Notably, Khayyam's construction generates a single line segment ¯LB which solves the cubic equation of the form x3+C1x2+C2x=C3, where C1,C2,C3>0. In Figure 6 and Figure 7, for an example, we have fixed the coefficients to be x3+7.9x2+1.32x=2.13, and in Figure 10, below, our example is x3+4.8x2+1.42x=1.73. In Figure 10, we have reflected Khayyam's construction about the line ¯HB, the vertical asymptote of the hyperbola. Then, superimposing a Cartesian graph of a cubic shows clearly that BL corresponds exactly to the positive root of the cubic.

Figure 10. A reflection of Khayyam's construction for x3+4.8x2+1.42x=1.73, superimposed with a Cartesian graph of the equivalent cubic function. We take line segment ¯HB, an asymptote for the hyperbola, as a vertical axis and line segment ¯BL to have length x, which is exactly the value of a root to the cubic polynomial.
The slider bars in the GeoGebra applet in Figure 11 (below) enable students to explore how Khayyam's construction works for cubics with a range of coefficients. This dynamic exploration facilitates an appreciation of the general nature of the proof – Khayyam's method is not simply constructing one root to one specific polynomial. His construction technique constructs one root for any polynomial of the form f(x)=x3+C1x2+C2x−C3, where C1,C2,C3>0. The applet also presents in an accessible visual form the relationship between Khayyam's line segment ¯LB and a root of the cubic graph familiar to many students.
Figure 11. A reflection of Khayyam's construction for x3+ax2+b2x=c3, with various positive values of a,b, and c, superimposed with a Cartesian graph of the equivalent cubic function. We take line segment ¯HB, an asymptote for the hyperbola, as a vertical axis and line segment ¯BL to have length x, which is exactly the value of a root to the cubic polynomial.
In Figures 10 and 11, superimposing a Cartesian graph of a cubic shows clearly that BL corresponds exactly to the positive root of the cubic. But what of the other roots?