Solving a quartic equation using a cool idea! #math #mathcontests #polynomials #olympiadmath

How do we solve this really seemingly complicated fourthderee polomial equation? Well, there's actually a really beautiful way to solve this and I'm going to show you most of the steps to get there. We're going to start by recognizing something very interesting about the left hand side. If we let f ofx be x^2 - 3x + 1, we notice that we're actually squaring that quantity, subtracting 3 * it, and adding one. So, the entire left hand side is actually f composed with itself. So we're solving the equation f of f ofx= x. Now this is going to let us do something really elegant. I'm going to let a be the argument inside of here which is f ofx itself. Then you notice since a is the argument of this second composition, we have f of a is actually x. So we can put these two observations in a table and actually plug in values to see what happens. Since a is f ofx, we'll get a is this quantity right over here x^2 - 3x + 1. But similarly x is f of a. So x is a^2 - 3 a + 1. Now subtracting one of these from the other will give us something really interesting. We'll get a - x is x^2 - a 2 - 3x + 3 a. And you notice a common factor of x - a everywhere. So if we move this left hand side to the right, we'll actually get the quantity x - a * x + a - 2. So x and a are actually related directly. Either x is a or x is 2 minus a in order for this quadratic to work out. So we can actually figure out what x is directly. Since x is a, we can replace x here and solve a quadratic in x to figure out what x is. And similarly, since x can be 2 - a, we can do the same thing with that value. If you plug in values, you end up with two different quadratics, each with two different solutions for four solutions in total. But the beautiful part of this solution is the functional composition that we had allowed us to use symmetry to get our solution.