 Hi, I'm Zor. Welcome to Unisor Education. Today we continue solving problems. In this particular case, we will go into Algebra. So this is Algebra number 5. This is part of the course called Mass Plus End Problems presented on Unisor.com. So the whole course is dedicated to solving problems. There is a prerequisite course, which is Mass for Teens. On the same website. By the way, all courses are totally free, no advertisement, no strings attached. So obviously, all these problems, which I present in Mass Plus End Problems course, are related to theoretical material presented in this previous course, Mass for Teens. Now, I have divided problems into different categories. Basically Algebra, Arithmetic, Logic, Geometry, etc. Now, the problems are not really related to each other. So you can solve all these problems in the course, Mass Plus End Problems, in whatever order you want. There is no theoretical consequence from one to another. So let's just solve these couple of Algebra problems. My suggestion is, go to the website Unisor.com and each lecture is presented in two formats. The video format, like you are watching right now, and there is a textual format right next to it, next to the video. So on the same screen, you basically see a video presentation and textual part, which is basically like a textbook. And I present exactly the same problems, sometimes with the solution, sometimes without the solution, maybe with some hints. I present them in the textual problem. So I suggested to read the textual problem first and think about it. What's most important is to think about these problems. It's good if you can solve it yourself. In any case, if you just listen to whatever I'm explaining like a solution to problems, well, I mean, I'm sure it's good practice, but that's not the purpose of the course. The purpose of the course is for you to solve these problems. And I'm just helping you with some hints, maybe, etc. Now, if you are watching the lecture, it makes sense actually to make a pause right after I explain what is the problem. Before I started explaining its solution. And again, try to solve the problems yourself. Whether you succeed or not, doesn't really matter. But it does matter, but it's not really important. What's important is you to think about these problems. And that's what actually is the purpose of the whole course. To encourage you to think about the problems. Attempting to solve it, obviously. Okay. I have three algebra problems. So let me just start from the first one. So there is a system of two equations. A is some number. X, Y and Z are unknown variables. So their sum is equal to A. And some of their reciprocations is equal to reciprocations of A. Kind of a strange combination. Now, I have only two equations. And I'm not asking you to solve, because there are only two equations. With three unknowns, it cannot be solved. However, there is something which we can say about variables X, Y and Z. You have to prove, basically, that one of them X or Y or Z is equal to capital A. So basically, just looking at this, you have to really make such a logical connection. That one of them X, Y or Z is supposed to be equal to A. Okay. So, again, that's the basically time where you can pause. Now, let me start solving the problem. And again, you can stop the video at any point and just think about further saying basically, oh, I know how to continue solving this problem. So feel free to do that. So what I'm doing is, first of all, let's just assume that one of them is not equal to A. Obviously. And we will come to some kind of result which you will see. Now, first of all, none of them is equal to zero, obviously, because they are in the denominator. So we know that. I mean, that's just from the condition. X not equal to zero, Y not equal to zero, Z not equal to zero, and A not equal to zero. Okay, fine. So otherwise, we will not be able to do this reciprocity. Now, next what I will do is, well, let's assume Z is not equal to A. I mean, if all of them are equal to A, by the way, no, no, it's impossible for all of them to be equal to A, obviously, right? Because then it just invalid conditions would be. But, okay, let's assume that Z is not equal to A. So what I will do, I will put Z on the right. So next system would be X, Y. X and Y is A minus Z, and 1X plus 1Y is 1A minus 1Z. Okay, now, this is something which I don't really like. I will just do this common denominator thing. So instead of this, I will put common denominator is X, Y. This would be Y plus X, or X plus Y equals to this common denominator AZ. It will be Z minus A. Okay, so I will replace this with this. It's equivalent. Now, I know this condition. So I can substitute it into this. And I will have A minus Z divided by X, Y equals to Z minus A divided by A times Z. Right? Now, A minus Z and Z minus A, right? So I can always have reduced them. I assume Z is not equal to A. So I will put 1 here, minus 1 here, and plus here. Okay? So I just reduce it by A minus Z. It was A minus Z here, and Z minus A here, so that's why it's minus 1. Now I can reverse it, and I will have that X, Y is equal to minus AZ. And that's basically everything I need right now. That's why it's equal to minus AZ. Okay. Now, here is a very interesting observation which I would like to make. If you have a quadratic equation, let's call it X square plus PX plus Q equals to 0. It has two roots. Now, there is a very easy to prove theorem that two solutions X1 and X2. Their sum is equal to minus P, and their product is equal to Q. Now, I did address this particular property of quadratic equations in the prerequisite course, mass for teens, where I discussed the quadratic equations. Now, in the textual part of this lecture, I specify exactly what is the lecture name and whatever the menu items you have to go through to get to this lecture. But it's very easy actually to prove even if you will take the formula for solutions of a quadratic equation of this type, and you will see that their sum is equal to minus, sum of the solutions would be minus P, and the product would be equal to Q. So, I assume that this is the property which is known. It was discussed in one of the earlier lectures. So, I'll just use this property. Now, what follows from this? Well, what follows is that if you have another equation, let's say Y, square plus PY plus Q is equal to 0. It also has two solutions, Y and Y2. Now, solutions to these two are supposed to be exactly the same, right? So, it's either X1 equals to Y1 and X2 equals to Y2, or X1 is equal to Y2 and X2 is equal to Y1, right? This is true. So, if you have the same quadratic equations, they have to have exactly the same roots. And the roots are completely defined by this property. So, let me just go back to this. What follows from this is the following. This is a very important logical kind of thing. If you have X plus Y is equal to A plus B, and X, Y is equal to A times B. Now, what actually it means? It means that X and Y are roots of the same equation as A and B. Since their sound is the first coefficient with a minus sign, and the product is the free member of the quadratic equation, which I was using like minus P and Q. So, they are supposed to be roots of the same equation, which means it's either X is equal to A and Y is equal to B, or X is equal to B and Y is equal to A. It's one of those two variations, right? So, that's what's very important. Now, if again, if you have this property, you have this property, because both X and Y and A and B are solutions to the same quadratic equation, which we can put X plus P, X plus Q is equal to 0, where P is equal to minus A plus B and Q is equal to AB, which is the same as minus X plus Y and X times Y. So, since some of them is the same and product is the same, they are solutions to the same quadratic equations, and that's why they are supposed to be equal, either first to first and second to second, or either first to second and second to first. Okay, great. This is basically the end of the proof, because as you see, if I will put A is equal to capital A and B is equal to minus Z, I will have that X plus Y is equal to A plus B, A plus B, right, A and Z, and XY is equal to A times B, A minus Z product. Which means that either X is equal to A and Y is equal to D, and A is capital A and B is minus Z, or X is equal to minus Z and A is equal to capital A, which is exactly what we wanted to prove, that one of those, either X or Y, are equal to A. In the beginning, I suggested that maybe Z is not equal to A. Well, if some other variable is not equal to A, we will do exactly the same thing instead of Z, we will use another variable. But basically, the whole meaning actually, of the whole problem is this. So if you have some of two variables equal to some of the others and product is equal to product, then they are supposed to be the same. I mean, their pair, either this is equal to this and this to that, or vice versa. So that's what's very important. The sum and the product of two variables totally define them, at least as a pair. Okay, number two. Number two is to prove the inequality. Okay. So we have to prove that X plus Y to the fourth less than or equal to X four plus eight Y to the fourth. Now, we have to prove this inequality. All right. So how can we prove it? Well, first of all, I would like to simplify it. If Y is equal to four, sorry, to zero, then it will be X to the fourth minus eight times X to the fourth. It will be obvious inequality because X is a positive. X to the fourth is a positive or not negative actually, the number. So if Y is equal to zero, what follows is obvious inequality. This is not negative. So obviously if it's zero, then it's obvious equality. And if it's not zero, we just reduce by X four and we will have one less than eight. That's obvious. So that's okay. Now, so we assume that Y is not equal to zero. Now, the plan which I'm going to go through is, I will transform this equation into equivalent one and then another and another and another until I will see the obvious, obviously true equation or in equation. Now, since all my transformations were equivalent, they are going in both directions from A follows B and from B follows A. So I will do only equivalent transformations. Then I can say that my end statement, which is obviously true, can be used as a starting point and then I go all the way up back to the original statement. So it's analysis and the proof. Analysis is from whatever you have to prove, you go down to something which is obvious. That's analysis and then the real proof is to start from the obviously true statement and then using logical transformation, equivalent transformation, you go back to your statement which you have to prove. So that's the plan. All right, so my first transformation is Y is not equal to zero. I can divide both sides by strictly positive Y is equal to fourth. So what will I have? I have X over Y plus one to the fourth. I have to prove that this is X to the Y to the fourth plus eight. Now, why did I do this? It just seems to be a little easier because I can always say that X over Y is equal to T and basically this is T plus one to the fourth less than eight T to the fourth plus eight. It looks simpler because it's only one argument rather than two X and Y. So that's easier for me. All right, that's kind of the thing which I did it without knowing what will be further on. Okay, next obviously my point is I have to somehow transform it into more, well, adequate kind of a form. So more adequate is just polynomial of T. So I will just open the parenthesis and it would be T plus one to the fourth. I'll just multiply T plus one to the second which is T square plus two T plus one and then multiply these two by themselves and I will give you the result. Now, everything from the left part I will go to the right. So I will have something on the right which I will get from here. So I'm not going to do this in detail. So I'll just give you this final formula. Okay. Now, T plus one to the fourth would be T to the fourth plus four T to the cube plus six T square plus four. And then I will subtract this from this part and that's what I will get here. Right. T to the fourth plus four T to the cube plus C, T square plus four T plus one and I will transfer it to this one. And the first and the second would be seven and seven. Everything else is left with a minus sign. Okay, correct. So I have to prove this. Well, again, it's not obvious. However, and this is something which you really have to come up with from some kind of an inspiration. Look at this. All of these coefficients equal seven minus four. It's three minus six minus three minus four is minus seven plus seven zero. Now, what does it mean that the sum of the coefficients is equal to zero? Well, it means that if I will put T is equal to one, then the result of substituting would be zero, right? If T is equal to one, I will have just some of these coefficients. Zero. Okay, that's great. So T is equal to one is a root of the equation. This one is equal to zero. Now, there is something which is called fundamental theorem of algebra. And again, in the previous course, Mass for T and so I explained it. If some kind of a polynomial P of M degree of, let's say X has root X equals to A, then this polynomial is divisible by X minus A. And the polynomial of degree one less than this. So this is one of the consequences of the fundamental theorem of algebra. Fundamental theorem of algebra actually says that any polynomial of the N's degree has N complex roots. And one of the first colorary of actually of this theorem is that if A is a root, it's divisible by X minus A. Now, T is equal to one is a root. So this thing is divisible by T minus one. So it can be represented as T minus one times something. So let me just do it. Seven minus minus four T cubed minus three T squared minus four T plus seven. If I will divide it by T minus one. Seven T four, so it's seven T cubed. It's seven T four minus seven T cubed times one. Subtract will be three T cubed. Minus six T squared would be plus three T squared, right? Am I right? Let me check. So it will be three T cubed minus three T squared. Subtract will be minus three T squared minus four T. It will be minus three T minus three T times three T squared plus three T. It will be minus seven T plus seven and it will be minus seven. So as a result, this is equal to T minus one times seven T cubed plus three T squared minus three T minus seven. This is supposed to be greater than or equal to zero. So I'll just represent this as a product of these two. Now, again, seven plus three ten minus three minus seven minus ten, zero again. So again, T is equal to one is the root of this one. And I can divide it again by T minus one. And so what will be? It will be T minus one squared and this is T minus one. That's why this is squared and some quadratic equation. And I will give you the result because I have already done this. It's just a simple division. Okay. So I represent this as a product of these. How? Because I notice that the one is the root of this equation and then root of the result of the division by T minus one, which means it's divisible by T minus one again. That's why it's T minus one squared. Great. Now this is greater than equal to zero because it's a square, right? Now this is a quadratic polynomial. It's discriminant, which is B square minus four AC. B square is one hundred minus four times seven and seven, four times forty nine. Whatever it is. It's minus ninety six. So this is negative. When the discriminant is negative, there are no real solutions to a quadratic equation, which means it's always either above or below the x-axis on the graph. Well in this case it's above because the first coefficient is positive seven. So it's a parabola which goes with its horns up and it goes above the x-axis because there are no solutions, no roots. So this is also positive. So this is positive, well this is not negative and this is just positive basically. And that's why this last equation is obvious. Since it's obvious, then all my transformations are totally equivalent. So I can say that if this is consequent, if this is obvious, then this is true and this is true and this is true and this is true and this is true and this is true. So that's the end of the proof. Again, what I did before was analysis. Proof is from here back to the original. From obviously true statement to the one which you would like to prove. Okay, and the next problem also proved the inequality. That's x to the 12 minus x to the 9 plus x to the 4 minus x plus 1 greater than, no, just strictly greater than zero. We have to prove it. Okay, fine. Let's just consider this particular polynomial. Now, this is x to the 9 times x to the cube minus 1, right? x to the 9 factor out, so I will have 12 minus 9 is 3 and this is 1 plus x times x to the cube minus 1. x factor out, x cube is minus 1. Now x to the cube minus 1 again. Well, I can actually, instead of x 9 times 6, I can put x 8 plus 1 and x, right? Okay, now, let's talk about a graph of this thing. Graph is something, whatever. Now, these are roots of the polynomial where it's equal to zero, right? Now, what are the roots of this? This doesn't have any roots under no circumstances. This one is equal to zero. This is equal to zero at zero and this is equal to zero at one. So basically we have only two roots. So my graph is supposed to go something like this. This is zero and this is one. Now, the greater than one and less than zero, it's supposed to be positive, obviously. If you will put a large positive number, it will be positive. So it's somewhere here, somewhere here and somewhere here. So that's basically the graph. It's supposed to be in this particular area. Well, okay, fine. So it doesn't really matter what the behavior of this function less than zero and greater than one. We have to only consider this particular interval from zero to one, right? Okay. So outside of this interval, I know that this thing is positive. So plus one will be definitely positive, not negative, and then plus one will be definitely positive. So there is no problem with outside of this interval. So let's consider only within this interval from zero to one. Now, that's easy. From zero to one, I will transform it a little bit differently. X12 plus X to the fourth, one minus X to the fifth. X to the fourth by X to the ninth combined together. X to the fourth outside of parenthesis. So I'll have one, which is X4, and minus X5 would be minus X9. Plus, and this is in brackets, in parenthesis. Now, I'm talking about only interval from zero to one, right? Which means this is non-negative and this is non-negative. And obviously X12 is also non-negative. So basically everything is non-negative here, here and here. This is non-negative, one minus X minus five on this interval, if X is less than one, and the same is here. So that's how this is greater than zero. I mean, if you wish, the only thing which kind of a iffy is endpoints, zero and one. Well, if it's zero, then it's one is greater. And if it's one, that would be one minus one, one minus one, and one again, one. So in both cases, our inequality has been proven. Okay, that's it. Thank you very much, and I suggest you to read the notes for this lecture. There are some solutions presented, and think, think and think. Try to think about each problem again, even before you read the solutions. Okay, good luck.