 So let's take a look at some of al-Qayami's solutions to cubic equations. So we might take a look at a cube and sides equal to number. In modern terms, we might read this as x cubed plus ax equals b. However, this is geometric nonsense. This is saying we have a cube, a solid object, plus a bunch of segments equal a number. And so we need to interpret these as the same type of geometric objects. So we have a cube, so that's a three-dimensional object. And so we can interpret ax as a three-dimensional figure with unknown length x. Now, one phase could be square or rectangular, and we'll let it be a square base. And so this is going to be p squared x. Now, b, our constant, also has to be a three-dimensional figure. And since we've already established a square phase p squared, let b have the same square phase and some length q. And so the problem is to make these two figures equal to the third figure. So let's consider our equation. And one way we might approach this is that if we move this p squared to the other side, then all of our terms on the right will have a factor of p squared, which is convenient. And I can break apart the factors and rewrite this as a proportionality. And what's interesting is this ratio x to q minus x shows up when we have a circle of diameter q. And so if I have a circle with diameter q and I cut off the diameter with a perpendicular segment, then x is to y as y is to the remaining part of the diameter. And our theory of ratio and proportion says that the first term is to the last, x is to q minus x, as y squared is to q minus x squared. So since p squared is to x squared is x is to q minus x, and x is to q minus x as y squared is to q minus x squared. And so that tells us p squared is to x squared as y squared is to q minus x squared. or if we reduce each of these, p is to x as y is to q minus x, but since y is to q minus x as x is to y, that says that p is to x as x is to y, or p y equals x squared. And also from x is to y as y is to q minus x, that corresponds to our circle x q minus x equals y squared. So the intersection of the circle x q minus x equals y squared with the parabola p y equals x squared gives us our solution. And so we can create our circle and we know the diameter is q, we can create our parabola and we know the parameter is p, and their intersection point will give us the value of x, which solves our equation. So Al-Qayami also proved that the intersection solves the equation. So in modern terms, the equation of the parabola is p y equals x squared, while the circle has equation x times q minus x equals y squared. So the parabola can be untangled into the rational equation p over x equals x over y, and similarly we can disentangle the equation for the circle into x over y is y over q minus x. And that gives us one of these continued proportionalities that are so very useful. So from p over x is equal to x over y, well since p over x and y over q minus x are equal, we can multiply one side by the one, and the other by the other, multiply across, and simplify. And so the x coordinate of the intersection point of the circle and the parabola also solves the equation we started with. But wait, there's more. Notice that regardless of the value of p and q, the two curves will always meet. If q changes, that just changes the diameter of the circle. If p changes, that just changes the width of the parabola, and no matter what p and q are, the two curves have to meet at some point. And Kayame recognized that this was true, and so he noted that this type of problem always has a solution. It's worth pointing out that this isn't the only way to solve the problem. Although Kayame focused on a single approach to finding the solution, there are many possibilities. We'll switch to modern notation for a little more insight into the process. Generally speaking, we want to define two curves, conics or circles, whose intersection point x, y gives us the solution to the equation. So let's take a look at our equation again, x cubed plus p squared x equals p squared q, and this time we might try the following. We can factor the left hand side, and this expression x squared plus p squared, we might note that x squared plus p squared is the expression for a vertically shifted parabola. So we might let q y equals x squared plus p squared, substituting into our equation, and we get x y equals p squared, and that says we can take the parabola q y equals x squared plus p squared, and the hyperbola p squared equals x y, their intersection point x y satisfies the equation, and as before, regardless of the value of p and q, these two curves will intersect. And Al-Kayame considered and solved every possible type of cubic equation except the impossible one, which has a whole bunch of solid figures together equal to nothing. Now while Kayame did not include this in his treatise, it turns out there's also examples of fourth-degree equations solved this way. For example, you might have seen that Abbasol in the 10th century constructed a heptagon by finding the intersection of the conics we'd describe as x squared equals 1 plus y, a parabola, and y squared equals x times x plus 1, a hyperbola. Now Abbasol didn't point this out, but if we solve the first equation for y, then substitute into the second, we get a quartic equation. And in general, quartic equations, equations involving a fourth power, can be solved by the use of intersecting conic sections.