The Greeks thought of a product of two things as an area. For example, ab was thought of as a rectangle with base a and height b. This idea is used in Euclid's "proof" of the quadratic equation a(a - x) = x2 (or x2 + ax - a2 = 0). Note, Euclid's "proof" was more of a construction of the positive root of the equation, followed by a verification. Euclid's Proposition 11, from Book II of Elements is:
To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.
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The goal is to find a point H on the line AB so that AB × HB equals AH × AH. Letting x = AH, and a = AB, then a - x = HB, and the equation to be solved for x is a(a - x) = x2 or x2 + ax - a2 = 0. AB, or a, is the given segment. From this, a square with side a is constructed. This being square ABDC in the figure above. AC is then bisected at point E. EB is drawn, and CA is extended to a point F so that EF = EB. Next draw square FGHA. Now H is the desired point so that x = AH is the positive root of x2 + ax - a2 = 0.
Below on the left is Euclid's verification. On the right modern notation and explanations are given.
| By proposition II, 6 | Prop. II, 6 is basically the identity | |
| CF × FG + AE2 = EF2 | (y + z)(y - z) + z2 = y2 |
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| (y + z)(y - z) = y2 - z2 | ||
| where y = x + a/2 and z = a/2 so that | ||
| By construction EF = EB; | y + z = x + a and y - z = x. | |
| thus | This gives us | |
| CF × FG + AE2 = EB2 | (a + x)(x) + (a/2)2 = (x + a/2)2 in (1) above. | |
| By Pythagorean Theorem, | By Pythagorean Theorem, | |
| CF × FG + AE2 = AB2 + AE2 | (a + x)(x) + (a/2)2 = a2 + (a/2)2 | |
| CF × FG |
(a + x)(x) |
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| AH2 = DB × HB | x2 = a(a - x) | |
| AH2 = AB × HB |
Thus H is the required point so that AH, or x, satisfies (2) on the left above.
You can try this by letting AB = a = 3 to get x2 + 3x - 9 = 0. If you do the above construction, you should find that AX = x = 1.2426, which agrees with the positive root, x = -3 +3 squareroot 2, from the quadratic formula.