Question 621183
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ =\ 16]


The Babylonian method is nothing more than using geometric figures to perform the operation of completing the square.


Factor the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x\ +\ 6)\ =\ 16]


This equation can be represented geometrically by a rectangle that measures *[tex \LARGE x] by *[tex \LARGE x\ +\ 6].


The rectangle described can be divided into two parts, one that is a square that mesures *[tex \LARGE x] on each side, and the other a *[tex \LARGE 6] by *[tex \LARGE x] rectangle.


If the *[tex \LARGE 6] by *[tex \LARGE x] rectangle is further divided in half to form two *[tex \LARGE 3] by *[tex \LARGE x] rectangles, then one of those smaller rectangles can be moved to either adjacent side of the *[tex \LARGE x] by *[tex \LARGE x] square.


The figure that results is a figure that has an area of 16 and is almost a square in shape except for a *[tex \LARGE 3] by *[tex \LARGE 3] square in one corner.  Adding that square adds an area of 9 to the overall figure, giving the overall figure an area of 25.  We can describe this algebraically by summing the areas of the four pieces that make the overall square:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 3x\ +\ 3x\ +\ 9\ =\ 25]


Combining terms


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ +\ 9\ =\ 25]


We recognize a perfect square trinomial in the LHS, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 3)^2\ =\ 25]


Taking the root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 3\ =\ \pm5]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -8]


Though I suspect the geometrically minded Babylonians may well have discarded the negative root as non-sensical in a world where quantities are considered measures of length. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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