Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Omar Khayyam is known for his significant progress in solving cubic polynomial equations. For example, his biography on www-history.mcs.st-andrews.ac.uk says

(...) This problem in turn led Khayyam to solve the cubic equation x^3 + 200x = 20x^2 + 2000 and he found a positive root of this cubic by considering the intersection of a rectangular hyperbola and a circle.

(...) Indeed Khayyam did produce such a work, the Treatise on Demonstration of Problems of Algebra which contained a complete classification of cubic equations with geometric solutions found by means of intersecting conic sections.

But I still can't see the big picture of those days. I'm possibly omitting something about the idea of geometric solutions of algebraic equations, but why were they trying hard to find intersections of conic sections, and building large classification schemes for it?. If the idea was to get a numerical value out of these constructions by measuring lengths on paper, they could just have prepared a careful template for the function $y = x^3$, and then solved all the cubic equations by intersecting it with a parabola, like in the figure below for the mentioned equation.

I would appreciate answers that would clarify my confusion. Was it that they did not conceive $y=x^3$ as a curve, if they were interested in getting a numerical value? Or was it a conceptual challenge to show that all cubic equations can be represented as an intersection of two conic sections?


share|improve this question
Before Descartes, the concept of writing down an arbitrary equation and looking at its zero set did not exist. –  Qiaochu Yuan Nov 25 '10 at 18:54
See "Episodes in the Mathematics of Medieval Islam" by Lennart Berggren for a discussion of Omar Khayyam. I also recommend "The Age of Genius: 1300 to 1800" by Michael Bradley (in the series "Pioneers of Mathematics") for a very good description of Jamshid Kashi (1380-1429) who did things such as: "Using polygons with more than 800,000,000 sides and an efficient algorithm for estimating square roots, he accurately determined the value of pi to 16 decimal places," which was unsurpassed until 1596 when Ludolph van Ceulen calculated pi to 20 decimal places using polygons with 60 x 2^33 sides. –  Marko Amnell Nov 25 '10 at 21:06
By "rectangular hyperbola" I believe it's supposed to be something like $xy=c$ (or a translation of it); eliminating variables from that and the equation of a circle (in our current notation, that is) can lead to a quartic, and maybe a cubic for special enough configurations. –  J. M. Nov 25 '10 at 23:02
As for the numerics, I'm pretty sure the "method of exhaustion" was the only tool they had at the time. –  J. M. Nov 25 '10 at 23:03
@Qiaochu Yuan: I think it did, Al Kharazmi did talk about solutions of algebraic equations (and in fact the name of the field algebra is derived from his book "al-Kitab al-mukhtasar fi hisab al-jabr wa'l-muqabala") but Muslim scientists are not usually well-credited in later European works. –  Kaveh Nov 26 '10 at 7:02

2 Answers 2

up vote 8 down vote accepted

There's a brief note in this book on how Khayyam bumped into having to solve a cubic.

I'll only make the note that you should remember the context of the time: there was no concept of negative, much less complex, solutions. Corresponding to our current Cartesian system, Khayyam only looked at intersections in the first quadrant.

Another note should be made that the curves of the time were constructed with geometric tools (straightedge, compass, and a bunch of other contraptions), and $y=x^3$ isn't really a sort of curve that easily lends itself to such a construction (but is now easily constructed thanks to our current knowledge of coordinate geometry).

Here is a more explicit mention of the hyperbola-circle intersection problem Khayyam studied and was mentioned in the OP.

Here is a (more or less) complete table of all the intersection cases Khayyam studied. (The book has an appendix containing a (translated) section of Khayyam's work.)

Here is yet another reference.

(I'll keep updating this answer as I comb through more books; watch this space! As an aside, it's funny that my attempts to look for answers to this question are leading me to references for this question!)

share|improve this answer
Thanks for your time! Looking at the sketches I saw before, I falsely thought that they knew about the coordinate plane. It is interesting that they actually constructed conics with geometric tools. –  AgCl Nov 26 '10 at 1:30
@AgCl: Just so you get an appreciation on how the Greeks and Arabs managed all this geometry, I encourage you to take a look at Lockwood's [ A Book of Curves ](books.google.com/books?id=pDCzyUCj90IC), most especially the sections on the conics. –  J. M. Nov 26 '10 at 1:39

I don't think they had the idea of $y=x^3$ as a curve. For one thing, there was no analytic geometry until Descartes. With coordinates, some problems became much simpler.

share|improve this answer
Thanks! So it seems they didn't consider $ax^2 + b x + c$ as a parabola either. It is becoming more clear now. –  AgCl Nov 26 '10 at 1:27
@AgCl: Well, the Greeks and Arabs did consider the intersection of two parabolas in their attempt to duplicate the cube (which is of course one of the problems of antiquity that naturally leads to the solution of a cubic). –  J. M. Nov 26 '10 at 1:41
@J.M. I mean I guess they didn't associate the algebraic form $a x^2 + b x + c$ with the corresponding parabola, if they didn't know about the Cartesian plane. –  AgCl Nov 26 '10 at 2:20
@AgCl: They didn't even have the concept of an algebraic form! All they were working on essentially were lengths. So the solutions to them were essentially the lengths from some convenient starting point (either the origin or a point on the horizontal or vertical axes in our current notation) to the point of intersection of the two curves being considered. –  J. M. Nov 26 '10 at 2:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.