 Hi, I'm Zor. Welcome to Unisor Education. This is our second lecture, second problem actually about equations. The previous problem was dedicated primarily to invariant transformations. This one is slightly more difficult and again it's not really difficult, but it presents a little bit more insight into what invariant and non-invariant transformations really are. So in this particular problem we will use non-invariant transformations, which means there is always a danger of something. You have to be very careful what you are doing with non-invariant transformations and ultimately checking the solutions which you have is really mandatory. And another very careful consideration should be applied for every step where you are using non-invariant transformations to prevent losing certain solutions or in case you gain something that's not as bad because at the end checking can actually help you to concentrate only on the real solutions. But to lose solutions is very, very important danger and that should be always watched for. Okay, in this case I have as three different equations, I put it here and I will consider them separately. So let's do it one by one. Equation number one is 2 times square root of x plus 4 minus 10 equals 0. Okay, now the first invariant transformation is obvious. This is plus 10 basically. We convert any number into this number plus 10. Apply to both sides of the equation, we get 2 square root of x plus 4 equals 10. Transformation is invariant by the way, which means we have not lost anything, we have not gained anything. Solutions of this equation are exactly the same as solutions of this equation. Now, we can simplify it further by another invariant transformation between divide by 2 just to get rid of this 2. And we get square root of x plus 4 equals 5. Next transformation, again obviously we have to apply it, we square the number. Now, this transformation is not invariant. Why? Because from let's say 7 you go to 49. From minus 7 you also go to 49 by squaring it. Which means there is no one-to-one correspondence between argument and the function in this particular case. We have 2 different arguments, 7 and minus 7, which are transformed into the same number 49. So, this is not invariant transformation. But let's just think about what this transformation can actually do as far as its negative effect. You can gain more roots, more solutions in equations which are transformed than the ones which you had originally. So, right now if I apply this transformation I will get square root of x plus 4 squared will be x plus 4 equals 225. And in theory I could have gained some new solutions. So, the mandatory check should always be applied at the end of set of transformations if there are some non-invariant transformation which can introduce extra solutions to your original equations. Now, other solution to this equation is after transformation of minus 4 you get x equals to 21. Now, let's check it out, it must be checked since we have supplied in one particular case a non-invariant transformation. So, 21 plus 4 is 25, square root of 25 is 5. Again, it's not minus, 5 is only 5 because when we are writing something like this we always mean a positive value. Square root, if it doesn't say anything it means the positive value of whatever the number is under the root. So, it's 5 times 2, 10 minus 10, 0, check is fine, so that's the true solution. We use this transformation which in theory can introduce extra solutions which are not really solutions to original equations. But in this case, this danger actually passed and we didn't really gain any non-existing solutions. In this case the solutions we got is the one which we are supposed to get solving the original equation. Since we have not lost any solutions, this is one and only solution to the original equation. Okay, let's go to the next one. Next one is x plus 4 divided by 2 minus 10 equals to 0. Actually, this one is not much different from the previous one. The only difference is division by 2, not multiplication by 2. So, the first and the obvious transformation is, let me do it this way, plus 10. I will use a shorter notation. So, I will get square root of x plus 4 divided by 2 equals to 10 multiplied by 2. We get square root of x plus 4 equals to 20. Power of 2, x plus 4 equals 400 minus 4, x equals 396. Now, this is a non-invariant transformation. So, that's why checking is mandatory. 396 plus 4, 400 square root is 20. Again, only 20, not minus 20, even if minus 20 square root is 400, because this always means the positive value. Divide it by 2, 20 divided by 2 is 10 minus 10, 0, everything checks, solved, fine, no problem. And the third one is slightly more involved. It's square root of x plus 4 divided by 2 minus 10, x plus 1 equals to 0. Now, what's interesting about this particular equation is that obvious transformation which we would like to apply, which is plus 10 square root of x plus 1, contains an unknown variable x, which was not the case before. But let's analyze what we're doing. We are adding something, maybe unknown number. We are adding some unknown number to the equation we are transforming using this adding of unknown number. But if you remember, addition is always invariant transformation. So, we don't actually lose anything, we don't gain anything, it's an invariant transformation, even if it contains x unknown variable. Since it's just addition, and addition is always transformation t from y to y plus a is always invariant regardless of the a. So, in this particular case, a is equal to 10 square root of x plus 1. And since regardless of a, this is invariant transformation, we can apply this transformation without any fear. And we get square root of x plus 4 divided by 2 is equal to 10 square root of x plus 1. So, since we apply this transformation, the left part will be minus 10 square root of x plus 1 plus 10 square root of x plus 1, reduced to 0 and on the right we have this 10 square root of x plus 1. Next up is transformation, again invariant is times 2. So, I can do it this way, times 2. And I'm getting square root of x plus 4 equals to 100, oops, sorry, 20 square root of x plus 1. Next, obviously non-invariant transformation is a square. Okay, if we square both sides, we will have x plus 4 equals to 400 times x plus 1. Now, obviously we have to open these parenthesis to using distributive law of multiplication and addition, getting x plus 4 is equal to 400x plus 400. And now the solution is, we will just apply two different transformations. One is minus x and this is invariant transformation. And I will get, even if it contains unknown variable x, it's still invariant because it's just plain addition or subtraction. So, I will get 4 equals to 400x plus 400 minus x. Now we will subtract 400 from here. So, 400 disappears from the right side and will appear as minus 400 on the left. We'll have minus 396 is equal to 400x minus x399. From which equation is solved, it's minus 396, 399. And this is our solution. So, what's important now is the following. Among all the different transformations we have, there was one which is non-invariant. And it might have introduced additional solutions. So, we have to make, we have to proceed with checking, mandatory checking. And I perfectly understand that nobody likes checking, especially if numbers are as crazy as this one, minus 396, 399. But, again, we must do it. Otherwise, non-invariant transformations can actually do this service to us and introduce some additional solutions which do not really exist. All right. So, first let's calculate square root of x plus 4. Now, in this case, it's square root of minus 396 over 399 plus 4. Not that I like it quite frankly, but in any case, I don't think we have a choice. Equals to, that would be square root of 399 in the denominator. And in the numerator, it will be square root of 4 times 399 minus 396. 4 times 399, well, the best way to do it is, without the calculator, obviously, it's multiplied 4 by 400, which is just 1 more than 399 and we will get 1600. And we have basically increased it by 4, right? Since 399 is 1 less than 400 times 4, it will be 4 less than 1600. So, it will be 1600 minus 4 and minus 396 divided by square root of 399, which is equal to square root of 1600 minus 400, it's 1200 divided by 399. And to simplify this, since this is 400 times 3 and 400 can be extracted, the square root will be 20. 1200 is 400 times 3 and square root from the 400 will be 20 and from 3 is square root of 3, we don't have any simplification of this. Okay, so we got this expression for the first thing. And we got to divide it by 2, so it will be 10 divided by this. So, that's the left part of this. Now, let's talk about this one. I hope the square root of x plus 1 is equal to square root of 3 divided by square root of 399, then by 10 it will be this and everything is okay. So, let's check if that's true. That's true. Square root of x plus 1 is equal to square root of minus 396 300 plus 1. Oh, sorry, 399. 399. Okay, which is equal to square root of minus 396 plus 399 divided by square root of 399 equals to great. That's exactly what we need, because times 10 it will be exactly the same thing as this one and it will be reduced to zero. So, we have correctly solved our equation. Checking was fine and basically that concludes these two little examples. Now, I will definitely provide more examples as we will move along of different kinds of equations like linear equations, quadratic equations, etc. What is important in all the cases is to understand that solving equations actually means applying certain transformations to this equation and transformations can be either invariant or non-invariant. In case of invariant transformations like plus, minus, multiplication by non-equal to zero or division by non-equal to zero, we have no fear, no problem. If it's a non-invariant transformation, then we have a problem. We have to really think about what exactly we are doing and checking is mandatory. Sometimes non-invariant transformations can introduce new solutions like whenever we square the numbers or expressions with unknown variable. That's not such a big deal because checking at the end will actually provide us with the correct solutions. Select out all those which are extraneous. If, however, the transformation is of such a nature that we lose some particular solution, that actually should be very carefully watched for. And we will have a few examples in certain exercises and problems which will be presented on the website. So far, thank you and good luck.