 So let's take a look at a few more equations reducible to quadratics. So let's try to solve x minus 2 squared of x minus 8 equals 0. So again, there's no algorithm, no formula, no set of steps, no divine light shining down from the heaven that will tell us what the best substitution is. We just have to try out different things and see what works. But there are a couple of things that we might try first. In this case, because we have a square root, we might try the substitution y equals square root of x. Now, since we only know how to solve quadratic equations, we also want to look at what y squared is, so y squared will be, and that's promising, because all the variable parts in our original equation are either square root of x or x, and so this substitution will eliminate all of those things. So let's make the substitution y equals square root of x, and that gives us a quadratic equation. Since the universe is a kind and gentle place, and I am a kind and gentle math teacher, every quadratic you'll ever encounter is factorable, so we'll try to factor. Yeah, I don't think so. Since the quadratic formula works on every quadratic equation, we'll use the quadratic formula to solve. And this gives us our two solutions, 4 or negative 2. And again, if it's not written down, it didn't happen. We should write down that we have solutions y equals 4, y equals negative 2. Well, again, the original equation was in terms of x, so y equals 4 is not a solution, but equals means replaceable. So our two equations here become... So now let's try to solve these equations. So the first one, we have square root of x equals 4, so we can undo the square root by squaring, and we get a solution x equals 16. We also have the second equation, square root of x equals negative 2. But remember, square root of n is never a negative number. So this equation, square root of x equals negative 2, we're saying a non-negative number is equal to a negative number, and this has no solution. And again, if it's not written down, it didn't happen. If we just leave our answer like this, it looks like we ignored this second solution and didn't do anything with it. So let's cross it out lightly so that we know we actually checked it and found that we couldn't get a solution from it. And maybe we'll emphasize that fact with a short note. How about x to the minus 2 plus 4x to the minus 1 minus 5 equals 0? This time, because our exponents are minus 2 and minus 1, we might try the substitution z equals x to the power minus 1. And since we're allowed to have a z squared in a quadratic equation, let's see what z squared is, which is x to the power minus 2. And again, it looks like this will work because our variable parts are either x to the minus 1 or x to the minus 2, and a substitution will allow us to get rid of both of them. So making the substitution z equals x to the power negative 1 gives us, and, well, maybe the universe is kind and gentle, and we can solve this by factoring. And we got lucky this does actually factor. And so now we have product equal to 0, so we know that one of the factors must be 0. So either z minus 1 is 0, or z plus 5 is 0. Solving these gives us our two solutions, which we write down. So again, our original equation was in terms of x equals means replaceable, so instead of z, I can write x to the power minus 1. And now we can try to solve both of our new equations. So x to the power minus 1 equals 1. Let's solve that. So remember a negative exponent corresponds to 1 over. So x to the power minus 1 is really 1 over x. Now, our equation is a quotient, 1 divided by x. So our first step is going to be to multiply. So we'll multiply both sides by x. If we simplify, we get our first solution, x equals 1. Similarly, x to the power negative 1 equals negative 5. We can rewrite this negative power. We can then multiply both sides by x and solve our equation. And we get our solution, x equals negative 1 fifth. Or how about x to the power negative 2 thirds minus x to the power negative 1 third minus 2 equals 0. Well, how about let's make our substitution u equals x to the power minus 1 third. Then u squared will be x to the power minus 2 thirds. And now our variable expressions are all taken care of, so we can make this substitution and our equation will become u squared minus u minus 2 equals 0. We'll solve this using the quadratic formula, giving us solutions of 2 or negative 1, and again, since our equation was in terms of x, we'll have to solve in terms of x. So x to the power negative 1 third. Again, a negative exponent is a 1 over the same base, same power, but not negative. So we can rewrite our equation as 1 over x to the power 1 third equals 2. Now we have a quotient, so let's simplify this by multiplying by the denominator x to the power 1 third. So we'll multiply on the left and on the right. Over on the right-hand side, we have 2 times x to the power 1 third. And remember that a fractional exponent corresponds to a root. So this x to the power 1 third is really the cube root of x. So again, over on the right-hand side, we have a product, so we'll divide both sides by 2. Over on the right-hand side, we have a root, so now we can raise both sides to the third power, and that gives us x equal to 1 eighth as one of our solutions. This other equation, x to the power minus 1 third equals negative 1, we can go through the same steps, giving us x equals minus 1 as a second solution.