 Hi, I'm Zor. Welcome to Unisor Education. We were talking about systems of equations in general and then in particular systems of linear equations. Well, this is kind of boring stuff because everybody knows how to solve them. It's not big deal. We have certain algorithms which always works. Like for instance, you do the substitution from one equation to another, you get another system of linear equations of a lesser number of variables and lesser number of equations and then etc etc you step by step you reduce it down to zero and then you solve the linear equation. No big deal. It's time to complicate your life a little bit. So after we were learning about linear equations with one variable, we switched to quadratic equation and then higher order equations. Same thing with systems of equations. We have learned about linear systems and now we will go into quadratic systems. So a couple of words related to definition, theory, etc. First of all, to define what is a quadratic system of quadratic equations, we will have to familiarize ourselves with the concept of univariate and multivariate polynomials. Now, univariate polynomial is something which you definitely know. It's something which looks like this. It's a function of one variable which is x in certain powers multiplied by certain constants, usually real numbers. Well, sometimes complex, but it's really very rare. So it's usually real numbers and then sums together. So this is a function of one variable and this particular function is a polynomial of nth degree because we assume that a0 is not equal to zero. So the maximum power x is raised into is n. Now, this is called, as I was saying, univariable polynomial because it has only one variable, x. I would like to extend this definition to multivariate polynomials. Now, what is this? Well, it's multivariate. It means there are multiple variables which are supposed to participate in this polynomial. Now, in what way? In some way, it's similar, but instead of one variable x, we have multiple variables. So let's say you have x1 to some power i1 times x2 to some power i2, etc. xn to power in. Now, this term will be used instead of this one. With some powers which each variable x1, x2, etc. xn is raised into. So this is one single element from which we will construct the multivariate polynomial. So we have to multiply this particular element to some constant, which I would put something like i, i1, i2, etc. It doesn't really matter what kind of indices I'm using. So we multiply it and then we add another term which looks exactly like this except these variables will be raised into different powers. So let's say it will be xi j1, x2, j2, etc. xn, jn. Again, multiplied by certain coefficient a, j, whatever. So some of these with these particular variables raised into different powers is a polynomial which we can call multivariate polynomial. Just as an example, let's say I have only two variables x1 and x2. Then an example would be x1 square times x2 plus 3x2 plus 4x1x2 to the third power. Something like this. It doesn't really matter. So the combination of product of variables raised into certain powers multiplied by constants, in this case the constant is one obviously, and sounds together. So anything which looks like this is a multivariate polynomial. Now it does make sense just to kind of to order thing in your mind. To really consider the polynomial of this type in a sequence of increasing or decreasing combined power of all the variables. Now what's the combined power? Combined power in this case is 2 plus 1 which is 3. Combined power of this is 1. Combined power of this is 1 plus 3. x1 to the first degree and x2 to the third degree, which is 4. So it makes sense to rewrite this as 4 x1 x2 cubed plus x1 square x2 plus 3x2. In the order of decreasing of the combined power, it's just to make it a little bit better, a little bit more orderly if you wish. So the maximum combined power will be in the first place, then it will be next less than that, etc, etc. Now if the combined power, the maximum combined power is equal to 2, then this particular multivariate polynomial will be called quadratic, which obviously is reasonable. So let's just talk about examples of quadratic polynomials, multivariate polynomials. For instance, x1 square plus x2 square. This is quadratic because the maximum power, combined power, is 2. Now x1 x2 plus x1 plus x2. This is also a quadratic polynomial because the maximum power is 1 plus 1, which is 2. These are less than that. So the maximum power is 2. So this is quadratic polynomial. And obviously there are many other examples, which you understand. Now in this particular case, I'm using x1 and x2 as variables. I can as well use x and y and z and u and v and whatever other letters doesn't really matter. So this is also a quadratic polynomial where a, b and c are constants and u and v are variables. Because in this case, the combined power of this particular term is 1 plus 1, which is true. So it's quadratic polynomial, multivariate, which depends on two variables u and v. All right, so that's all about the theory of what is quadratic polynomial. Now obviously, the system of quadratic equations is a system, which means more than one equation. Each one of them contains quadratic or less than quadratic, which is actually linear polynomial. But at least one equation in the system must contain the quadratic polynomial for the whole system to be quadratic. So this system is quadratic. Why? Because one of the polynomials on the left of the equations is quadratic. The second one is having to be linear. So again, the system of quadratic equations is a system of equations with the left part containing multivariate polynomials. At least one of them of the second degree and there is no more than second degree polynomial. So there is no third degree or fourth degree, etc. So it's either second degree or the first degree multivariate polynomial. So that's the definition. All right, so I have a couple of examples here and I will just put it on the board. Okay, this is the first system which I suggest as an example of something called this. This is a system of three equations. It's a quadratic system because each one of these equations is a multivariate polynomial on the left of no more than the second degree. And we do have actually the second degree polynomials on the left as well. One is a linear. So this is an example of a system which I am talking about. Now, it would be interesting to talk about solutions obviously, right? It's one thing to present a system of equations, another is to solve it. And this is actually much more problematic in case of quadratic systems than in case of linear. In case of linear system you can just learn how to solve the equation. Well, you just, for instance, using the substitution method. You take the first equation, the second equation, etc., substitute, no problem. There is no general methodology to solve quadratic equations. Why? Because let's say you want to do it using the substitution method. Well, look at this. You can actually find out what z, let's say, is from this. z is minus square, I mean, sorry, z square is minus xy. So z is square root of minus xy. And it can be actually plus or minus. Then substitute it into this. And what will you have? You will not have an equation which can be easily solved or to this. I mean, it's not easy. There is no general solution, general approach, general algorithm how to solve the quadratic equations. So it's your ingenuity, actually, which must really help you to find certain peculiarity in the system which will allow you to solve. Well, except one case. One case is what if in my system I have one equation of the second degree like this one and other equations or other equations are linear, something like this. In this case, the situation is easier and it can be actually solved like basically algorithmically. I mean, there is a way to do it, how to do it. Well, take a look at the linear equations. Now, if you have a system of n equations with n variables, n minus one of them are linear. Now, let's assume that you just take one particular variable out of these three as a base and just consider it's given. And you can solve this system of two equations with three variables as basically two variables in terms of the third one. So let's consider that z is something which is known to you. Then basically you can consider this as a system x plus y equals five minus z, like this is the first one, and x minus y equals five plus z. Now, if z is known, you can solve this equation of two variables, but it will be obviously in terms of z. So x will be some kind of linear function, linear, that's important of z, and y should be linear function of z. Well, in this particular case, the easiest way is just to add them together, it will be two x, ten, so x is equal to five, it's a constant actually. But if you subtract from this, you subtract this, you will have two y equals to minus two z, so y is equal to minus z. So this is a solution, but the solution in terms of z. Now, considering this, you can substitute both x and y into this first equation, which is of a second degree. It's a polynomial of the second degree, real quadratic polynomial. But we know that this is a polynomial of second degree, so if anything, all these variables are described as linear functions of z, then I will as a result have a quadratic equation of z. Alright, so let's write it down. Instead of x, I will put five, so it's 25, plus x, y, it's x times, so it's minus five z, x, y, so it's minus five z, plus x, z, x, z, which is five z equals to eight. Alright, in this case I have a contradictory equation, because I just made it up, I didn't really think about what kind of equation I said. So this is a system of equation which doesn't really have a solution, because I have 25 equals to eight. However, if I were somehow differently, let's say this is instead of y, z, x, z, what this is y, z, for instance. I don't know what happens, so let's see. In this case, instead of five z, I will have plus, y, z is minus z square, right, so it's minus z square. Now, what happens now? Is this an equation which does have solutions? Well, it's z square plus five z minus, what, seventeen, right? z square plus five seventeen minus seventeen, right? And this is an equation which does have a solution. z is equal to two minus five plus minus twenty-five plus thirty-four, sixty-eight, something like this. So whatever this is, this gives me z, and then since x is equal to five and y is equal to minus z, I have the solution for everything. So in this particular case, when the system contains only one real quadratic equation and all others are linear, then we can actually basically have a procedure which will definitely give you some result. You solve these which are linear in terms for all variables instead of one and then substitute. In all other cases, it's much more difficult. It needs some ingenuities, as I was saying. And the only way to approach learning something about this system of equations is, well, very simple. You solve as many as you can. You listen to lectures if I'm presenting something, you read whatever you can read. And the more practice you have in solving different equations, the more approaches you will have in your mind and the next one will be probably easier. Because there are only a certain number of different techniques which you can use. And obviously people who are creating systems of quadratic equations for you are also kind of restricted to a certain number of rules and techniques, etc. So the more you solve them, the easier you will feel. Now, in my particular case, I just wanted to relate you to the previous lecture which I had about relativity, which I included into the systems of linear equations. And for one particular reason, because it's actually of this type I was just mentioning, one particular equation was a quadratic equation, but the rest there are four equations all together. So one is quadratic and the three were linear. So I was basically using this approach which I was just talking about to solve linear equations in terms of one particular variable, the fourth variable, if you wish, or the first variable. And then the second, the third and the fourth are in terms of the first. And then substitute it into the quadratic equation, basically getting the solution. So let me just very briefly tell you what it is, since this is a good example. This is what I've had in this lecture about relativity and I will try to do it very quickly. For details, I do refer you to this lecture. This is an example of the system of four equations with four variables P, Q, R and S. Now C and V are constants. Now, as you see, this, this and this are linear equations where P, Q and R and S all are linearly related to each other. And only this one is really quadratic equation. So what I did in that lecture, and I don't want to repeat it actually, I expressed every variable in terms of P. Q is already expressed because V is a constant. Now here, these two, the easiest way to do it is add two variables and I will have Q equals RC square, right? If you add P, C will eliminate each other. This will be two Q and two R, C square. And since I know how the Q is expressed in terms of P, I know R. Now if I subtract, I will have similarly four S. So I have expressed P, Q, in terms of P, I have expressed Q, R and S, then substitute it to this. I can get P and then from P I can get all the rest of this. So that's one of the examples when it's really kind of a simple thing to do. Now what if you don't have this particular type of a system of equations which you know basically how to do. There is a standard approach, so to speak. Well, again, it's just your ingenuity and certain things you can come up with. And the more you solve different equations, finding these peculiarities which you can use to solve it, the richer will be your vocabulary, your repertoire of how to approach different problems. So let me just present you a couple of examples which probably might help you. Okay, this is the first one. Both equations are really quadratic equations of second degree. This is second degree because everything is squared. And this is also second degree because one and one and multiplied together, we are summing the powers. So it's not so easy, right? Well, let's think about it. Let me try to do the substitution. Maybe it will work. First of all, I have to observe that X is not equal to zero and Y is not equal to zero. Because otherwise their product would not be equal to six. If that is true, then I can express Y in terms of X from this particular thing. I divide by Y, left and right. So I will have X equals to six. And I can divide by X. So I have Y on the left and six over X on the right. And X is not equal to zero, so this is fine, no problem. And then I can substitute this into the first equation. So what do I have? X squared plus 36 X squared equals 13. Okay? What is this equation? Well, it's not quadratic, that's for sure. Well, let's do something about this equation. I will multiply by X squared, both parts. I will have X to the fourth. Plus 36 equals 13 X squared. Or X to the fourth minus 13 X squared. Plus 36 equals to zero. Okay? Now, this is an equation of the fourth degree. But if you remember, in lectures dedicated to high-order equations of one variable, this is a special case. Let's see, what's special about this? Well, this is the fourth degree polynomial. However, there are only fourth and the second degree here, which means if I will have a new variable, Z, which is equal to X squared, then X to the fourth will be Z squared minus 13 Z plus 36 equals to zero. And this is the square, this is the quadratic equation, right? So we can solve it for Z, whatever the solution will be. I will have two solutions. 13 plus minus, plus minus, square root 169 minus 412144, is that right? Divided by 2. So this is 25, so it's 5 divided by 2. So it's 9 and 4. So, roots of this equation are 9 and 4. Now, this is X squared. In case of 9, X might be equal plus or minus 3. In case of 4, X is equal plus and minus 2. Now, since Y is equal as you remember this, so I have four different solutions to my quadratic equations. 3 and 6 divided by 3, which is 2. This is one pair. Second, minus 3, 6 divided by minus 3, it's minus 2. Now, if it's 2, it will be 2 and 3, and if it's minus 2, it will be minus 2, minus 3. So these are four pairs. Each pair is a solution. By the way, if you see these are very symmetrical, relatively to exchange of the places between X and Y, and the solutions are asymmetrical. I have 3, 2, and I have 2, 3. I have minus 3, 2, and minus 2, minus 3. So solutions are also symmetrical, which basically kind of agrees with whatever we feel it's supposed to be. Now, obviously we can just check each one of them. Well, let's say this one, minus 3, minus 2. Minus 3 squared is 9, minus 2 squared is 4, so sum is 13 and the product is 6. So basically the checking is working, and you can check every one of those. So this is an example of how, well, non-trivial, if you wish, case can still be resolved. So in this particular case, we use substitution from this to this. As a result, we have an equation of the force degree, but the good equation of the force degree, which we can solve using the substitution Z is equal to X squared. Alright, and one more example. This is basically an introductory lecture about quadratic equations. It needs a lot more of examples with solutions, which I do intend actually to provide in other lectures dedicated to this topic. But since this is just about basically definitions and basic properties and basic approaches to quadratic equations, I will restrict myself to only one more example. And here it is, X squared plus Y squared plus Z squared equals to 29, X Y equals to 6, and X plus Y plus Z equals to 9. Well, again, this is not exactly the system which we can solve blindly using some approach, because I have two equations of the second degree and one equation of the first degree. However, you know, I might actually try to do something about it. Now, what can be done? Well, what can be done? Here is what I might actually suggest. Why don't you multiply this by 2? I will have 2XY equals 12, and I will add it to the first equation. So I will have X squared plus 2XY plus Y squared plus Z squared. 2XY, I inserted in between these equals to 41. Now, what is this? Well, this is obviously X plus Y squared. Plus Z squared equals to 41. And let me rewrite the bottom, but I will put parenthesis around it. What's interesting about this system? Well, if I will put X plus Y is equal to U, some kind of an intermediary variable, I will have a system which is U squared plus Z squared equals to 41. And U plus Z equals to 9. This is much easier. This is a system of two equations with two variables, granted it's quadratic equation, but only one of them is really quadratic, another is linear, which means what? Which means we can very easily solve it using the substitution. So, U would be equal to 9 minus Z. U squared would be equal to 81 minus 18 Z plus Z squared plus another Z squared equals to 41. 2 Z squared minus 18 Z plus 81 41 plus 40 equals to 0, reduced by 2 Z squared minus 9 Z plus 20 equals to 0. This is quadratic equation. We can very easily solve it. I just see that roots are 4 and 5. So, U is equal to, sorry, Z is equal to 4 or 5. Alright, that's much easier. Now, we can substitute it to, let's say, this and this, and basically, or actually we can use this one and this one. It doesn't really matter. Okay, so let's do it. We don't need this anymore. So, we'll use it here. So, let's say case number 1, Z is equal to 4, and I will consider only these two. So, X, Y is equal to 6, X plus Y is equal to 5. Well, this is again, substitute Y is equal to 5 minus X, put it into the first equation, and I will have minus X squared plus 5 X equals to 6, alright, or X squared minus 5 X plus 6 is equal to 0, X is equal to 2 or 3, and Y is equal to 3 or 2. So, what we have right now is, that's very important not to lose our ends. We have Z is equal to 4, and X is equal to 2, and Y is equal to 3, or 3, 2. That's what we have in this case. That's number 1. Number 2, when Z is equal to 5. Now, same thing here. I have X, Y is equal to 6, X plus Y is equal to 4. So, Y is equal to 4 minus X from here. Substitutes to this would be minus X squared plus 4 X equals 6. Or X squared minus 4 X plus 6 equals to 0. And this doesn't have any solutions, because 4 squared minus, which is 16 minus 4 times 6, which is 24, is negative. So, you will have negative under the square root. So, no solutions here. So, the whole system has only these two triplets of X, Y and Z as solutions. And if you can substitute it, you can definitely see that the checking works as well. All right, great. Now, I do recommend you to read notes for this particular lecture. It's on Unizor.com in the Algebra section. It's called System of Quadratic Equations. Now, I do plan to provide more examples of different systems of quadratic equations with non-trivial approaches how to solve them. And, again, what's very important is the more problem you will solve or you will listen to what I am solving, and then you will try it yourself, the richer will be your repertoire of techniques which you can use. And that's why what I was talking about as ingenuity would actually be much easier for you. You don't really have to invent something. It will be already inculcated into your minds that you can try this way, you can try that way. It's easier to, well, invent quote-unquote something if you had something similar in the past. That's it for today. Thank you very much.