 Hi, I'm Zor. Welcome to Unizor Education. Today, we will consider the third problem in the general topic of equations. It's not any specific to linear equations or quadratic equations. This is still about invariant and non-imperial transformations. So, the first two problems were a little simpler. The first one was actually dedicated to only invariant transformations, which is the simplest case. In the second problem, we have introduced certain non-imperial transformations which will, in theory, introduce some extra solutions. So, we have to make a checking at the end just to make sure that we didn't really gain any new solutions. The dangerous part is the one which is today's topic. It's when transformation might lose certain solutions. It's more dangerous than the previous case when we gain some solutions, which are not real solutions, because checking doesn't really help much. If you lose a solution, let's say equation had two different solutions. You lost one during the transformations and you still have the second one. So, the checking will define a problem because it is a solution, but you lost one. That's what is real danger and you have to be very, very careful when you're making transformations not to lose solutions. These three examples, which I have right now, are exactly about this particular topic. So, let's consider the first, which is x square minus 16 equals to zero. Okay, how can we solve this equation? Well, very first transformation is obviously plus 16 and that results in x square equals to 16. The second transformation is square root. Now, that's where the danger comes. Square root is not an invariant transformation. Primarily because, as I was saying before, two different numbers, positive and negative, being squared, will result in the same number. So, there is no reverse transformation. There's no one-to-one correspondence. Usually, when we are using the square root in algebra, we mean the positive value of square root. So, the square root of 16, we actually mean four, not minus four. But if you look carefully, if x is equal to four, if you just square root both sides using just a plain algebraic square root function, you will get that. And that's the solution. And obviously, you have lost another solution, which is x equals to minus four. So, the point which I wanted to make is when you are extracting the square root from some number, especially the number which contains an unknown variable, you really have to understand and you have to make sure that you consider all the cases. You might lose the negative part. So, whenever you are extracting a square root from both sides of the equation, you really have to write two different equations. Namely, you have to divide actually your domain, this is your domain, where x is defined in two pieces. For positive or zero, x and for negative x. And therefore, you have to put, for the positive x, you have to put x is equal to four, or x greater than or equal to zero. And for negative x, you have to put for negative x equals to four. Or rather, we'll do it this way. That's easier. So, we have split our domain in two pieces and basically solve our equation separately in one piece versus another. So, whenever you are extracting square root, this is always a necessary procedure. Otherwise, you might lose one of the solutions. In this case, solution x is equal to minus four. And again, why? Because square root is not an invariant transformation. Okay. So, whenever you are dealing with a square root, you always have to deal with positive and negative sides of it. Let's go to the next equation. Complex square minus 16 divided by x plus four squared equals to zero. Okay. Now, what transformation should be applied in this particular case? Well, obvious choice is, well, let's just apply both sides by x plus four squared. Just to get rid of this on the left, zero will give zero anyway. And so, what we have as a result, we will have x squared minus 16 on the left equal to zero on the right. Well, is it right? Well, not exactly. Why? Because this transformation of multiplication by something is not always invariant. As you understand, multiplication by some number is invariant only if that number is not equal to zero. Right? Which means that we really have to consider separately a case when we are multiplying by zero. Now, when is this particular expression is equal to zero? Well, it's equal to zero when x plus four squared equals to zero or x plus four is equal to zero or x is equal to minus four. So, this is a separate case. Now, let's go back to this equation which we got after the transformation. As before stated, we had two different solutions x equals four and x equals to minus four. Because both of them being squared will give 16. Well, again, we can do this only if x is not equal to minus four because then this particular multiplier becomes zero. And obviously, in this particular case, if x is equal to minus four, you have a denominator equal to zero, which is definitely the bad thing. So, x minus four cannot be actually the solution of this problem. You have to very, very carefully consider cases when you have something in the denominator and you just multiply by that denominator without thinking about what's the consequences of this. Now, the consequence of this is that something which is not really a solution, x minus four cannot be a solution because it's in denominator here, becomes a solution all of a sudden when you get rid of this denominator. So, that's one more point to consider. Whenever you have something in the denominator immediately, the first thing we should do is to restrict your values of unknown to those when denominator is not equal to zero because the domain of this function is definitely this, x not equal to minus four. Therefore, solutions which should belong to domain of this function, obviously, should be restricted as well. This is not a solution. Okay, got away with the second example. Let's go to the third one, x square minus one times x minus three plus x plus one equals to zero. Well, for those who like to do it straight and multiply, you will get the equation of the third degree, the third power, x to the third power, you see. Now, the equations of power of three are not easily solvable. Although some time ago, the mathematician with the name Cardano, I think, Italian mathematician, he really reduced a general formula for the rules of the equation of the power of three. But we're not talking about this. The formula is that big, so we don't really need that. Well, what can we say about solving this particular equation? Well, it's basically quite easily. Let's do it this way. This x square minus one can be represented as x minus one times x plus one. Well, indeed, actually, a square minus g square is always a minus b times a plus b. Because if you will multiply this, a times a will be a square, a times b will be plus a b, minus b times a will be minus b a, and minus b times b will be minus b square. And this is reduced, so you have a square minus b square. So this formula is very easy, and it should be actually remembered by Cardano quite frankly. It's a very convenient one. So using this formula with a is x and b is one, I can represent it this way. So next let's continue. I have x minus three plus, and let me put x plus one in parenthesis for better visibility. Now, look at this. Now, you remember there is a derivative law of multiplication relative to addition, which means we can actually have x plus one outside of the parenthesis. And in the parenthesis, I will have what's left, x minus one and x minus three in this case. And in this case, it's equals to zero, right? So far, I did not do any transformations in variant or non-variant. I'm just changing the formula on the left, changing this particular expression to get this. Okay, now we can do the transformation. What's the transformation? Divide by x plus one. You see, both sides can be divided by x plus one. If you divide the left part by x plus one, you will get rid of this one and the zero divided by x plus one will give zero, right? So without thinking much, we can write this equation. What's wrong about this picture? The wrong about this picture is that when we divide it by x plus one, we actually made a non-variant transformation. Division by something is invariant only that something is not equal to zero. So again, we have to specifically consider the case when x plus one is equal to zero, so x is equal to minus one. Now, specifically considering this case, we can see the following. If you put x equals to minus one to this equation or this or this or this, it doesn't really matter because they're all equivalent. Obviously, we'll have this equal to zero. Doesn't matter what the value of this guy is because the multiplied by zero will be zero, so x minus one is a solution, which we lost when we had transformed this equation into this. So if we forget about x is equal to minus one, we basically lost the solution because this one has completely different solutions. So x minus one is one solution which we have lost if we just would have lost if we didn't really pay attention to this. But since we're smart, we did pay attention that x plus one should not equal to zero to make this the transformation invariant and x is equal to minus one is a separate case which we have to basically consider separately. Again, it's like if we will split our domain where this formula is defined, we split it into two pieces, not equal to others, the pieces. One is x equals to minus one and then all other x's. So x is equal to minus one is a separate case and it is a solution. We have found one solution. Now, considering x is not equal to minus one, now we can reduce it by x plus one. We can transform divide it by x plus one, getting this one. And now let's solve this equation. Well, let's just open all the parentheses and we'll see that this is x squared minus x and minus three x is minus four x plus three and plus one is equal to zero which is x squared minus four x plus four is equal to zero and this is x plus two square is equal to zero. x plus two square is equal to x squared minus four x plus four. We can check that. Now, the next transformation we can apply is square root. If we apply the square root, we again have to be very careful. In this case, however, it's easier because on the right we have zero. So plus zero or minus zero, it doesn't really matter. It's still zero. So the solution to this is x plus two equals to zero. So x is equal to minus two. That's the second solution. So there are two different solutions to these equations. Minus one, which we would have lost if we did not pay attention to the fact that we are dividing by something which contains an unknown variable x and the minus two, which is the result of the equation which is left after the transformation. Well, I just wanted to make sure that certain dangerous parts of invariant and non-invariant transformations are covered. This is still an introduction to equations in general. And again, my point was, yes, transformations can be applied, but if you are applying a non-invariant transformation, which is not multiplication by two or doing something or adding two or whatever, some very trivial transformation. So for all other transformations, you really have to be very, very careful. There are no, how should I say it, like one recipe which fits all the situations. You just have to understand that every action which you do, every transformation which you make with an equation, in case it's non-invariant and most of the transformations are actually non-invariant, you just have to be careful. When you divide by something, make sure you divide by not equal to zero or consider this member being equal to zero as a special case, same thing when you multiply it. If you apply the square root, for instance, again, consider positive and negative parts, there are many different special cases and these are only a couple which I wanted to cover in the introduction. There might be more. Well, obviously because there are more functions. There are exponential functions, logarithmic functions, trigonometric functions. Every function has its specifics and that's why you have to be very careful when you're applying any kind of transformation to equation. You should not lose solutions, you should not gain solutions. If you feel that this transformation might gain something, obviously checking at the very end would help. If you feel that something might lose the solution, if some transformation, then you just have to separate this case and consider it completely individually what happens if that particular thing is equal to zero or something. Well, that's it for today. Thank you very much. Good luck.