 I'm Zor. Welcome to Unisor Education. Today we will continue talking about equations. This is part two, and we will mostly concentrate on how to solve equations. Basically, from the part number one, you could learn that the basic method to solve equations is to use invariant transformations. That's kind of an obvious and relatively easy way if you have an equation at plus five equals seven. Convariant transformation is obviously transformation which transfers every variable into that variable minus five, and if you apply this transformation to the left part, you will have x plus five minus five, and if you will apply to the right part, it will be seven minus five, which actually results in x equals two. Now, what's important about this transformation being invariant is the fact that from not only from this, you can come to this, but also you can go backwards, and invariant transformation, which is reverse of this one, would be x plus five, so you can always from here by adding five to both sides of this equation. You can come to here. These three equations are completely invariant to each other. They have exactly the same solutions, and this is a solution already, so that's exactly the solution of the first equation. So this is an easy part. Now, let's just examine what kind of invariant transformations we can use, and also what non-invariant transformations exist and how they can be used. That's much more interesting than the difficult topic. Okay, so as far as the invariant transformations, number one is transformation when we add any number. Let's just assume that our equations are with real numbers, and we are looking for solutions among real numbers only, just to make our job a little bit easier. So this is an invariant transformation, and the inverse one is obviously x minus a. All right, now next. The obvious next choice is multiplication. Is this an invariant transformation? Well, almost. Obviously, if we want the reverse one to be correct, and the reverse one is obviously dividing by a, a should not be equal to zero. So now we have a transformation which is almost inverse. We have a small restriction of this multiplier not to be equal to zero. Now, how can that be used? Well, obvious. If you have something like x multiplied by 5 equals to 20, you can apply a transformation, g of x equals to x divided by 5, or if you wish. Since we're talking about multiplication here, we can say x multiplied by one-fifths. So if we will apply this transformation to both sides of this equation, we will have x times 5 times one-fifths equals 20 times one-fifths, which is x equals to 4. And since our multiplier is not equal to zero, this is invariant transformation. So from here we can go back, and that proves that x is equal to 4 is a solution. By the way, instead of applying reverse transformation, which is in this case x times 5, we can just substitute this 4 into an original equation, and that will be just a plain checking, which is an mandatory thing, by the way, when you solve any equation. Checking is always mandatory. Now, that's fine. That's easy. Now, as far as invariant transformations, there are many different examples of invariant transformations which we can just talk about right now. But what's the most important about every invariant transformation is that it represents a one-to-one correspondence between the domain where the equation actually is defined. In this particular case, if we will go back to this equation, x times 5 equals to 20. It's defined basically on the set of all real numbers, and this particular transformation is also defined on the set of all real numbers, and that's exactly where it is reversible. It represents a one-to-one correspondence between the domain and co-domain of every function. The one-to-one correspondence can be established by the fact that from every real number, we can multiply it and get some other number, and then backwards from that other number, we always go back to the original one. If we have two different numbers as origination, it will have two different results, and if we have two different results, we can always go back into two different original numbers. That's what one-to-one correspondence actually means. Every invariant transformation is a one-to-one correspondence between the domain and co-domain. As you know, it can be represented graphically if you have something like two coordinate axes which represent the function t of x, then any graph which will establish this one-to-one correspondence would actually constitute some kind of invariant transformation. Why is it one-to-one? Well, from every point on the x-axis, you get one and only one point on the y-axis, right? And vice versa. From every point on the y-axis, you can get one and only one point on the x-axis. So that's what one-to-one correspondence is. Now, obviously, functions like x plus a are graphically represented as a line which has an angle of 45 degrees and shifted upwards or downwards, vertically depending on the sign of a. Function x times a very close to this is actually the graph of this function goes through the center and the angle depends on the a. If a is positive, the angle is positive. If a is negative, the angle will be negative. But in both cases, any straight line represents a one-to-one correspondence. And in this particular case, it's very easy to see that the line with a equal to zero does not represent the one-to-one correspondence because here it is. If you have a not equal to zero, let's say a is equal to two, that would be the graph. This is definitely a one-to-one correspondence. From every x, you get the t of x and from every t of x, you get back to x. But if a is equal to zero, the graph will be t of x is equal to zero, which is coinciding with this horizontal x-axis. Now, is this a one-to-one correspondence? Well, let's think about it. From every x, we can get y, which is always zero, right? But from every y, we cannot get the corresponding x because from these ys, there is no x at all. There is no prototype. There is no domain point, which will result in non-zero in this case. And if y is equal to zero, if t of x is equal to zero, again, there is no way to find out from which x it actually came to zero because any x will give zero. So this is definitely not a one-to-one correspondence between domain and codomain. So that's why a is not equal to zero in the graphical to the graphical reference infection. But anyway, as I was drawing before the graph, an example, by the way of that graph, would be t of x equals to x to the power of 3. Now, this is also one-to-one correspondence because for every x, we can find a concrete t of x and vice versa. It's always the three-root, the cubical root exists for every number, every real number, and it's always defined, well-defined, without any problems. And obviously, you can use this particular transformation to solve equations like cubical root of x is equal to 5. Now, since you can apply this invariant transformation to both parts, so let's use the power of 3. In this case, it will be this. Now, by definition of the three-root, it's basically some variable, some number, which if you're using the power of 3, it will give original number x, right? That's just from definition of the root. And this will be 125. So that's the solution. And since it's invariant transformation, it's obviously the correct solution checking its cubical root of 125 will give 5. However, there are some seemingly simple but very treacherous other functions. So now we are moving towards the category of non-invariant transformations. And the question is, can we use them, because sometimes they seem to be quite useful, and how can we use them safely? Okay, the first non-invariant transformation will be x where? Well, first of all, y is this transformation non-invariant. It looks very much close like x3, but however, let's do it graphically. Remember, if you have the function x to the power of 3, the graph looks like this. So it's one-to-one correspondence between domain and codomain, and the correspondence is established through this graph. How? You do this, and you get y. So from every x, you get only one y. And from every y, you get only one x. So it's always one-to-one correspondence. The situation is quite different with x2. Why is it different? Well, what was the graph of x2 in this? This is parabola. Sorry for my drawing, not exactly perfect parabola. But anyway, for every x, we can find the corresponding y. That's no problem. How about the reverse? Definitely not, because from this particular y, we have two different axes which correspond this particular value of the function. So if you have, for instance, y equals to 4, x can be 2, or x can be minus 2, because both being squared give you 4. So transformation is not invariant. Does it mean we can't use it? Well, we can't. But we really have to do it very carefully. And let's see how. Let's say you have an equation that goes very much like in the case of x to the power of 3. So you have a square root of x equals to, let's say, 9. Well, what do we do in this case? Well, we apply the transformation x2 to both sides. You will get x equals to 81. Well, is it correct? Everything seems to be fine. If you will go back, the square root of 81 is 9. Everything seems to be fine. Okay, let's do a different equation. How about square root of x is equal to minus 9? Again, apply transformation t. We get x equals to 81. Exactly the same thing. Now, is it a solution? Well, obviously not. Because again, you substitute to the square root of 81. You get 9. 9 equals to minus 9. That's nonsense. So this is not a solution. So apparently, in one case, we get the correct solution. And in another case, we are getting not the correct solution. And why? Because this is not an invariant transformation. So how can we do it, let's say, more accurately? And how could we use the whole thing with a certain level of assurances that we have the right solution? Well, there is only one answer in this particular case. Do the checking. After you solve the equation, in case you're using all invariant transformations, you have to do checking, but it's not really mandatory because if your invariant transformations are really invariant, you are absolutely sure about this. And all calculations which you do, also done quite accurately, there is no way you get the wrong solution. Everything will be fine. There will be no problem. We still encourage people to do the checking, but it's not because there is something wrong if they don't. In case of non-invariant transformation, as you see, we might get into the wrong solution and think quite correctly the transformation which we wanted to use. And the incorrect solution, actually, is not something because we made a mistake. We didn't make any mistake. We squared the whole thing quite correctly. It's because the transformation we are using to square both sides is not invariant. So if you're using certain non-invariant transformation, rule number one is always checking your solution. Well, actually, it's rule number one, even in case of invariant transformation, but over there it's just to protect us against some mistake which we made in calculations. In this case, it's not really a protecting against mistake. If you do everything very accurately, you still can get the wrong answer. And again, the reason is it's non-invariant transformation. So sometimes, if you're using certain invariant transformations, you can get extra solutions which are not really solutions to the original equation. That's why you have to check it. Well, that's not it. Unfortunately, the story doesn't end here. As we know, right now, sometimes we are getting extra solutions which are not the real solutions to our original equation. But sometimes, unfortunately, we can lose the right solutions and that we have to avoid, obviously. And here is the next example. Another transformation is square root of x. Okay? Let's think about this. And let's start with a solution with equations like this. First of all, let's guess what's the solutions to this particular equation. Well, obviously, it has two solutions. Six and minus six, both being square root of v with 36. So we know in advance we are very smart. We guessed the solution to this equation and there are two of them. Now, let's apply this particular transformation and let's do it very, I would say, immaturely, blindly, wrongly, if you wish. Okay, square root of both sides. Well, square root of x squared is x. Square root of 36 is 6. We got the solution. Well, did we get the really right solution? Well, this is the right solution, by the way, because if you will substitute it back and you will check everything is correct. But don't forget that we have lost this solution on the way. And why? Again, we have lost something because the transformation is not really invariant. So this is an example that we lost a correct solution. The previous one was that we have actually came up with a solution which is not really a solution. So when we are using invariant transformations, we can either gain more solutions and some of them will be wrong, obviously, although we did not lose anything. But in some other situations, we might lose the right solution and that also should be somehow dealt with. Now, to protect us against gaining wrong solutions, as I said before, you just check it. That was in the previous case. In this case, check really confirms that we came up with the right solution. So somewhere in between, we have lost this solution and didn't really pay attention to this. And that's where there is a difference between art and science, if you wish. Checking is just a straight science. This is by the book. There is no problem with this. Square root, this particular transformation, is a little bit on the verge of art. You really have to know what you are doing. You have to know under what circumstances you can lose something because if you lost it, then there is no checking or whatever which can help you to regain that solution you have lost. So in this particular case, there is no general recipe, unfortunately. That's why it's art and not science. You really have to know what you are doing. And square root is such an animal that you really have to understand the intricate details of square root. That's why I would like to devote a few minutes just explaining what the square root actually is. And then we will return to this equation. If we are talking about square root of something, of x, whatever. Obviously, the definition of the square root is, well, this is the number which is being squared gives the original number from which we extracted that root. But in this particular case, square root of 36 is not just 6. It's plus 6 or minus 6. And that's where we were wrong. So we have to definitely know the definition of this thing. And there is actually a better terminology, not just square root of x, but arithmetic value of square root of x. And in most cases, when we are talking about equations or something like this, mathematicians prefer to deal with arithmetic value of the square root. And here is the difference. It's very easy to see graphically. If I'm saying that the value of square root of, let's say, 36 is plus 6 or minus 6, then how can I draw the graph of the function square root of x? If I have 36 here. So if I'm saying that we have positive or negative 6 as the value of square root, there is no way I can draw a graph in the way the real functions are defined in mathematics, because if I will have 6 and minus 6, it looks like one value of argument has two different values of function. Now, if this is, let's say, 9, I have 3 and minus 3, which is this. Which is this point and this point. If it's 0, it's only 1. So it looks like the graphically, it looks like this, which is not a function in a traditional definition of this world. Traditional function is a rule which converts one particular argument into one particular value. Argument should be from the domain, value should be from the color domain. But in this case, we have two values, which means it's not really a function. It's much easier to say that square root of x is first of all defined only on a positive, non-negative rather real number. So that's the domain. And as far as its values are concerned, for all non-negative values of x, we only take the positive number, square of, which gives you the original argument. So the value of, a arithmetic value of square root of 36 is only 6, not minus 6, only 6. But this is a real function, why? Because again, for every x in the domain, and domain is only non-negativity, you get a concrete value, one and only value of the function. So now we have one-to-one correspondence between domain, which is all non-negative numbers, and co-domain, which is also all non-negative real numbers. And this graph represents the correspondence. For every non-negative y, you can have only one, non-negative x. So if we are talking about equations, you really have to understand what is the square root. And in most of the cases, it's assumed that the square root is arithmetic square root, which means arithmetic variable square root, which is only the positive value. In which case, we obviously have to consider this thing in a following way. What is square root of x square? It's not x. This is wrong, because we assume that this is arithmetic root. And usually it is assumed, which means on the left you always have the positive number, which is being squared, give you x square. Now, what is x? You have no idea. X is an unknown variable in this particular equation. So how can we make sure that we have something like this, but the right one, the right equation, the right correspondence between the square root of x and something? Well, obviously the right thing is absolute value of x. So no matter what x is, if x is equal to minus 6, square would be 36. A arithmetic value of the square root of 36 is 6. And absolute value of minus 6 is also 6. So this is a correct equation, and not this one, because it contains the unknown variable x, sign of which you don't really know. Now, if we are using this type of square root as a transformation, and in this case, we can say that this is invariant transformation. So if we assume that this square root, t of x is equal to square root of x, is invariant transformation when we also take the arithmetic value of the square root, then it's invariant and there is no problem because square root of x square is equal to, again, a arithmetic value of square root of 36 is plus 6. This is absolute value of x, and this is an equation solution of which we will have to find. And this is obviously x equals 6, and minus x equals to 6, which is x equals to minus 6. One solution and another solution. Because what is absolute value of x? How did I derive this? There's just a little more detail. What's the definition of absolute value? Absolute value of x is x with x is greater or equal to 0, and minus x with x means less than 0, right? That's the definition of the absolute value. For positive x, that's x itself. Like for 5, it's 5. For 25, it's 25. For negative x, for negative x, let's say minus 100. The absolute value is minus minus 100, which is plus 100, and that's what the absolute value of x is. So from this definition, we can say that basically for positive x, x should be equal to 6, and it is definitely a positive x. 6 is a positive x, so this is a solution. For negative x, let's look in the negative x, this function. In the negative x, we are saying, okay, if negative x is equal to 6, x is equal to minus 6, which is negative, so it's also a solution. Great, so that's how we get both solutions, plus 6 and minus 6 for the original equation. So as you see with square roots, the situation is a little bit more complicated because you really have to think about the square root, not as any number which being squared gives you the value under the sign of radical, but only the positive or zero, non-negative value. That's very important. And that restriction actually makes the square root of x an invariant transformation, but only within its domain, the domain is non-negative numbers, and the codemain as well. Okay, now we know these little tricks, and let me just say that basically any transformation which involves the even power, like x to the fourth, x to the sixth, etc., they will all have the same property of not being really invariant transformations because the power is even. It's like in case of a square you have two different values which you can gain, one of them is right, another is wrong, same thing with x to the fourth, x to the sixth, etc. And the same thing actually is with using radicals, with using roots. Square root or four root or six root, etc., all the even numbers have the same property as a square root, and we always have to consider the arithmetic value of that root, the value which is always non-negative. Now, obviously there are many, many different examples of how these transformations can be applied. Now, in elementary mathematics, the transformations of this type, power and root and multiplication and addition, are the most important transformations which you have to use, but you always have to think about how to use it. And as a final example, let me just go back to this type of transformation, x times a, but in this case, I will do it in a slightly different fashion. Yes, we can multiply both sides of equation, but as I was saying before, the multiplier should not be equal to zero. Well, that's actually very important, because let's say your multiplier contains an unknown variable. So if it's unknown, you don't really know whether this is equal to zero or it's not equal to zero, because sometimes it might be equal, sometimes not. Like in the very simple situation, if you have something like 1 over x equals to 5, well, what do you do? You apply this particular transformation with a equals to x, right? So you multiply both sides by x. And what do you do in this particular case? Well, you have something like 1 over x times x is equal to 5 times x. Now, you don't think twice. You reduce this to 1. This is equal to 5 times x. Now you multiply both sides again by 1 fifths and you get x equals to 1 fifths. Is it a solution? Well, it is if you substitute it there. That's fine, that's no problem. Now, let's do it slightly differently. Let me just think about an example where you might be in a little bit worse situation because when you're multiplying by something, you might actually have an incorrect solution. Well, how about this? If you have 1 over x equals 2 over x. What happens in this particular case? Now, that's not a good solution. How about 1 over x? Let me think. What I would like to actually accomplish here, I would like to have an equation, you see x cannot be equal to 0 here, but I would like to have an equation which will have the root equal to 0 when you multiply it by x. How can you do that? When you multiply it by x and if x contains 0, you basically, because in this particular equation x cannot be equal to 0, right? Oh, very simple. How about if I will do something like this? Under this circumstance, x cannot be equal to 0 because basically it's in the denominator and we cannot divide by x. So if you will try to substitute 0 into this equation, you will definitely say that, okay, if it's impossible, x equal to 0 cannot be a solution to this equation. But let's try to multiply by x using this transformation and we do not pay attention for the second that x is a variable and it's an unknown variable and we actually can do something wrong by multiplying by 0. So if you multiply both sides by x, you will have back x equals x times x. You do not think twice and you basically do this type of thing and you have x equals to x squared, okay? Now, solutions to this are very easy to obtain by using this transformation. I just brought this out of the brackets and I will get that. Now, this equation obviously has two solutions. x equals to 1 because then the product will give 0 and x equal to 0 because again the product will be equal to 0. So if we blindly use this type of transformation which contains this type of transformation which contains where a contains unknown variable and we don't know in advance what's the value of this unknown variable, we can actually come to a solution which is not the right one. The x equals to 0 is not the right solution. x equals to 1 is 1, 1, 1, that's the right solution. With 0 you cannot substitute here and that's actually the wrong thing. Now, you can say that I can reduce this right now but if I will do something a little bit more complicated, you will not really be able to do this something like x minus 1 divided by x equals to x minus 1, let's say, something like this. There is nothing which can be reduced right now and if you will multiply both sides by x you will get x minus 1 equals to this multiplied by x would be x times x minus 1. Obviously, x is equal to 1 you can reduce and you will get x is equal to 1 which is a correct solution. It's not always quite visible whether you can or you cannot simplify the equation from the very beginning and so if you blindly multiply both sides by an expression which contains variable you can actually gain the root which you did not have before. That's basically my point. Even if certain invariant transformation seems quite obvious, it's not really obvious whether it is an invariant in case when some number you are multiplying by contains the unknown variable. You might gain something or lose something based on the value of this variable and you really have to very carefully check the solutions in these particular cases. So probably the best way to approach solutions to how to solve the equations using invariant or non-invariant transformation is just to solve equations. I will try to put on the website as much different equations as possible and maybe some of them will be a little bit more tedious as far as calculations are concerned. That's quite all right actually. I think solving equations even in case when it involves certain tedious calculations is also good because it teaches you some patience and accuracy, etc. But the most important thing is that you really have to do it very accurately and on every step you have to think about whether transformation which you are applying to an equation is invariant or not. Let's say you are squaring both sides of the equation. But immediately the red light should go into your brain. Okay, square is not always an invariant transformation. We can actually gain some extra solutions which we did not have before. So that's what's important. Solving more and more equations of different kinds, obviously, which I will try to put as exercises. That's exactly what will teach this accuracy and the correct approach to choosing the way how you fight your fight. That basically concludes this particular lecture. Try to solve as many equations as you can among those which you will find as exercises which follow this particular lecture. Well, thank you very much and good luck in solving these equations.