Quadratic equations have been considered and solved since Old Babylonian times (c. 1800 BC), but the quadratic formula students memorize today is an 18th century AD development. What did people do in the meantime?
The difficulty with the general quadratic equation (ax2+bx+c=0 as we write it today) is that, unlike a linear equation, it cannot be solved by arithmetic manipulation of the terms themselves: a creative intervention, the addition and subtraction of a new term, is required to change the form of the equation into one which is arithmetically solvable. We call this maneuver completing the square.
In Old Babylonian Mathematics
The Yale Babylonian Collection’s tablet YBC 6967, as transcribed in in Neugebauer and Sachs, Mathematical Cuneiform Texts, American Oriental Society, New Haven, 1986. Size 4.5 × 6.5cm.
The Old Babylonian tablet YBC 6967 (about 1900 BC) contains a problem and its solution. Here is a line-by-line literal translation from Jöran Friberg, A Remarkable Collection of Babylonian Mathematical Texts, (Springer, New York, 2007).
The igi.bi over the igi 7 is beyond. The igi and the igi.bi are what? You: 7 the the igi.bi over the igi is beyond to two break, then 3 30. 3 30 with 3 30 let them eat each other, then 12 15 To 12 15 that came up for you 1, the field, add, then 1 12 15. The equalside of 1 12 15 is what? 8 30. 8 30 and 8 30, its equal, lay down, then 3 30, the holder, from one tear out, to one add. one is 12, the second 5. 12 is the igi.bi, 5 the igi
The Old Babylonians used a base-60 floating-point notation for numbers, so that the symbol corresponding to 1 can represent for example 60 or 1 or 1/60. In the context of YBC 6967, the reciprocal numbers, the igi and the igi.bi, have product 1 0. Their difference is given as 7.
Here is a diagram of the solution to the YBC 6967 problem, adapted from Eleanor Robson’s “Words and Pictures: New Light on Plimpton 322” (MAA Monthly, February 2002, 105-120). Robson uses a semi-colon to separate the whole and the fractional part of a number, but this is a modern insertion for our convenience. The two unknown reciprocals are conceptualized as the sides of a rectangle of area (yellow) 1 0 [or 60 in decimal notation]. A rectangle with one side 3;30 [=3(1/2)] is moved from the side of the figure to the top, creating an L-shaped figure of area 1 0 which can be completed to a square by adding a small square of area 3;30 × 3;30 = 12;15 [=12(1/4)]. The area of the large square is 1 0 + 12;15 = 1 12;15 [=72(1/4)] with square root 8;30 [=81(1/2)] . It follows that our unknown reciprocals must be 8;30 + 3;30 = 12 and 8;30 − 3;30 = 5 respectively.
In modern notation, the YBC 6967 problem would be xy = 60, x − y =7, or x2 − 7x − 60 = 0. In this case the term to be added in completing the square is b2 /( 4a2 ) = 49/4 = 12(1/4) corresponding exactly to the area of the small square in the diagram.
This tablet, and the several related ones from the same period that exist in various collections (they are cataloged in Friberg’s book mentioned above), are significant because they hold a piece of real mathematics: a calculation that goes well beyond tallying to a creative attack on a problem. It should also be noted that none of these tablets contains a figure, even though Old Babylonian tablets often have diagrams. It is as if those mathematicians thought of “breaking,” “laying down” and “tearing out” as purely abstract operations on quantities, despite the geometrical/physical language and the clear (to us) geometrical conceptualization.
In Islamic Mathematics
Solving quadratic equations by completing the square was treated by Diophantus (c.200-c.285 AD) in his Arithmetica, but the explanations are in the six lost books of that work. Here we’ll look at the topic as covered by Muhammad ibn Musa, born in Khwarazm (Khiva in present-day Uzbekistan) and known as al-Khwarizmi, on his Compendium on Calculation by Completion and Reduction dating to c. 820 AD. (I’m using the translation published by Frederic Rosen in 1831). Negative numbers were still unavailable, so al-Khwarizmi, to solve a general quadratic, has to consider three cases. In each case he supposes a preliminary division has been done so the coefficient of the squares is equal to one (“Whenever you meet with a multiple or sub-multiple of a square, reduce it to the entire square”).
“roots and squares are equal to numbers” [x2+bx=a]
“squares and numbers are equal to roots” [x2+a=bx]
3.“roots and numbers are equal to squares” [x2=bx+a]
Case 1. al-Khwarizmi works out a specific numerical example, which can serve as a template for any other equation of this form: “what must be the square which, when increased by ten of its roots, amounts to thirty-nine.”
“The solution is this: you halve the number of the roots, which in the present case equals five. This you multiply by itself; the product is twenty-five. Add this to thirty-nine, the sum is sixty-four. Now take the root of this, which is eight, and subtract from it half the number of the roots, which is five; the remainder is three. This is the root of the square which you sought for; the square itself is nine.”
Note that this is exactly the Old Babylonian recipe, updated from x(x+7)=60 to x2+10x=39, and that the figure Eleanor Robson uses for her explanation is essentially identical to the one al-Khwarizmi gives for his second demonstration, reproduced here:
“We proceed from the quadrate AB, which represents the square. It is our next business to add to it the ten roots of the same. We halve for this purpose the ten, so it becomes five, and construct two quadrangles on two sides of the quadrate AB, namely, G and D, the length of each of them being five, as the moiety of the ten roots, whilst the breadth of each is equal to a side of the quadrate AB. Then a quadrate remains opposite the corner of the quadrate AB. This is equal to five multiplied by five: this five being half of the number of roots which we have added to each side of the first quadrate. Thus we know that the first quadrate, which is the square, and the two quadrangles on its sides, which are the ten roots, make together thirty-nine. In order to complete the great quadrate, there wants only a square of five multiplied by five, or twenty-five. This we add to the thirty-nine, in order to complete the great square SH. The sum is sixty-four. We extract its root, eight, which is one of the sides of the great quadrangle. By subtracting from this the same quantity which we have before added, namely five, we obtain three as the remainder. This is the side of the quadrangle AB, which represents the square; it is the root of this square, and the square itself is nine.”
Case 2. “for instance, 'a square and twenty-one in numbers are equal to ten roots of the same square.”
Solution: Halve the number of the roots; the moiety is five. Multiply this by itself; the product is twenty-five. Subtract from this the twenty-one which are connected with the square; the remainder is four. Extract its root; it is two. Subtract this from the moiety of the roots, which is five; the remainder is three. This is the root of the square you required, and the square is nine.
Here is a summary of al-Khwarizmi’s demonstration. The last of the four figures appears (minus the modern embellishments) in Rosen, p. 18.
The problem set up geometrically. I have labeled the unknown root x for modern convenience. The square ABCD has area x2, the rectangle CHND has area 10x, the rectangle AHNB has area 21, so x2+21=10x.
The side CH is divided in half at G, so the segment AG measures 5−x. The segment TG parallel to DC is extended by GK with length also 5−x. Al-Khwarizmi says this is done “in order to complete the square.”
The segment TK then measures 5, so the figure KMNT, obtained by drawing KM parallel to GH and adding MH, is a square with area 25.
Measuring off KL equal to KG, and drawing LR parallel to KG leads to a square KLRG. Since HR has length 5−(5−x)=x the rectangles LMHR and AGTB have the same area, so the area of the region formed by adding LMHR to GHNT is the same as that of the rectangle formed by adding AGTB to GHNT, i.e. 21. And since that region together with the square KLRG makes up the square KMNT of area 25, it follows that the area of KLRG is 25−21=4, and that its side-length 5−x is equal to 2. Hence x=3, and the sought-for square is 9.
Al-Khwarizmi remarks that if you add that 2 to the length of CG then “the sum is seven, represented by the line CR, which is the root to a larger square,” and that this square is also a solution to the problem.
Case 3. Example: “Three roots and four simple numbers are equal to a square.”
Solution: “Halve the roots; the moiety is one and a half. Multiply this by itself; the product is two and a quarter. Add this to the four; the sum is six and a quarter. Extract its root; it is two and a half. Add this to the moiety of the roots, which was one and a half; the sum is four. This is the root of the square, and the square is sixteen.”
As above, we summarize al-Khwarizmi’s demonstration. The last figure minus decoration appears on Rosen, p. 20.
We represent the unknown square as ABDC, with side-length x. We cut off the rectangle HRDC with side-lengths 3 and x. Since x2=3x+4 the remaining rectangle ABRH has area 4.
Halve the side HC at the point G, and construct the square HKTG. Since HG has length 1(1/2), the square HKTG has area 2(1/4).
Extend CT by a segment TL equal to AH. Then the segments GL and AG have the same length, so drawing LM parallel to AG gives a square AGLM. Now TL = AH = MN and NL = HG = GC = BM, so the rectangles MBRN and KNLT have equal area, and so the region formed by AMNH and KNLT has the same area as ABRH, namely 4. It follows that the square AMLG has area 4+2(1/4)=6(1/4) and consequently side-length AG = 2(1/2). Since GC = 1(1/2), it follows that x=2(1/2)+1(1/2)=4.
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https://www.ams.org/publicoutreach/feature-column/fc-2020-11
A Simple Proof of the Quadratic Formula
The Quadratic Formula was a remarkable triumph of early mathematicians, marking the completion of a long quest to solve arbitrary quadratic equations, with a storied history stretching as far back as the Old Babylonian Period around 2000–1600 B.C. . Over four millennia, many recognized names in mathematics left their mark on this topic, and the formula became a standard part of a first course in Algebra. However, it is unfortunate that for billions of people worldwide, the quadratic formula is also their first (and perhaps only) experience of a rather complicated formula which they must memorize. Countless mnemonic techniques abound, from stories of negative bees considering whether or not to go to a radical party, to songs set to the tune of Pop Goes the Weasel. A derivation by completing the square is usually included in the curriculum, but its computations are somewhat messy, and challenging for first-time Algebra learners to follow. Indeed, the concept of completing the square itself is a significant leap of insight, discovered by ancient masters. This article introduces a surprisingly simple derivation of the quadratic formula, which also produces a computationally-efficient, natural, and easy-to-remember algorithm for solving general quadratic equations. The author would actually be very surprised if this approach has entirely eluded human discovery until the present day, given the 4,000 years of history on this topic, and the billions of people who have encountered the formula and its proof. Yet this technique is certainly not widely taught or known (the author could find no evidence of it in English sources), and so this article seeks at the very least to popularize a delightful alternative approach for solving quadratic equations, which is practical for integration into all mainstream curricula.
I don’t have time to read, let alone understand this, but, very cool!
Whoa!
