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[Algebra] Omar Khayyam's Cubic

2021-08-29 10:12 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng (鄭濤)

【Problem】

Omar Khayyam (1048 – 1131 AD) solved cubic equations of the form x%5E3%20%2B%20ax%20%3D%20b%20 by transforming the problem into finding the intersection of a circle and a parabola:


%5Cbegin%7Bcases%7D%0Ax%5E2%20%2B%20y%5E2%20%3D%20qx%20%5C%5C%0Ax%5E2%20%3D%20py%0A%5Cend%7Bcases%7D


where a%2Cb%2Cp%2Cq are positive numbers.

Part 1: Determine p%2Cq in terms of a%2C%20b.
Part 2: Use Omar Khayyam’s method to solve the cubic equation x%5E3%20%2B%204x%20%3D%2016%20 by sketching the two conic sections and locating the intersection points.



【Solution】

Solution for Part 1
Write the first equation as y%5E2%20%3D%20x(q-x)%20 and the second equation as y%3D%5Cfrac%7Bx%5E2%7D%7Bp%7D. Substitute the second equation into the first equation to eliminate y, and obtain %5Cfrac%7Bx%5E4%7D%7Bp%5E2%7D%20%3D%20x(q-x).

Consequently,

x%5E3%20%3D%20p%5E2%20(q-x)

or

x%5E3%20%2B%20p%5E2%20x%20%3D%20p%5E2%20q


Matching terms with the cubic equation x%5E3%20%2B%20ax%20%3D%20b%20 gives


%5Cbegin%7Bcases%7D%0Ap%5E2%20%3D%20a%20%5C%5C%0Aq%20%3D%20%5Cfrac%7Bb%7D%7Ba%7D%0A%5Cend%7Bcases%7D


Since a%2Cb%2Cp%2Cq%20 are positive numbers,


%5Cbegin%7Bcases%7D%0Ap%20%3D%20%5Csqrt%7Ba%7D%20%5C%5C%0Aq%20%3D%20%5Cfrac%7Bb%7D%7Ba%7D%0A%5Cend%7Bcases%7D


Solution for Part 2
For the equation x%5E3%20%2B%204x%20%3D%2016, a%20%3D%204%20 and b%20%3D%2016. Thus,


%5Cbegin%7Bcases%7D%0Ap%20%3D%20%5Csqrt%7B4%7D%20%3D%202%20%5C%5C%0Aq%20%3D%20%5Cfrac%7B16%7D%7B4%7D%20%3D%204%0A%5Cend%7Bcases%7D

and we need to graph


%5Cbegin%7Bcases%7D%0Ax%5E2%20%2B%20y%5E2%20%3D%204x%20%5C%5C%0Ax%5E2%20%3D%202y%0A%5Cend%7Bcases%7D



Intersection points: (0,0);? (2,2). When the values of x is substituted into x%5E3%20%2B%204x%20%3D%2016, we find only one solution to this cubic x%20%3D%202.


[Algebra] Omar Khayyam's Cubic的評(píng)論 (共 條)

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