On this page
In JEE maths you learn to solve quadratics with a formula and to treat parabolas and circles as a separate chapter. Omar Khayyam, writing nearly a thousand years ago, used one to do the other: he solved cubic equations by finding where two curves meet. This post covers who he was, how his method works on a real example, and what it can teach you about connecting algebra and coordinate geometry.
Who was Omar Khayyam?
The MacTutor biography gives his birth as 18 May 1048 in Nishapur, Persia. Sources differ on his death: MacTutor gives 4 December 1131, while Behnaz Hashemipour's entry in the Biographical Encyclopedia of Astronomers gives about 1123. When reliable references disagree, the honest move is to say so.
Around 1070 he moved to Samarkand, where a jurist called Abu Tahir supported him while he wrote his main algebra book, the Treatise on Demonstration of Problems of Algebra. He was then called to Isfahan by Sultan Malik Shah and the vizier Nizam al-Mulk to set up an observatory, and MacTutor says he led the scientists there for 18 years until 1092.
What did Khayyam do for the calendar?
Hashemipour writes that by 1079 a new solar calendar was ready, named the Jalali or Maliki calendar after its patron. MacTutor records that Khayyam was one of the eight scholars on the reform, and that he gave the length of the year as 365.24219858156 days. Today's value is about 365.242190 days. Quoting a result to eleven decimal places was bold, and the first five decimal places match.
How did Khayyam solve a cubic equation?
Before Khayyam, mathematicians such as al-Mahani had turned geometric problems into cubic equations but had no general method for them. MacTutor describes his treatise as "a complete classification of cubic equations with geometric solutions found by means of intersecting conic sections." He wrote equations in words rather than symbols, calling the unknown "the thing", and arranged each one so that every term was positive, as in the equation x³ + 200x = 20x² + 2000 that MacTutor quotes from his work. In an earlier book he had promised to give "all these fourteen forms with all their branches and cases".
The idea: if you cannot solve an equation directly, find two curves you can draw whose meeting point has the right x-coordinate. Turning geometry into equations at all built on the algebra begun by al-Khwarizmi, covered in our post on al-Khwarizmi and the birth of algebra.
Worked example: solve x³ + 2x = 12 with a parabola and a circle
This is Khayyam's type "cube plus things equals a number", x³ + cx = d, with c = 2 and d = 12. Set a = √c and b = d/c. Here a = √2 and b = 6.
Draw the parabola x² = a·y, that is, x² = √2·y.
Draw the circle with diameter b along the x-axis, from (0, 0) to (6, 0). Its centre is (3, 0) and its radius is 3, so its equation is (x − 3)² + y² = 9, which expands to x² − 6x + y² = 0.
Find where they meet. From the parabola, y = x²/√2, so y² = x⁴/2. Put this into the circle: x² − 6x + x⁴/2 = 0. Divide by x (we want x ≠ 0): x − 6 + x³/2 = 0. Multiply by 2: x³ + 2x = 12. The meeting point's x-coordinate is exactly the root we want.
Read off the answer. Try x = 2: 2³ + 2×2 = 8 + 4 = 12. So x = 2, and the curves meet at (2, 4/√2) = (2, 2.83). Check on the circle: (2 − 3)² + 2.83² = 1 + 8 = 9. It works.
In general, the same steps turn x² − bx + y² = 0 and y = x²/a into x³ + a²x = a²b, which is x³ + cx = d. That one line of algebra is the proof that the construction always works.
What mistakes do students make with this method?
Forgetting the origin. The parabola and circle always meet at (0, 0). The value x = 0 satisfies x² − 6x + x⁴/2 = 0, but it is not a root of the cubic, because it appeared only before we divided by x. Discard it.
Mixing up a and b. The parabola uses a = √c; the circle's diameter is b = d/c. Swapping them gives a different cubic. Re-derive the substitution each time instead of memorising.
Trusting the sketch for exact values. A drawing shows roughly where the root is. Khayyam himself then used tables to get a numerical value. Always confirm by substituting back.
Did Khayyam think cubics could be solved by formula?
MacTutor reports that he hoped "someone else who comes after us" might find arithmetic solutions, meaning a formula. That came in the 16th century from del Ferro, Tartaglia and Ferrari. MacTutor also notes that he realised a cubic can have more than one solution, though he did not find a case with three.
What can you learn from Khayyam's way of working?
Khayyam's method links two chapters you probably study separately: equations and conic sections. When a problem looks impossible in one language, translate it into another. Our post on how to approach conic sections gives practice in the coordinate side of this.
A practice routine for this week:
Solve x³ + 3x = 14 with Khayyam's construction (c = 3, d = 14). Check your answer by substitution.
Factor the cubic and use the discriminant to explain how many real roots it has.
Sketch both curves on graph paper and see if the picture matches your algebra.
If you want to see how a parabola and circle move as you change c and d, the Fermi Simulation Lab has interactive simulations for ideas easier to see than read.
Start with step 1 tonight: write down a and b for x³ + 3x = 14 before you draw anything.
