the simple form Ax2 = k.  I have not worked out for myself yet whether such an approach is feasible for a common trinomial quadratic \(Ax^2 + Bx  = C\) . 

In the same book they describe a problem from a Mesopotamian Tablet (C 1700 BC (see bottom of post)) that is the first known quadratic to be solved by the method of completing the square. I think this is still my favorite of all the ways to solve quadratics because of it's geometric elements. The table is YBC4663 and the eighth problem describes in part, the problem states that in rectangle the length exceeds the width by 3 1/2 units and the area is 7 1/2 square units. The problem is to find the lengths of the two sides of the rectangle. The scribe uses the method of least squares in purely geometric methods to solve the challenge.
from Math is (Let b= 3 1/2)
He first divides the rectangle into a square and another rectangle by subtracting 3 1/2 from the longer side and dropping a perpendicular to the short side, thus creating a square of x2 units and another that is 3 1/2 x square units.
Then he divides the rectangle in half (a process of geometric manipulation by cut and paste that seems to be unique to the Mesopotamian mathematics) and matching the two pieces against the x by x square to make an l-shaped gnomen (square with a square cut out of one corner). Knowing the edges of the two small rectangles forming this "missing square" is \( \frac{7}{4})^2= \frac{49}{16}, the scribe realizes that if he added that area to the original 7 1/2, he would have an area of 10 9/16 or 169/16 .
He takes the square root to determine that the completed square has sides of 3 1/4 units, and that that if he adds the side of the small square to the large one, (3 1/4 + 1 3/4 = 5) he will get the length of the original rectangle, and by subtracting it, (3 1/4 - 1 3/4 =1 1/2) he gets the width. 
My search continues, so if you know of an approach unique to all these, be assured, I'm looking for solution method number 24 .
My 2007 document has a statement that completing the square was later in Mesopotamian Math, around 400 BCE. This seems to be contradicted by several assertions that the above Tablet was from the Old Babylonian period (c.1792 - 1595 BC)
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