I was reading a bit of math history, which says that the ancient Greeks treated geometrically problems we would treat algebraically. It gives the example of the equation x[sup]2[/sup] + ax = b[sup]2[/sup], which could be stated as:
My comments in the brackets, in case anyone needs them.
Upon reading this, I realize that I haven’t the slightest how to accomplish this without algebra, especially in a general form. I realize that I can put a geometric interpretation on actions I take to solve the equation (e.g., completing the square seems particularly appropriate to this problem), but I can’t see what would motivate me to undertake these actions from the geometric point of view (e.g., why would I add an area of a[sup]2[/sup]/4 to both figures, and then reconstruct the areas to create squares of side x+a/2 and sqrt(b[sup]2[/sup]+a[sup]2[/sup]/4))
Any resident math experts that can show me the way?
This sort of thing gives me a headache, and my thin book on the history of equations only gives a geometric solution to ax - x[sup]2[/sup] = b[sup]2[/sup].
But it does say you need proposition 6 to do x[sup]2[/sup] - ax = b[sup]2[/sup] or x[sup]2[/sup] - ax = b[sup]2[/sup]
That however, is not something I can do right now. But at least you are one step closer to enlightenment.
Let’s try it with a specific example: x[sup]2[/sup] + 6x = 16 (so that a = 6 and b = 4). This will then generalize to other a’s and b’s.
Imagine an x by x square with an x by 6 rectangle stuck onto the side of it, for a total area of x[sup]2[/sup] + 6x. This has to equal, in area, a 4 by 4 square.
+--------+------+
| | |
| | |
| | | x
| | |
| | |
+--------+------+
x 6
Chop off half of the x by 6 rectangle and stick it on top. (You’re just rearranging things, so the area is still x[sup]2[/sup] + 6x.)
Now, in order to make this a perfect square, you have to fill in the corner with a 3 by 3 square (in general, a square of side a/2).
This new square will have the same area as the sum of your 4 by 4 square plus an identical 3 by 3 square (i.e. 25 square units), since it used to be equal to the 4 by 4 square before you added in the 3 by 3 square.
So each side of the new square (which has length x + 3) would be the same as the side of a square of area 25 (i.e. length 5). Since x + 3 = 5, x = 2. (The original equation also has another, negative solution, but anyone thinking geometrically wouldn’t have recognized negative numbers.)
Sorry, Thudlow Boink, but that’s not a geometric solution. A geometric solution starts with the information known, in this case a and b, and ends up with x as a result of geometric operations with ruler and compass. I’ve spent the entire morning on this instead of doing what I should have done, so I hope everyone will appreciate my hard work.
Admittedly it wasn’t as hard once I properly understood what the book did with ax - x[sup]2[/sup] = b[sup]2[/sup] and Euclid’s proposition 5, but my head now hurts anyways.
Also, I’m going to use algebraic notation here, since you’d have to be an ancient greek mathematician to follow the geometric notation.
We start with the equation ax + x[sup]2[/sup] = b[sup]2[/sup] which in geometric terms is equal to the OP’s quote:
To solve this we need to know Euclid’s proposition 6:
Moving those around, in either notation, we get:
ax + x[sup]2[/sup] = (a/2 + x)[sup]2[/sup] - (a/2)[sup]2[/sup]
Inserting this right side for the left side in our original equation we get:
(a/2 + x)[sup]2[/sup] - (a/2)[sup]2[/sup] = b[sup]2[/sup]
We move the negativ part across and get:
(a/2 + x)[sup]2[/sup] = b[sup]2[/sup] + (a/2)[sup]2[/sup]
Because we known our Pythagoras we now construct the right triangle with legs b and a/2, and with the compass mark a segment of it of length a/2, the remainder of the hypothenuse has got to be the segment we were looking for, the x in our algebraic notation.
Not so hard when you know it, and when you can transform it to algebra to convince yourself it’s right, but I’m thoroughly impressed by the guys who thinking only in geometry, could not only develop such propositions, but use them to solve such problems.
Sheesh, I just had to put in a typo, didn’t I. For “and with the compass mark a segment of it of length a/2”, read “and with the compass mark a segment of the hypothenuse of length a/2”.
Also, in case you hadn’t noticed (you probably did, but I’m a bit slow and only just realised), Euclid’s proposition 6 is the geometric equivalent of:
(x + c)[sup]2[/sup] = x[sup]2[/sup] + 2cx + c[sup]2[/sup]
Also, everything up to the last paragraph of Thudlow Boink’s solution is geometric, and actually parallel to what I did, only possibly more comprehensible
The steps in the last paragraph though are, I think, a translation back into algebra and something the greek wouldn’t have done. He’s absolutely right they weren’t into negative numbers though.
