 Hi, I'm Zor. Welcome to your new Zor education. I'd like to solve a couple of non-trivial equations. We all know how to solve some really simple things like linear or quadratic equations. That's really not very interesting, but boring quite frankly. You have the formula. If you remember it, it's good. If you don't remember it, it's easy to derive it and just use the formula. So as long as you have an equation with all its coefficients, just use this particular formula and you'll get the solution. So what if you don't have the formula? Excuse me. In the real life, you don't really have the formulas all the time. So you have to invent something interesting, something new, something which will help you to solve a particular problem. That's what actually the whole course of mathematics which I'm offering here is about, to develop these qualities of creativity, logic, analytics, which will help you to solve something which you don't know how to solve before, and nobody taught you how to do this. So these are examples of this type of problems where you have to really think about it and use some non-trivial non-standard methodology to come up with a solution. All right. Let's go. Number one, x square minus 4x square plus 7x minus 2 square equals 16. Okay. Obviously, this is an equation of the force degree, x square squared, right? So it will be x to the fourth. Obviously, you don't know nobody knows the formula. So we have to invent something, something peculiar for this particular case, and it does have certain peculiarity. Here is what I can suggest. You see, here is a very interesting thing, x square minus 4x. What does it remind? Well, it reminds me x minus 2 square, right? Because x minus 2 square will be x square minus 4x plus 4. So x minus 4x is part of something which I kind, well, it's my guess. I mean, maybe it's good, maybe it's not. But it looks like x minus 2. But this is x minus 2 square. You see, this is exactly the peculiarity which you have noticed, or I have noticed. So let's just try to use it. Now, how can we use it? Well, there are many ways. For instance, if I will just open this up, I will have the same x square minus 4x square plus 7x square minus 4x plus 4 equals 16. And now I see that these are common things. So I can just substitute y is equal to x square minus 4x as a new variable. And now I have a quadratic, basically, equation of y. Now, this is y square. This is 7y plus 28. And there is a 16 here, so it's plus 12. And solutions to this are obviously minus 3 and minus 4. Because their product is the free member. And their sum is this coefficient with a minus sign. But you can actually do it properly. It's minus 7 plus minus 49 minus 48. So it's minus 7 plus minus 1 divided by 2 plus is minus 6. So it's minus 3 and minus minus 4. Yes, that was right. So minus 3 and minus 4. Now, knowing that, I can use this expression and basically find x. Now, that's easy. So x square minus 4x. This is minus 3, so it will be plus 3 equals to 0. That's one equation. And another is x square minus 4x plus 4 is equal to 0. This is from minus 3. And this is from minus 4. Now, here I will have two roots which are 3 and 1. 1 minus 4, 3 and 9, minus 12, minus 3 plus 3. Correct. And here, this is actually x minus 2 squared. So x is equal to 2, and this is the double root. So these are my solutions. Now, this is actually a double root, as I was saying, because it's x minus 2 squared. So I have four solutions. Two of them are the same, which corresponds to the degree of this particular equation. It's the fourth degree, so it should have four solutions. And here they are. All right. Now, the checking. Well, let's just do a quick checking, please, for something, back to 3, for instance. 3 squared 9 minus 12 minus 3 squared 9. So this is 9. This is 3 minus 2, 1, so it's 7, 9 plus 7, 6, correct. And you can check the others. I'm sure they're relative. OK. So what was an interesting part in this particular equation? It's just to notice that these two have something in common. That x minus 2 squared would look like this one plus some free coefficient, all right? So this is just a guess, if you wish. But the problem is that if you solve hundreds of problems like this, you will probably see these peculiarities, well, more or less easily. OK, let's go on. x plus 2 over x, x plus 16 over x plus 7 equals 4. Well, it might look a little scary, and it's obviously not linear, not quadratic, et cetera, et cetera. But there are two very important things to say about this. First of all, if we would just use the common denominator, which is x times x plus 7. Now this would be x plus 2 times x plus 7, which is a polynomial of the second degree. This would be x times x plus 16, also second degree. And this would be x times x plus 7, x times x plus 7 times 4, which is also a polynomial of second degree. So if you will just use this common denominator and put everything above it, we will have a quadratic equation on the top. And we can basically get rid of the denominator in this case. Now there is one very important thing, and it's very important not only to this particular equation, but for many others. You see, all these non-trivial equations might have certain restrictions on the various of the solutions which we are looking for. Now think about this. If I would really like just do whatever I just said, which is just use the common denominator. So this would be from this plus this plus this equals and divided by common denominator equals to 4 times x plus 7 times x divided by x x plus 7. Now my intention is just to get rid of this common denominator because if these fractions are equal to each other and denominators are the same, then the numerators must be the same. And equalizing these numerators, they just have a quadratic equation because this is x squared, this is x squared, and this is x squared. Now, but this is a very important thing. I said just get rid of the denominator and equalize the numerators. Yes, it can be done. However, existence of denominator right now before I drop them actually restricts certain values of the solution. x cannot be equal to 0 and x cannot be equal to minus 7 because in both cases, this common denominator would be 0 and the whole thing would have no sense at all. Same thing if you look at the original equation. It cannot be 0 here and it cannot have 0 there. So x not equals to 0 and x not equals to minus 7 are initial restrictions on the solutions. So you have to really realize whenever you see something like this which might really have certain restriction on the values of the solutions, you have to upfront do this and then look for solutions because during your looking for solutions, whenever you do this, you immediately lose this particular restriction. There is no restriction like this in this case, but in the original equation it will. So if by any chance I will get a solution to this quadratic equation, x equals to 0 or x equals to minus 7, I should really immediately discard it as the one which is outside of the domain of allowed solutions. I don't know whether I will or will not have these solutions among solutions of this quadratic equation, but if I will, I should discard them. So let's just solve this particular thing. That should be easy. So this is x squared 2x and 7x and 9x plus 14 plus x squared plus 16x equals 4x squared plus 28x. This is my equation. So x squared, x squared, and 4x squared. So subtracting 2x squared from both sides on the right, I will have 2x squared. 9x, 16x, it's 25. This is 28. So I subtract 25, and I will have 28 minus 25, 3x. And finally, the free member is 14, but it would be with a minus sign, right? Is this right? OK, looks like it. And solutions are divided by 2 minus 3 plus minus where root of 9 plus 8 times 14 is agency 12, which is, so x is equal to minus 3 plus minus 128. That's 11. So it's 11 over 2. So I will have x equals to 8. Oh, I'm sorry. It should be 4, not 2. It's a double first coefficient. It was 2x squared something, so it should be 4 here. So it should be 4 here, right? So it's 8 divided by 4. It's 2. And the next one is minus 14 over 4. So it's minus 7 second. OK, so these are solutions. Now, none of them is this. So it's OK. No problem. They're not within the restricted set of solutions. So let's just check one of them. But 2 is easier, right? So 2 plus 2 is 4 divided by 2 is 2. This is 18 divided by 9. It's also 2 plus 2 4, correct? And I'm sure minus 7 seconds also works fine. So what was interesting about this particular example? Well, the interesting thing is that although you can really very easily bring the equation to the square, to the quadratic equation, you should not forget that there is a denominator, which you really have to be very careful about. And you have to really make some kind of exceptions. Your set of solutions is restricted to those which do not really put the denominator to 0. That's very important. OK, the last one, x plus 1 to the minus second degree, x plus 2 to minus second equals 1.25. OK, well, let me rewrite it in a little bit better format. And one more thing, which I probably can really make my life a little bit easier. You see, x plus 1 and x plus 2, I think it would be easier if I will use substitution. This would be 1 over x plus 1, and this would be square. And it would be 1 over x plus 2 square. Now, it would be easier if instead of x plus 1 and x plus 2, I would make a substitution y is equal to x plus 1. And then x plus 2 would be y plus 1, right? Now, why is it easier? Because it's easier to deal with the denominator y than denominator x plus 1. So instead of this, I will use this substitution. And I will put here y and y plus 1, where y is equal to x plus 1, OK? And what's equal to 1 and 1.25, let me deal with rational numbers, fractions like this. That would be easier. And obviously, I will try to do something similar to what I did before. I will get rid of the denominator, and I will put everything on the top to have some kind of polynomial equation. Polynomial is easier than dealing with this thing. Obviously, I have to immediately restrict my set of allowed solutions to y not equals to 0 and y not equals to minus 1 to prevent these two denominators to be equal to 0. Now, after I have restricted myself this way, now let me just try to basically bring everything to the polynomial format. So I will bring everything to a common denominator, which is y squared times y plus 1 squared, right? I can always rewrite it using this, right? And same thing here. It looks a little bit easier, right? So the common denominator is y squared times y1, y plus 1 squared. Now, what's good and what's bad about it? What's good is that on the left, I will have quadratic, y plus 1 squared here and y squared there. But on the right, unfortunately, I have the polynomial of the fourth degree, right? So it will be y squared times y plus 1 squared. Well, whatever it is, let's just deal with this. So on the left, I will have y plus 1 squared, and I will open it up, which is y squared plus 2y plus 1. And this will be y squared by fourth of y squared. So y squared times this, which is y squared times y to the fourth plus 2y plus squared will be to the cube plus 1 times y squared. It will be y squared. That's on the right, OK? Now, obviously, I have to put 4 here. So I will have only integer numbers. And what would I have? I have y squared plus y is 2y squared times 4. It's 8y squared plus 8y plus 4 on the left. On the right, I have 5y to the fourth plus 10y to the third, 5 times 2 plus 5y squared. I think that's right. All right, so let's move everything on one side of the equation. And that will be 5y to the fourth plus 10y cube. Now, 5y squared, 8 on this side. So it would be minus 3y squared minus 8y minus 4 equals to 0. That's my equation, the equation of the fourth degree. Well, you don't like it, neither. But let's just think about it this way. If I am trying to solve this particular equation that I'm giving to my students, and it has integer coefficients, then it does make sense to look for some solutions among integer numbers, which are divisors of the free member. If you remember when we were talking about equations of the higher order, it's always like this. If you have integer coefficients and you're looking for integer solutions, they must be among divisors of the free element. Now, divisors are 1 and 2, basically. Well, forget about 1 probably. Let's just check 1. 5 plus 10 is 15 minus 312. Now, 1 actually does solve the problem. So 1 is a solution. Great. So if 1 is a solution, it means we immediately can divide it by 1. We have to extract the factor x minus 1 from this particular polynomial. So how can we do it? Well, very simply. It's 5y to the force minus 5y to the cube. If you will multiply, if you will factor out 5y cube, you will have y minus 1 left, right? Now, but we need 10. So minus 5, it means we have to add 15y cube, right? Now, to again factor out y minus 1, we have to subtract 15y square. So this would be multiplication of factor of y minus 1 by something. This would be y minus 1 by something else. Now, I need minus 3. I have minus 15. So I have to add minus 12y square. Minus plus 15 minus 15, no, plus 12, I'm sorry, plus 12. Minus 15 plus 12, that gives me minus 3. Now, if I have plus 12, I have to have minus 12y to have y minus 1 from here. But I have minus 8, so I have to add 4y and minus 4. As you see, here is I have rewritten my equation by adding and subtracting certain things to every member so that in each pair of parentheses, I have y minus 1 as a factor, all right? Now, why do they do it? Because again, my divisor of 4, which is 1, fits as a solution. I basically guessed the solution. But since I have guessed the solution, I can factor out xy minus 1. If 1 is a solution, I can factor out from the polynomial 1 minus 1. And how to do it? By basically breaking each component into something which I can factor out y minus 1. So in this case, it would be y minus 1 times 5y to the third. In this case, it would be plus 15y square. In this case, it will be 12y. And in this case, it will be 4. So that's my equation. This is the same as this one. Since I have guessed my first solution, y minus 1, I can separately have y is equal to 1 as a solution. And by the way, it's a loud solution because it's not 0 and it's not minus 1. After which, I can solve this equation because when the product of two polynomials can be equal to 0, then either 1 is equal to 0 and that gives me this solution, or this is equal to 0. Now I have to find some other solution. Well, let's try to guess again. 1 we have already exhausted. 1 is definitely no more than a root. So it's not like a double root. How about minus 1? Well, let's check it out. This will be minus 5 plus 15. That's 10. This would be minus 12, which is minus 2, plus 4, 2. So minus 1 doesn't really fit. Now let's check 2. Well, 2 obviously will not fit because everything is positive. But how about minus 2? Minus 2 cubed with minus 8 times 5 is minus 40. So this is minus 40. Let's write it down. Now this is, if it's minus 2 square, it's 4 times 15 is 60. This is minus 2. It's minus 24. And this is plus 4. Minus 64 minus 2. OK, so minus 2 does actually fit. Great. If y equals to minus 2 fits, it means I can represent it as y plus 2 times something. So let's check what y plus 2 might be. Well, if this is supposed to be broken down, so I will have y plus 2. It should be 5, y cubed plus 10, y square. Then it would be 5, y square would be factored out. And what will be remaining is y plus 2. OK, now I have 15 y square. I have 10 here, so I have to add 5. Now, so this is factorable by y plus 1. Now, here I need 10 y. So if I will factor out 5 y from both, I will have remaining y plus 2 as well. Great. Now, I have 10, I need 12 plus 2, y plus 4. And this can be factored out 2. And I will also have y plus 1, y plus 2, sorry. So right now, I can rewrite this as a plus 5, sorry, a plus 2. In this case, it's 5 y square, right? 5 y square. In this case, it's 5 y. And in this case, it's 2. So this is my next representation of the same equation of the third now, power, as a product of 2, the first power and the second power. OK, now from this, I see that my new solution is also good because it's not part of the prohibiting values for the denominator. And what's left, so if the product is equal to 0, either this is equal to 0 and that gives me the solution minus 2 or this is equal to 0. So this is my, this time, a quadratic equation, which I already know how to solve. This is just the formula, which we all remember. However, it does not really have any real solution because 5 square minus 4 times 5 times 2 is less than 0. So we don't have any solutions in this case. So this is the end of it. I've got two solutions. So it's a quadratic equation. I mean, it's a fourth degree equation, which can be represented as quadratic equation and two multipliers. So one would be y minus 1, another is y plus 2, and another is 5y square plus 5y plus 2. This has no solutions. This gives me 1, and this gives me minus 2. Both are allowed. Well, let's just check the solution for x. Now, we know that x is related to y with this formula. So y is x plus 1, so x is equal to y minus 1, which is either 0 or minus 3. OK, 0. Well, this is 1, and this is 1 quarter, and that gives me 1.25. So 0 is really the solution. We just checked it out. That's fine. And minus 3. So this would be 1 over minus 2 square, which is 1 quarter. And this would be, if it's minus 3, then it's minus 1 square 1, so it's again 1 and 1 quarter. So it's 1.25. So yes, checking if it's fine. So this is a solution, both solutions. And that's the end of this particular equation. Equation seems to be a little bit complicated, and again, it's transferred into x to the fourth degree polynomial, which is pretty difficult thing to solve. However, with certain degree of ingenuity and creativity, we have really decided to look for integer solutions and we found them. Obviously, if again, let me just repeat it, if I'm giving this particular problem to students to solve, then I kind of specifically chose not just the general equation of the fourth degree, but the equation which can be solved in some reasonable way. So to look for integer solutions is a reasonable way which I have decided. OK, so that's it for this particular lecture. The purpose was to introduce you to some non-trivial examples of equations to solve. And I'm sure as the time allows, I will probably introduce more and more problems of this type. The number of available problems of this particular kind, like non-trivial equations to solve, is really unlimited. And the more time you spend to basically solve these non-trivial problems, non-trivial equations, well, the better your creativity and your logic actually will be developed. So I do recommend you to solve as many as possible non-trivial examples when there is no ready to use recipe, when you have to really invent something, you have to guess something. And that's what's very important. OK, thank you very much. As usually, you can find all the information on my website and notes for this lecture. Thanks again. Until next time.