 Hi, I'm Zor. Welcome to Unizor education. Today we will continue solving quadratic equations without using the formula, trying to simplify it to use the full squares and something like this. The only difference is we will do it this time in the domain of complex numbers. I have three problems here, three equations and two of them will have real solutions and the third one will have true complex solutions. Without much to do, let's just solve one after another. So the first one is this. Now, if you would like to solve quadratic equation without using the formula, the first thing which is usually recommended is get rid of this coefficient with x squared so you will have just plain x squared with a coefficient 1. So you divide everything by 9. So you will have x squared plus 12 over 9. You can actually reduce it by 3, it will be 4 over 3. x minus 4, 4 over 3 equals to 0. So divide by 9, everything, which is invariant transformation, and reduce in both cases by 3. Okay, now this piece is supposed to be part of x plus a squared. Now in this case, this is x squared plus 2ax plus a squared. So 2ax must be 4 third x, which means that a is equal to, if 2a is equal to 4 third, then a is equal to 2 third. So I will do this. That gives me x squared plus 4 third x, but it gives me something extra, which is the square of the last number, a squared, which is 4 ninths. So I will subtract 4 ninths, and now that's an invariant transformation of this piece into this. And now I still have minus 4 third, now it's equal to 0. Okay, now is it easier? Of course it is easier, because I have only once unknown x. Here I have it in two places, and they don't know what to do with this. In this case it's very easy. Just invariantly transform x with whatever expression to the left, and everything else, free member to the right, and you will have x plus 2 over 3 squared is equal to, 4 ninth plus 4 third. Okay, now what is this? Instead of 4 third, you can multiply by 3 both sides, it will be 12 over 3, 12 over 9, sorry. So 9 is common denominator, 12 over 9 plus 4 over 9, that's 16 over 9. So let me just put 16 ninths. Now, non-invariant but very carefully performed transformation is we can make a square root from, we can apply the square root to both sides of the equation. Now it's non-invariant, you know about that, so the correct way is to use the absolute value to have only the positive part. So the positive part is this, and the positive part of this is 4 third. Square root of 16 is 4, square root of 9 is 3, so that's 4 third. So this is the correct derivation from this, only the positive sides. Now from which we have obviously two solutions, X plus 2 over 3 is equal to 4 over 3, and X plus 2 over 3 is equal to minus 4 3. Now in both cases, the absolute value of X plus 2 third will be 4 third. Right? So this is the positive part, and this is the negative part. If you have minus here, it's equal to 4 third, I just transferred minus to the right side, multiply both sides equations by minus 1, and so I've got this. From which we have two rules, two solutions. Okay, what are these two solutions? One is if you subtract 2 third from both sides, it will be X equals to 2 third, and in this case it will be minus 2 third, minus 4 third, that's minus 6 third, which is minus 2, and X is equal to minus 2. Let me make a very small observation here. Let's go back to the original equation and reduce it by 9, so it will be X square plus, what was it, 4 third? X minus 4 third. This was an equation with 1 as a coefficient with X square. Now, here's an interesting thing. Multiply these two things, these two solutions. 2 third times minus 2, it will be minus 4 third, you see, minus 4 third. Add these two solutions, minus 2, 2 third, minus 2 is equal to 2 third, minus 6 third, it's equal to 4 third. Now, this is 4th minus 4 third. So, this coefficient is the same as this one except that the sign is opposite. So, let me just repeat it. The product of two solutions is equal to the free member of this equation, and the sum of two solutions is equal to the coefficient with X with an opposite sign. Just make an observation because we will have exactly the same situation in other cases as well. Alright, so that's the three solutions. Oh, we wanted to make a check, right? What was it? Minus 2 and 2 third, right? Minus 2 square is 4 times 9, 36, 12 times minus 2 is minus 24, 36 minus 24 minus 12, that's 0, good. And 2 third square will be 4, 9 times 9 is 4. 2 third times 12 will be 12 and 3, that would be 4 times 2, it's 8, and minus 12 also 0. Everything is fine, solution is correct, let's move on. Next equation, 25 X square plus 70 X plus 49 equals to 0. I hope the numbers don't scare you. Okay, again as usually we reduce by 25 to get 1 as a coefficient with X. So we will have X square plus 70, 25th can be reduced by 5. 70 is 5 times 14, so it will be 14 over 25 reduced by 5 will be 5X plus 49, 25. Alright, as usually remember this coefficient is double what we have to put here. So half of it will be 7th square. So that would give me X plus 2 times X times 7, so it's 14 over 5, X plus square of the free member here which is 49, 25th. So I have to subtract 49, 25th to make it invariant. And I still have my plus 49, 25th. Now what's interesting about this equation? Well obviously that's what's interesting and we don't have anything on the right side of the equation. We have X plus 7, 5th square equals to 0. And here we have only one solution obviously because if you extract the square root from both sides you will have 0 on the right. So X plus 7, 5, absolute value is equal to 0. But if it's equal to 0 it doesn't really matter whether it's an absolute value or not absolute value. It's still 0 plus 0 or minus 0 it's still the same 0. So that's the solution X equals to minus 7 fifths. Now is it only one solution? Well as I told you before it's actually supposed to be considered as a double solution. It's just two different solutions coincide with each other. It's a very convenient formulation in this particular case. Now let me return back to that little observation which I made in the first place. That the product of two solutions is supposed to be the free member of the equation. What if the coefficient is 1 here? Well product of two solutions, one solution is minus 7 fifths and another solution is also minus 7 fifths. So their product is 7 fifths times 7 fifths minus times minus so it's 4925th which is exactly this. And the sum of these two solutions should be equal to the second coefficient with an opposite sign. Well minus 7 fifths and minus 7 fifths again because it's a double solution as you see. That's why it's very convenient to consider it as a double solution. So minus 7 fifths and minus 7 fifths is minus 4 gene fifths which is opposite with an opposite sign that corresponds to the second coefficient. Alright so everything checks. Great let's move on. We have equation number 3 where as I promised the solutions will be non real. There will be a true complex numbers with this square root of minus 1 which is i. Alright so what's the equation? The first equation is 16x squared plus 24x plus 57 equals to 0. First step is usually we reduce by 16 just to get rid of the coefficient with x squared that's much easier. And here we have x we have 24 sixteenths which is reducible by 8 so it's 3 over 2x plus 57 16 is equal to 0. Now next thing is to make it a full square of something. x squared plus 3 to 3 second x it will be x plus half of this coefficient which is 3 over 4 squared. Now we satisfied x squared plus 3 4 times x times 2 which is 3 over 2x that's satisfied as well. Now this also has the square of this number which is supposed to be subtracted to make it a variant. So now we have transformed this piece into this piece and we still have 57 sixteenths. Alright x plus 3 4 squared equals 2. Okay 57 and this is minus 9 so it's 4 to 8 but it goes to the right. Now 4 to 8 divided by 16 by the way it's 3 so on the right I will have minus 3. Now you remember that square root of minus 1 is in the complex numbers is i. So when we will use the square root from both sides what we have to do is obviously the same thing. x plus 3 fourths x plus 3 fourths is equal to sorry square root of 3 i right. If I am using the square root of minus 3 square root of 3 is square root of 3 and minus 1 will be i. So that's in the area of complex numbers represents this particular solution. And in the area of complex numbers since we are just using the square root from both sides we obviously have to use the second one minus square root of 3 i. Because in both cases when you square this plus square root of 3 i squared will be minus 1. And minus square root of 3 i squared will be minus 3. In both cases they represent both solutions. And that's what actually gives us the final complex solutions of this equation. So let's write it down here. Let me do it this way that will be simpler. Output x is equal to minus 3 4 plus or minus. So I transport it 3 fourths to the right side of the equation by subtracting from both sides. Okay that's two solutions. Now before I move any further let's again check if that little observation which we made before holds in the area of complex numbers that the product of these two solutions is the free member of this equation and the sum of these two solutions is the second coefficient with a minus sign. Okay now the product. So we have to multiply them. Now you remember that you multiply i by i it will be square root it will be minus 1. i squared is minus 1 right? i squared is minus 1. Now keeping that in mind let's multiply these solutions. This times this will give you 9 sixteenths. This times this will give you 3 fourths and square root of 3. It will be plus 3 fourths square root of 3 i. Now this times this it's minus 3 fourths square root of 3. And then this times this first of all there is a minus here but then there is i squared which is minus 1. So it will be plus and square root times square root it will be just 23. And as you see i goes out as it should actually and we have 9 sixteenths plus 3. Well you can always write instead of 3 4G 8 sixteenths 9 sixteenths plus 4G 8 sixteenths that's 57 sixteenths. So we have that. So multiplication is fine. So product of these two solutions is exactly the free member. Now sum again i is completely reduced so we have minus 3 fourth and minus 3 fourth. So it's double 3 fourths so it will be 3 seconds but it's minus and this is plus. So I told you from the beginning that this is supposed to be with a negative sign with an opposite sign. So that holds as well. So as you see in this particular case we have two solutions in the area of complex numbers. It's a true complex numbers but still this little observation which we made before that the product is equal to the free member and the sum is equal to the second coefficient with a minus sign this still holds. Well basically I can very easily prove that that's supposed to be the case always and here is why. Let's consider our equation has two solutions. Now if x is equal to a and x is equal to b are two solutions of the equation x square plus what letters should we use let's say px plus q is equal to 0. So if these are two solutions of this equation it means that this polynomial is equal to the product of these two linear expressions. Only this will give you x equals a as a solution to equation when the whole thing is equal to 0 or x is equal to b will give you this. So if the product is equal to 0 it means one of these is equal to 0 or vice versa. But now let's open the parenthesis here and what do we see? This is equal to x square minus ax minus dx plus ab or x square minus a plus b x plus ab. As you see the product of these two is equal to the free member q. So q is equal to ab and some of these two solutions gives you the second coefficient but with an opposite sign. So p is equal to minus a plus b. Well what it means it means that sometimes you can even guess solutions to quadratic equation if the equation is really like simple enough. Just as an example let me just make an example of this. Now I am talking only about equations with one as a coefficient with x square. So let's just make a very simple equation. Now if I as a teacher I want to come up with an equation which has simple solutions. Let's say I want to have solutions x is equal to minus one and x is equal to two. How can I come up with an equation which has these roots? Well very simply it's x square then the sum of these two solutions is minus one and two it's one. So I have with an opposite sign x and the product is minus two. Okay this is the equation and I claim that this equation has these roots. Well let's try it. Two square it's four minus two minus two is zero correct. Minus one square one minus one and minus it will be plus one so one plus one minus two zero again. So you see just using the property that the sum of these is supposed to be equal to this coefficient with an opposite sign. Minus one plus two it's one opposite sign is minus one and the product is supposed to be the free member the third coefficient. Here you go but sometimes if you just look at this equation you might consider okay somebody gave you this equation just maybe to check your skills or whatever and it's supposed to be a simple equation. The solution is supposed to be some small number so if the product of these two solutions is supposed to be minus two then what choice do we have? If you have plus minus two plus minus one you can just check one of these solutions and you will get exactly what you need. And if you get one of them immediately you get another. So that's basically one of the interesting applications of this particular property of the solutions of this quadratic equation with a coefficient one with x square. Well that would be it for today. We will probably have some more problems less trivial ones and well good luck and thanks for listening to me. Don't forget to check Unisor.com where you will have much more lecture and material and lots of other things. Thank you very much.