,
,
, if c < (b/2)2, and
.
Notice: This material will be included in a forthcoming (summer 2000) book with the tentative title Experiencing Geometry in Euclidean, Spherical, and Hyperbolic Spaces. This new book will be an expanded and updated version of Experiencing Geometry on Plane and Sphere. This material is in draft form and may not be duplicated or quoted without the author's written permission, except for purposes of review or trying out the material with students. As always comments are welcome and will affect the final draft. Send comments to dwh2@cornell.edu.
Chapter 15
Whoever thinks algebra is a trick in obtaining unknowns has thought it in vain. No attention should be paid to the fact that algebra and geometry are different in appearance. Algebras (jabbre and maqabeleh) are geometric facts which are proved by Propositions Five and Six of Book Two of [Euclid's] Elements.
— Omar Khayyam, a paper [A: Khayyam (1963)]
In this chapter we will see how the results from Chapter 14 were used historically to solve equations. Quadratic equations were solved by "completing the square" — a real square. These in turn lead to conic sections and cube roots and culminate in the beautiful general method from Omar Khayyam which can be used to find all the real roots of cubic equations. Along the way we shall clearly see some of the ancestral forms of our modern Cartesian coordinates and analytic geometry. I will point out several inaccuracies and misconceptions that have crept into the modern historical accounts of these matters. But I urge you to not look at this only for its historical interest but rather also look for the meaning it has in our present-day understanding of mathematics. This path is not through a dead museum or petrified forest; it passes through ideas which are very much alive and which have something to say to our modern technological, increasingly numerical, world.
Problem 15.1. Quadratic Equations
Finding square roots is the simplest case of solving quadratic equations. If you look in some history of mathematics books (e.g., [G: Joseph] and [G: Eves]), you will find that quadratic equations were extensively solved by the Babylonians, Chinese, Indians, and Greeks. However, the earliest known general discussion of quadratic equations took place between 800 and 1100 AD in the Muslim Empire. Best known are Mohammed Ibn Musa al'Khowarizmi (who lived in Baghdad and from whose name we get our word "algorithm") and Omar Khayyam (the Persian geometer who is mostly known in the West for his philosophical poetry The Rubaiyat). Both wrote books entitled Al-jabr w'al mugabalah (from which we get our word "algebra"), al'Khowarizmi in about 820 AD and Khayyam in about 1100 AD. An English translation of both books is available in many libraries, if you can figure out whose name it is catalogued under (see, [A: al'Khowarizmi] and [A: Khayyam, 1931]). We previously met Khayyam in Chapter 12.
In these books you find geometric and numerical solutions to quadratic equations and geometric proofs of these solutions. But the first thing that you notice is that there is not one general quadratic equation as we are used to it:
ax2 + bx + c = 0.
Rather, because the use of negative coefficients and negative roots was avoided, they list six types of quadratic equations (we follow Khayyam's lead and set the coefficient of x2 equal to 1):
,
,
, if c < (b/2)2, and
.
a. Show that these are the only types. Why is x2 + bx + c = 0 not included? Explain why b must be a length but c an area.
The avoidance of negative numbers was widespread until a few hundred years ago. In the sixteenth century, European mathematicians called the negative numbers that appeared as roots of equations "numeri fictici" — fictitious numbers (see [A: Cardano, page 11]). In 1759 Baron Francis Masères, mathematician and a Fellow at Cambridge University and a member of the Royal Society, wrote in his Dissertation on the Use of the Negative Sign in Algebra:
...[negative roots] serve only, as far as I am able to judge, to puzzle the whole doctrine of equations, and to render obscure and mysterious things that are in their own nature exceeding plain and simple.... It were to be wished therefore that negative roots had never been admitted into algebra or were again discarded from it: for if this were done, there is good reason to imagine, the objections which many leaned and ingenious men now make to algebraic computations, as being obscure and perplexed with almost unintelligible notions, would be thereby removed; it being certain that Algebra, or universal arithmetic, is, in its own nature, a science no less simple, clear, and capable of demonstration, than geometry.
More recently in 1831, Augustus De Morgan, a famous professor of mathematics at University College, London, wrote in his On the Study and Difficulties of Mathematics:
The imaginary expression 
and the negative expression - b have this resemblance, that either of them occurring as solution of a problem indicates some inconsistency or absurdity. As far as real meaning is concerned, both are equally imaginary, since 0 - a is as inconceivable as 
.
b. Speculate about why these mathematicians avoided negative numbers and why they said what they said?
To get a feeling for why, think about the meaning of 2 x 3 as two 3's and 3 x 2 as three 2's and then try to find a meaning for 3 x (-2) and -2 x (+3). Also consider the quotation at the beginning of this chapter from Omar Khayyam about algebra and geometry. Some historians have quoted this passage but have left out all the words appearing after "proved." In my opinion, this omission changes the meaning of the passage. Euclid's propositions that are mentioned by Khayyam are the basic ingredients of Euclid's proof of the square root construction and form a basis for the construction of conic sections — see below. Geometric justification when there are negative coefficients is at the least very cumbersome, if not impossible. (If you doubt this, try to modify some of the geometric justifications below.)
c. Find geometrically the algebraic equations which express all the positive roots of each of the six types. Fill in the details in the following sketch of Khayyam's methods for Types 3-6.
For the geometric justification of Type 3 and the finding of square roots, Khayyam refers to Euclid's construction of the square root in Proposition II 14, which we discussed in Problem 14.1.
For Type 4, Khayyam gives the following as geometric justification:
Figure 15.1. Type 4.
and thus, by "completing the square" on x + b/2, we have
(x + b/2)2 = c + (b/2)2.
Thus we have x = ...? Note the similarity between this and Baudhayana's construction of the square root (see Chapter 14).
For Type 5, Khayyam first assumes x < (b/2) and draws the equation as:
Figure 15.2. Type 5, x < (b/2).
and note that the square on b/2 is (b/2 - x)2 + c.
Figure 15.3. Type 5, x < (b/2).
This leads to 
. Note that if c > (b/2)2, then this geometric solution is impossible. When x > (b/2), Khayyam uses the drawings:
Figure 15.4. Type 5, x < (b/2).
For solutions of Type 6, Khayyam uses the drawing in Figure 15.5.
Figure 15.5. Type 6.
Do the above solutions find the negative roots? Well, first, the answer is clearly, No, if you mean: Did al'Khowarizmi and Khayyam mention negative roots? But let us not be too hasty. Suppose -r (r, positive) is the negative root of x2 + bx = c. Then (-r)2 + b(-r) = c or r2 = br + c. Thus r is a positive root of x2 = bx + c! The absolute value of the negative root of x2 + bx = c is the positive root of x2 = bx + c and vice versa. Also, the absolute values of the negative roots of x2 + bx + c = 0 are the positive roots of x2 + c = bx. So, in this sense, Yes, the above geometric solutions do find all the real roots of all quadratic equations. Thus it is misleading to state, as most historical accounts do, that the geometric methods failed to find the negative roots. The users of these methods did not find negative roots because they did not conceive of them. However, the methods can be directly used to find all the positive and negative roots of all quadratics.
d. Use Khayyam's methods to find all roots of the following equations: x2 + 2x = 2, x2 = 2x + 2, x2 + 3x + 1 = 0.
Problem 15.2. Conic Sections and Cube Roots
The Greeks noticed that, if a/c = c/d = d/b, then (a/c)2 = (c/d)(d/b) = (c/b) and thus c3 = a2b. Now setting a = 1, we see that we can find the cube root of b, if we can find c and d such that c2 = d and d2 = bc. If we think of c and d as being variables and b a constant, then we see these equations as the equations of two parabolas with perpendicular axes and the same vertex. The Greeks also saw it this way but first they had to develop the concept of a parabola!
To the Greeks, and later Khayyam, if AB is a line segment, then the parabola with vertex B and parameter AB is the curve P such that, if C is on P, then the rectangle BDCE (see Figure 15.6) has the property that (BE)2 = BD · AB. Since in Cartesian coordinates the coordinates of C are (BE,BD) this last equation becomes a familiar equation for a parabola.
Points of the parabola may be constructed by using the construction for the square root given in Chapter 14. In particular, E is the intersection of the semicircle on AD with the line perpendicular to AB at B. (The construction can also be done by finding D' such that AB = DD', then the semicircle on BD' intersects P at C.) I encourage you to try this construction yourself; it is very easy to do if you use a compass and graph paper.
Figure 15.6. Construction of parabola.
Now we can find the cube root. Let b be a positive number or length and let AB = b and construct C so that CB is perpendicular to AB and such that CB = 1. See Figure 15.7. Construct a parabola with vertex B and parameter AB and construct another parabola with vertex B and parameter CB. Let E be the intersection of the two parabolas. Draw the rectangle BGEF. Then
(EF)2 = BF·AB and (GE)2 = GB·CB.
But, setting c = GE = BF and d = GB = EF, we have
d2 = cb and c2 = d. Thus c3 = b.
If you use a fine graph paper, it is easy to get three-digit accuracy in this construction.
Figure 15.7. Finding cube roots.
The Greeks did a thorough study of conic sections and their properties which culminated in Apollonius's book Conics which appeared in 200 BC. You can read this book in English translation; see [A: Apollonius].
a. Use the above geometric methods with a fine graph paper to find the cube root of 10.
To find roots of cubic equations we shall also need to know the (rectangular) hyperbola with vertex B and parameter AB. This is the curve H, such that if E is on H and ACED is the determined rectangle (see Figure 15.8), then (EC)2 = BC·AC.
The point E can be constructed using the construction from Chapter 14. Let F be the bisector of AB. Then the circle with center F and radius FC will intersect at D the line perpendicular to AB at A. From the drawing it is clear how these circles also construct the other branch of the hyperbola (with vertex A.)
Figure 15.8. Construction of hyperbola.
b. Use the above method with graph paper to construct the graph of the hyperbola with parameter 5. What is an algebraic equation that represents this hyperbola?
Notice how these descriptions and constructions of the parabola and hyperbola look very much like they were done in Cartesian coordinates. The ancestral forms of Cartesian coordinates and analytic geometry are evident here. Also they are evident in the solutions of cubic equations in the next section. The ideas of Cartesian coordinates did not appear to Descartes out of nowhere. The underlying concepts were developing in Greek and Muslim mathematics. One of the apparent reasons that full development did not occur until Descartes is that, as we have seen, negative numbers were not accepted. The full use of negative numbers is essential for the realization of Cartesian coordinates.
Problem 15.3. Roots of Cubic Equations
In his Al-jabr wa'l muqabalah, Omar Khayyam also gave geometric solution to cubic equations. You will see that his methods are sufficient to find geometrically all real (positive or negative) roots of cubic equations; however; in his first chapter Khayyam says (see [AT: Khayyam (1931), page 49]):
When, however, the object of the problem is an absolute number, neither we, nor any of those who are concerned with algebra, have been able to prove this equation — perhaps others who follow us will be able to fill the gap — except when it contains only the three first degrees, namely, the number, the thing and the square.
By "absolute number," Khayyam is referring to, what we call, algebraic solutions as opposed to geometric ones. This quotation suggests, contrary to what many historical accounts say, that Khayyam expected that algebraic solutions would be found.
Khayyam found 19 types of cubic equations (when expressed with only positive coefficients). (See [AT: Khayyam (1931), page 51].) Of these 19, 5 reduce to quadratic equations (for example, x3 + ax = bx reduces to x2 + ax = b). The remaining 14 types Khayyam solves by using conic sections. His methods find all the positive roots of each type although he failed to mention some of the roots in a few cases; and, of course, he ignores the negative roots. Instead of going through his 14 types, I will show how a simple reduction will reduce all the types to only 3 types in addition to types already solved such as, x3 = b. I will then give Khayyam's solutions to these three types.
In the cubic y3 + py2 + gy + r = 0 (where, here, p, g, r, are positive, negative, or zero), set y = x - (p/3). Try it! The resulting equation in x will have the form x3 + sx + t = 0, (where, here, s and t are positive, negative, or zero). If we rearrange this equation so all the coefficients are positive then we get four types that have not been previously solved:
(1) x3 + ax = b, (2) x3 + b = ax,
(3) x3 = ax + b, and (4) x3 + ax + b = 0,
where a and b are positive, in addition to types previously solved.
a. Show that in order to find all the roots of all cubic equations we need only have a method that finds the roots of Types 1, 2, and 3.
Khayyam's solution for Type 1: x3 + ax = b.
A cube and sides are equal to a number. Let the line AB [see Figure 15.9] be the side of a square equal to the given number of roots [that is, (AB)2=a, the coefficient]. Construct a solid whose base is equal to the square on AB, equal in volume to the given number [ b ]. The construction has been shown previously. Let BC be the height of the solid. [That is,BC·(AB)2 = b.] Let BC be perpendicular to AB ... Construct a parabola whose vertex is the point B ... and parameter AB. Then the position of the conic HBD will be tangent to BC. Describe on BC a semicircle. It necessarily intersects the conic. Let the point of intersection be D; drop from D, whose position is known, two perpendiculars DZ and DE on BZ and BC. Both the position and magnitude of these lines are known.
Figure 15.9. Type 1 cubic.
The root is EB. Khayyam's proof (using a more modern, compact notation) is: From the properties of the parabola (Problem 15.2) and circle (Chapter 13) we have
(DZ)2 = (EB)2 = BZ·AB and (ED)2 = (BZ)2 = EC·EB ,
thus
EB·(BZ)2 = (EB)2·EC = BZ·AB·EC
and therefore
AB·EC = EB·BZ
and
(EB)3 = EB·(BZ·AB) = (AB·EC)·AB = (AB)2·EC.
So
(EB)3 + a(EB) = (AB)2·EC + (AB)2·(EB) = (AB)2·CB = b.
Thus EB is a root of x3 + ax = b. Since x2 + ax increases as x increases, there can be only this one root.
Khayyam's solutions for Types 2 and 3:
x3 + b = ax and x3 = ax + b.
Khayyam treated these equations separately but by allowing negative horizontal lengths we can combine his two solutions into one solution of x3 ± b = ax. Let AB be perpendicular to BC and as before let (AB)2 = a and (AB)2·BC = b. Place BC to the left if the sign in front of b is negative (Type 3) and place BC to the right is the sign in front of b is positive (Type 2). Construct a parabola with vertex B and parameter AB. Construct both branches of the hyperbola with vertices B and C and parameter BC.
Figure 15.10. Types 2 and 3 cubics.
Each intersection of the hyperbola and the parabola (except for B) gives a root of the cubic. Suppose they meet at D. Then drop perpendiculars DE and DZ. The root is BE (negative if to the left and positive if to the right). Again, if you use fine graph paper, it is possible to get three digit accuracy here. I leave it for you, the reader, to provide the proof which is very similar to Type 1.
b. Verify that Khayyam's method described above works for Types 2 and 3. Can you see from your verification why the extraneous root given by B appears?
c. Use Khayyam's method to find all solutions to the cubic
x3 = 15x + 4.
Use fine graph paper and try for three-place accuracy.
Problem 15.4. Algebraic Solution of Cubics
A little more history: Most historical accounts assert correctly that Khayyam did not find the negative roots of cubics. However, they are misleading in that they all fail to mention that his methods are fully sufficient to find the negative roots as we have seen above. This is in contrast to the common assertion (see, for example, [SE: Davis & Hersch]) that Girolamo Cardano (sixteenth-century Italian) was the first to publish the general solution of cubic equations when in fact, as we shall see, he himself admitted that his methods are insufficient to find the real roots of many cubics.
Cardano published his algebraic solutions in his book, Artis Magnae (The Great Art) which was published in 1545. For a readable English translation and historical summary, see [AT: Cardano]. Cardano used only positive coefficients and thus divided the cubic equations into the same 13 types (excluding x3 = c and equations reducible to quadratics) used earlier by Khayyam. Cardano also used geometry to prove his solutions for each type. As we did above, we can make a substitution to reduce these to the same types as above:
(1) x3 + ax = b, (2) x3 + b = ax,
(3) x3 = ax + b, and (4) x3 + ax + b = 0.
If we allow ourselves the convenience of using negative numbers and lengths, then we can reduce these to one type: x3 + ax + b = 0, where now we allow a and b to be either negative or positive.
The main "trick" that Cardano used was to assume that there is a solution of x3 + ax + b = 0 of the form x = t1/3 + u1/3. Plugging this into the cubic we get
(t1/3 + u1/3)3 + a(t1/3 + u1/3) + b = 0.
If you expand and simplify this, you get to
t + u + b + (3t1/3u1/3 + a)(t1/3 + u1/3) = 0.
(Cardano did this expansion and simplification geometrically by imagining a cube with sides t1/3 + u1/3.) Thus x = t1/3 + u1/3 is a root if
t + u = - b and t1/3 u1/3 = -(a/3).
Solving, we find that t and u are the roots of the quadratic equation
z2 + bz - (a/3)3 = 0
which Cardano solved geometrically (and so can you, Problem 15.1) to get
and 
.
Thus the cubic has roots
x = t1/3 + u1/3
= {
}1/3 + {
}1/3.
This is Cardano's cubic formula. But, a strange thing happened. Cardano noticed that the cubic x3 = 15x + 4 has a positive real root 4 but, for this equation, a = -15 and b = -4, and if we put these values into his cubic formula, we get that the roots of x3 = 15x + 4 are
x = {
}1/3 + {
}1/3 .
But these are complex numbers even though you have shown in Problem 15.3 that all three roots are real. How can this expression yield 4?
In Cardano's time there was no theory of complex numbers and so he reasonably concluded that his method would not work for this equation, even though he did investigate expressions such as 
. Cardano writes ([A: Cardano, page 103]):
When the cube of one-third the coefficient of x is greater than the square of one-half the constant of the equation ... then the solution of this can be found by the aliza problem which is discussed in the book of geometrical problems.
It is not clear what book he is referring to, but the "aliza problem" presumably refers to al'Hazen, an Arab, who lived around 1000 AD and whose works were known in Europe in Cardano's time. Al'Hazen had used intersecting conics to solve specific cubic equations and the problem of describing the image seen in a spherical mirror — this later problem is in some books called "Alhazen's problem."
In addition, we know today that each complex number has three cube roots and so the formula
x = {
}1/3 + {
}1/3
is ambiguous. In fact, some choices for the two cube roots give roots of the cubic and some do not. (Experiment with x3 = 15x + 4.) Faced with Cardano's Formula and equations like x3 = 15x + 4, Cardano and other mathematicians of the time started exploring the possible meanings of these complex numbers and thus started the theory of complex numbers.
a. Solve the cubic x3 = 15x + 4 using Cardano's Formula and your knowledge of complex numbers.
Remember that on the previous page we showed that x = t1/3 + u1/3 is a root of the equation if t + u = - b and t1/3 u1/3 = -(a/3).
b. Solve x3 = 15x + 4 by dividing through by x - 4 and then solving the resulting quadratic.
c. Compare you answers and methods of solution from Problems 15.3c, 15.4a, and 15.4b.
So What Does This All Point To?
So what does the experience of the last two chapters point to? It points to different things for each of us. I conclude that it is worthwhile paying attention to the meaning in mathematics. Often in our haste to get to the modern, powerful analytic tools we ignore and trod upon the meanings and images that are there. Sometimes it is hard even to get a glimpse that some meaning is missing. One way to get this glimpse and find meaning is to listen to and follow questions of "What does it mean?" that come up in ourselves, in our friends, and in our students. We must listen creatively because we and others often do not know how to express precisely what is bothering us.
Another way to find meaning is to read the mathematics of old and keep asking, "Why did they do that?" or "Why didn't they do this?" Why did the early algebraists (up until at least 1600 and much later, I think) insist on geometric proofs? I have suggested some reasons above. Today, we normally pass over geometric proofs in favor of analytic ones based on the 150-year-old notion of Cauchy sequences and the Axiom of Completeness. However, for most students and, I think, most mathematicians, our intuitive understanding of the real numbers is based on the geometric real line. As an example, think about multiplication: What does a ´ b mean? Compare the geometric images of a ´ b with the multiplication of two infinite, nonrepeating, decimal fractions. What is 
?
There is another reason why a geometric solution may be more meaningful. Sometimes we want a geometric result instead of a numerical one. As an example, I shall describe an experience that I had while a friend and I were building a small house using wood. The roof of the house consisted of 12 isosceles triangles which together formed a 12-sided cone (or pyramid). It was necessary for us to determine the angle between two adjacent triangles in the roof so we could appropriately cut the log rafters. I immediately started to calculate the angle using (numerical) trigonometry and algebra. But then I ran into a problem. I had only a slide rule with three-place accuracy for finding square roots and values of trigonometric functions. At one point in the calculation I had to subtract two numbers that differed only in the third place (for example, 5.68 - 5.65); thus my result had little accuracy. As I started to figure out a different computational procedure that would avoid the subtraction, I suddenly realized — I didn't want a number, I wanted a physical angle. In fact, a numerical angle would be essentially useless — imagine taking two rough boards and putting them at a given numerical angle apart using only an ordinary protractor! What I needed was the physical angle, full-size. So I constructed the angle on the floor of the house using a rope as a compass. This geometric solution had the following advantages over a numerical solution:
Meaning is important in mathematics and geometry is an important source of that meaning.