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Geometric Approaches to Quadratic Equations from Other Times And Patricia R. Allaire and Robert E. Bradley involved more meaningful. Even those who are Geometric Approaches to comfortable with a symbolic conception of algebra will probably find different points of view to be Quadratic Equations from enriching. In addition, a historical survey of geo- metric techniques can be used to introduce both Other Times and Places completing the square—it even explains the origin of the name—and deriving the quadratic formula. s mathematics evolves, the techniques that survive Finally, our classroom experience suggests that a are those that have the greatest power and general- historical approach always makes the subject more ity. The reason that we teach symbolic algebraic appealing to the typical student. A What follows is a collection of geometric tech- manipulation has everything to do with its efficien- cy and nothing to do with the historical develop- niques from ancient Babylonia, classical Greece, ment of the problems being solved. medieval Arabia, and early modern Europe that We can consider the following problem: can enhance the quadratic-equation portion of an algebra course. The purpose is not to replace the A field has a perimeter of 40 yards and an area symbolic approach but to offer alternative perspec- of 96 square yards. What are the dimensions of tives. Although the topics are interrelated, teachers this field? do not need to present all of them. Rather, we A modern solution would involve two variables, encourage teachers to sample and experiment with one of which is eliminated to derive the quadratic as many techniques as seem appropriate to the equation teaching schedule and the interests and abilities of the students. x2 – 20x + 96 = 0. This equation is then solved by factoring or by GEOMETRICAL ALGEBRA We teach using the quadratic formula. Geometrical algebra is the branch of mathematics algebraic This procedure would seem arbitrary to someone that is concerned with using geometrical concepts manipula- who did not know modern algebraic techniques. and proofs as the underpinnings of algebraic tech- The skills involved, that is, representing quantities niques. Although the ideas of geometrical algebra tions because by letters and performing algebra on these symbols, were clearly present in Asia, Mesopotamia, and of efficiency, seem to have little relevance to the practical prob- Egypt centuries earlier, classical Greek mathemati- not because lem of finding the dimensions of a physical object. cians are usually credited with its development. We teach these skills because of their broad applic- When we geometrically solve a quadratic equation, of historical ability not only to quadratic problems but also to we are doing geometrical algebra. development linear and cubic problems, as well as to problems In the world of geometrical algebra, simple quan- tities, whether fixed numbers or unknowns, are of the involving rational functions, square roots, and more complicated forms. represented by physical objects, almost always a problems In this article, we concentrate exclusively on line segment whose length, relative to some fixed quadratic problems, that is, on problems whose unit, is the quantity in question. A product of two twentieth-century solution involves setting up a quantities is thus interpreted as the area of a rec- quadratic equation in one unknown. Before symbol- tangle, and a product of three quantities is inter- ic manipulation became the lingua franca of mathe- preted as the volume of a right rectangular prism. This interpretation is the origin of our use of the matics, mathematicians could solve quadratic prob- lems. Their techniques usually involved some geometric feature of the situation. Pat Allaire, [email protected], teaches at Queens- Even if we admit the greater efficiency of the borough Community College, Bayside, NY 11364. She is symbolic approach, many good reasons remain for interested in the history of mathematics, especially in the studying geometric solutions of the quadratic. For development of algebra in nineteenth-century Britain, and in community theater. Rob Bradley, [email protected], those who are visually, as opposed to symbolically, teaches at Adelphi University, Garden City, NY 11530. oriented, studying geometric solutions can make His interests include probability theory, ergodic theory, both the problems themselves and the algebra the history of mathematics, and acoustic guitars. 308 MATHEMATICS TEACHER Copyright © 2001 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved. This material may not be copied or distributed electronically or in any other format without written permission from NCTM. words square and cube for second and third powers. the following property: when a square of side x is Because lengths, areas, and volumes are positive added to a rectangle with sides of length 5 and x, quantities, geometrical algebra deals exclusively the result is a rectangle with an area of 36 square with positive numbers and positive values of units. unknowns. Figure 1 is an example of geometrical algebra, illustrating the identity (a + b)2 = a2 + 2ab + b2 for positive a and b, and the figure could even be Geometrical called a proof of the identity. x2 5x = 36 algebra requires that Fig. 3 x2 + 5x = 36. coefficients b ab b2 and roots be positive, so cases must be a a2 ab CLASSIFICATION OF considered QUADRATIC EQUATIONS If we divide a quadratic equation by the coefficient a b of x2, the equation assumes the form x2 + bx + c = 0. Fig. 1 However, since geometrical algebra by its nature An example of geometrical algebra: requires that coefficients and roots be positive, we the identity (a + b)2 = a2 + 2ab + b2. must work by cases when solving them geometri- cally. The five possible cases in which b and c are always positive are the following: A generalization of this figure can be used to (1) x2 = bx illustrate the FOIL technique for multiplying two (2) x2 = c binomials. The following example is illustrated in (3) x2 = bx + c figure 2: (4) x2 + c = bx 2 (x + 3)(x + 2) = x + 2x + 3x + 6 (5) x2 + bx = c = x2 + 5x + 6 Because zero is not an acceptable solution in geo- metrical algebra, x = b is the only solution in case (1). In case (2), the positive square root of c is the only solution; it must be constructed in a geometri- cal sense before we can claim to solve this problem 2 2x 6 geometrically. We turn our attention to this con- struction before considering cases (3), (4), and (5). √ x x2 3x CONSTRUCTION OF c In The Elements, written about 300 B.C., Euclid expressed virtually all his mathematical reasoning x 3 in a geometrical way. This work is a compilation of the basic mathematics of the Greeks beginning Fig. 2 (x + 3)(x + 2) = x2 + 2x + 3x + 6 with Thales (ca. 625–547 B.C.). The construction = x2 + 5x + 6. shown in figure 4 is seen in The Elements (II.14). We can construct a line segment, the length of which is the square root of the length of an arbi- trary line segment, given that segment and a seg- The fact that geometrical algebra deals only with ment of unit length, as follows: positive quantities has certain implications for the Construct AB with AB = c. way that quadratic problems are set up. For exam- • ple, if we want to solve the equation x2 + 5x – 36 = 0 • Extend AB through B to C so that BC = 1. geometrically, we must consider instead the form • Bisect AC at M. x2 + 5x = 36, shown in figure 3. If we read this • With center M and radius AM, construct a equation geometrically, it asks for a quantity x with semicircle. Vol. 94, No. 4 • April 2001 309 • Erect a perpendicular to AC at B. Label as E the point at which the perpendicular intersects the semicircle. Then BE = ¡c. C See figure 4. ¡c A E M B D b x ¡c Fig. 5 M Application of areas: Application B 2 A C x = bx + c. of areas is c 1 believed Fig. 4 Constructing a line segment, to have the length of which is the square root of tion in figure 5 as an “application of areas”? originated an arbitrary line segment The term bx in equation (3) represents the area of a rectangle. We are given the line segment b in with the setup of the problem. We “apply an area” to b Pythagoras when we erect perpendiculars of length x on it and We next demonstrate that this construction gives or his school complete a rectangle. Since the perpendiculars the desired length by showing that triangles BCE have length x, the applied area is bx. and BEA are similar, although Euclid did not use To show that x2 = bx + c, we must show that this the following method of proof: rectangle, along with another rectangle with area c, • ∠EBC and ∠EBA are right angles by construc- makes a square of side x, whose area is x2. See fig- tion and are therefore congruent. ure 6, in which AB has the given length b, and AD • ∠CEA is a right angle because it is inscribed in a semicircle. • ∠C and ∠BEA are both complementary to ∠CEB, so they are congruent. • Thus, the triangles are similar. Since BC = BE , BE BA 3 C 1 = BE , BE `c` ¡c 2 we obtain BE2 = c by cross multiplying. Therefore, A B BE = ¡c. M D APPLICATION OF AREAS The classical Greek method for solving quadratics 1 of types 3, 4, and 5 is called application of areas and is believed to have originated with Pythagoras KLb/2 (ca. 572–497 B.C.) or his school. We illustrate the x x – b 2 H technique as applied to case 3, x = bx + c; see JN figure 5.
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