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Lots of people have asked how to use Khayyam's method but I am studying for my dissertation so really need to understand the why. What I really don't understand/ can't find useful proofs for is how he separated cubic equations into two conic sections. A work I have been using to gain a preliminary understanding is 'Omar Khayyam: Geometric Algebra and Cubic Equations' by Siadat and Tholen (https://doi.org/10.1080/10724117.2020.1770495) which gives a quick breakdown of equations of the form x^3 + bx = c into a semicircle and a parabola but it does not explain how Khayyam came to centre the semicircle at (r,0) or why the parabola is of the form y = x^2/sqrt(b), or even why he chose a semicircle and parabola in the first place, apart from that the maths simply works. I know that he built off of the works of menaechmus if that is relevant?

Any good references anyone knows or if anyone knows how I should go about trying to prove equations of differing forms would be really helpful, thanks!

Bountifull
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2 Answers2

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It is a very simple trick. I’m not sure where you (or Khayyam) are getting a semicircle from but here is the idea how to get two conics.

Assume that you want to solve the cubic $ x^3 + ax^2 + bx + c = 0 $. Set $ y $ to be $ x^2 $. Then your equation becomes $ xy + ay + bx + c = 0 $ which is the same as $ (x+a)(y+b) = ab - c $ which is a hyperbola. Remembering that we have $ y = x^2 $ which is a parabola, we see that the roots of the original cubic are just the $ x $-coordinates of the points of intersection of that hyperbola with the parabola.

Cranium Clamp
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  • Thanks for the solution, I believe the semicircle comes from the fact that Khayyam only considered the positive quadrant so the circle is split in half along the x-axis :) – Bountifull Nov 26 '22 at 17:23
  • No problem. My answer is certainly correct but in the spirit of your specific question, I think you should consider accepting Intelligenti pauca’s answer. – Cranium Clamp Nov 26 '22 at 17:26
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    They are both very helpful Khayyam had 14 forms of the cubic equation, so having two examples to work from should help me get to some of the rest! – Bountifull Nov 26 '22 at 17:32
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Here's how to deal with the particular case $x^3+bx=c$, which was solved by Khayyam from the intersection of a parabola and a circle.

First of all, we can multiply the above equation by $x$ (this will introduce the spurious solution $x=0$) obtaining: $$ \tag{1} x^4+bx^2=cx. $$

If we set now $$ \tag{2} x^4= b y^2, $$ and substitute that into equation $(1)$, we get: $$ \tag{3} by^2+bx^2={c}x, $$ which is the equation of a circle, centred at $\big(c/(2b),0\big)$ and passing through the origin.

On the other hand, equation $(2)$ is the square of $$ \tag{4} y={x^2\over\sqrt b}, $$ (we must keep in mind that $b$, $c$ and $x$ are all positive quantities for Khayyam), which is the equation of a parabola, with vertex at the origin. Hence $(1)$ is equivalent to the system of equations $(3)$ and $(4)$.

Parabola and circle meet at $(0,0)$, which is the spurious solution introduced when we multiplied the original equation by $x$, and at another point, whose abscissa is the desired solution of the original equation.

EDIT.

Case $x^3+c=bx$ can be solved by repeating the same steps as above: the solution is then found from the intersection of parabola $(4)$ with a rectangular hyperbola whose equation is $bx^2-by^2=cx$.

Case $x^3+ax^2=c$ is solved by Khayyam again intersecting a parabola and a hyperbola. Set $h=\root3\of c$ and multiply the equation by $h$, obtaining: $$ hx^2(x+a)=h^4. $$ Setting $$ h(x+a)=y^2, $$ which is the equation of a parabola, and substituting into the preceding equation one gets: $$ x^2y^2=h^4, $$ which is the squared equation of a rectangular hyperbola.

Intelligenti pauca
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