 That koi, who got Cardano interested in the solution of the cubic, played another rule and posed the problem that led to the quartic and its solution. Divide 10 into three proportional parts, the product of the first and second is 6. So let's consider this problem, the requirement that the product of the first and second is 6. So we might make one x and the other 6 over x. It doesn't matter what order we take them, but Cardano made x the second term. And since the quantities are in proportion, we note that 6 over x times something gives us x, and x times the same thing gives us the last term. And so if we think about it, the something has to be x squared over 6, and so that tells us the last term will be x cubed over 6. And so we might begin by saying let the parts be 6 over x, x, and x cubed over 6, and we require that these three add to 10. And if we multiply through by 6x, we'll be able to eliminate the fractions and get the equation. Now notice the left hand side looks like it could be a perfect square. And so we can try to complete the square on the left, but notice that if we add a constant, the right hand side will be 60x plus a number, which isn't obviously a square. So let's switch things around a bit. Instead of adding a constant term to make a perfect square, let's add an x squared term. And so the question we have to ask ourselves is if the square and constant of a perfect square are x to the fourth and 36, then what's the remaining term? The remaining term must be 12x squared. And so that means we'll want a 12x squared over on the left hand side, and we can get that by adding 6x squared. And so as promised, the left hand side is now a perfect square, and the right hand side, well, it isn't. But let's see what we can do about that. So the idea is that we want to add y inside the parentheses over on the left, which will keep the left hand side a perfect square, so that we'll also have a perfect square on the right. And to proceed, it's helpful to note that a plus b squared is a squared plus 2ab plus b squared. So if we add a term inside the parentheses, we'll get twice the product of the added term and the other terms inside the parentheses and the square of the added term. So if my left hand side is x squared plus 6 plus y, my right hand side will be 6x squared plus 60x plus 2 times y times x squared plus 6 plus y squared. And we can do a little bit of rearranging. So now remember our goal. We want the right hand side to be a perfect square. So let's think about this. Suppose we have px squared plus qx plus r, and suppose this is a perfect square. If we compare this to ax plus b squared, which expands to a squared x squared plus 2abx plus b squared, so p is a squared, q is 2ab, and r is b squared. And we'd like to find a relationship between them. And so here's one we could use, 2ab over 2 squared is a squared b squared. But 2ab is q, so this is really q over 2 squared. And since p is equal to a squared and r is equal to b squared, then a squared b squared is pr. And so we could read this as saying the square of half the coefficient of x must be the product of the other coefficients. So if we want our right hand side to be a perfect square, the square of half the coefficient of x, that's 60, has to be the product of the other two coefficients, 6 plus 2y times 12y plus y squared. And we can expand, and that gives us a cubic equation. Now this is all a very long and involved process, so let's go ahead and summarize all of our steps. McCoy's problem requires us to solve the cubic 450 equals y cubed plus 15y squared plus 36y. The reason we're trying to solve this cubic equation is that this y value will make the right hand side of this equation into the square of ax plus b for some values a and b. And since both sides will be perfect squares, then taking the square roots of both sides gives us x squared plus 6 plus y equals ax plus b. And since we know y, remember solving for y was our first step, and in fact this was the motivation for Cardano publishing Tartaglia's solution to the cubic. This is a quadratic equation which we can solve using our standard method. We leave the rather long and tedious details to the viewer.