 Okay, so homogenization is the concept which is useful in solving a lot of locus questions okay. Now what is this concept? It says that if there is a second degree equation, if there is a second degree equation which actually can represent either a conic or a pair of straight lines, okay. So let's say there is a conic, I am just drawing a rough structure of it and there is a straight line Lx plus My plus n equal to 0. So this is a conic or a pair of straight lines, so this may be a conic or it could be a pair of straight lines and we have Lx plus My plus n equal to 0 as another straight line, okay. And these two intersect, okay something like this. So this yellow line is your Lx plus My plus n equal to 0, okay. So let's say this point of intersection is p and this point of intersection is q, okay. And let's say there is origin over here, 0 comma 0 and from this origin I draw a pair of straight lines connecting origin to p and origin to q, okay. So let me explain the event once again for the people who joined in late. So now I am dealing with the concept of homogenization, okay. I will tell you why we call it as homogenization. In this concept we are dealing with a second degree equation which may represent a pair of straight lines or it may represent a conic, okay. And there is another line Lx plus My plus n equal to 0. So what is this line doing? This line is cutting this conic, okay and is cutting it at p and q, is that fine? Now there is an origin over here. From this origin I am drawing two lines, let me name it L1 and L2. Connecting origin to p and q, p being the p and q being the points of intersection of the line Lx plus My plus n equal to 0 and this conic, okay. Now this L1 and L2 combined equation is obtained by homogenizing the equation of this conic with the help of this line. Let me repeat what I said. The combined equation, let me write it also, the combined equation of L1 and L2 is obtained by homogenizing the equation of the conic with the help of this line. And how is that done? All of you please pay attention to that, okay. So those who have joined in, please settle down, okay. So now I am discussing how to find out the equation of a scenario where there is a conic cut by a line and from the origin you are joining the points of intersection. So I am trying to find out the combined equation of L1 and L2, okay. Good afternoon to everyone who has joined in. There is still more to join in, I don't know why they are delaying it. You should never miss the initial part of the class, okay. Now the combined equation is obtained by homogenizing. Now why do I need to homogenize first of all? Now go back to your understanding of the previous class. A pair of straight lines which is passing through the origin is a homogeneous second-degree equation. Isn't it? Yeah? Do you recall that? A pair of straight lines passing through the origin has to be a homogeneous second-degree equation. Correct? That's why I have to homogenize this conic with the help of this line. How do we do that? For that step number one. So these are the steps that we follow. Step number one, we first write down this line equation like this, okay. Then we divide it by minus n throughout and we obtain this expression equal to 1, okay. Step number two. In step number two, we write down the conic equation, okay, in this manner. That is the second degree terms we do not do anything with it. While the first degree terms, let's say 2gx is a first degree term. So we'll put 2gx into 1. Again, 2fi is also a first degree term. So I'll put 2fi into 1 and c is a zero degree term. So I'll put into 1 square. Now you soon realize what will I do with these ones which I'm writing? What I'll do next is I will use this information. Let's say I call this 1 in the second equation, okay. So I will use 1 in 2, okay. So the second degree terms, they will be unaffected because in the process of homogenization, I'm converting everything to a second degree term. So this linear term 2gx into 1, 1 will be written as lx plus my by minus n. Get this clear, guys? This one has to be replaced with this expression lx plus my by minus n, okay. Similarly, this one also will be replaced with lx plus my by minus n, okay. And c into 1 square. 1 square will be basically the square of this expression. So lx plus my by minus n the whole square and equal to 0 will remain equal to 0. Okay, now you can see here that I have converted this to a homogeneous second-degree equation. To a homogeneous second-degree equation. And this will be actually we representing our pair of straight lines l1 and l2 equation. Is that fine? Now people who joined in late, first of all, good afternoon. Say something here. Okay. So how is the time going on in your house? I mean, I use studying hard or just sleeping off. Nafal lives at ending classes. Okay, what about R&R? Something is going on. I think Dhan told me that computer. Okay. Thanks to the NPS administration that they're doing so much for you guys. Okay. So we'll discuss this concept once again for the people who joined in late. First of all, you should not join in late. It's very important. You can leave early. That's fine. Don't don't join in late. Okay, the concept here is the concept of homogenization where we are going to learn how to find out the equation of the pair of straight lines connecting origin to the intersection of a conic and a line. So basically you can see this is a conic. Let me write it here. Conic. And this is a straight line. LX plus MY plus N equal to 0. So conic equation is given here. Straight line equation is given here. Okay. We are looking for the combined equation of the pair of straight lines L1 and L2. For that, what process we do, we do the process of homogenization. In homogenization, the first step is we write the equation of a line in this way. Okay. Divide by minus one. We get this expression. Then what do we do? We take the equation of the conic and whatever are the second degree terms, we don't disturb them. So this remains as it is. This remains as it is. This remains as it is. And in 2GX term, we put an extra one over here. Okay. Similarly, in 2FY term, we put an extra one over here. And in the constant term, we put C into one square. Okay. Now what we do, we replace these ones with this expression LX plus MY by minus F. So wherever I have written one, I am going to replace it with LX plus MY by minus F. And now I claim that this newly formed equation that you have here is the equation of L1 and L2 combined. Now how do we justify this? Why is this working? First of all, if it is a homogeneous, it must pass through origin. Correct? So it is taking care of the fact that whatever equation I have is passing through origin. Okay. That is number one. That's a necessary thing that we had to satisfy. Number two is if you see P point, P point is a point. Let's say I call this P point as alpha beta point. This P point simultaneously satisfies this line equation as well as this conic equation. Isn't it? Yes or no? So you will see that this P point is going to be satisfied even by this particular equation. Okay. So even this equation, which I am showing with a double tick mark, that will be satisfied by alpha beta. Let's check. So this will become A alpha square. This will become 2H alpha beta. This will become B beta square. This will become 2G alpha. And this, let me tell you since alpha beta also lies on the line, this will still remain a 1 for us. Okay. And again 2F beta. This will again be a 1 for us. And this will be C into 1 square. Now please note alpha beta will also satisfy conic and what you have is the conic equation, which will actually become 0. Same will be true even for the point 2. So what I have done, with one arrow, I have shot two birds. Okay. In Hindi we say it. So I have not only converted it to a homogeneous equation, I have also made it satisfy the line as well as the conic. And it is also passing through origin. That means it can only represent a pair of straight lines which is basically connecting O2P and O2Q. Is that fine? Any question regarding this? Please let me know. This is a very useful concept for solving locus questions. So I'll write it down over here. Useful for locus questions. Yes sir. But we don't know A, H and B, right? Like A, H, B and all. We don't know those. Sir it will be given. Yeah. These two will be given to you. Don't worry. This will be given. This will be given. Okay. Fine. And then you will be given, ask the condition. Is that fine? Any question here? Anybody? Yeah, sure Aditya. Don't waste time copying it. The notes will be shared with you. Don't worry. R it says 1 in the step 1. Sorry? R it says 1 in the step 1. How is the R it says 1 here? Yeah. No. Lx plus n by minus n is equal to 1. One thing you tell me. If you divide both sides by minus n. What would you write in there? Yeah. Yes. Okay. All right. So let's ask a question to understand this in a better way. Sir. Okay. Sir. Yes sir. Tell me. Sir, can the Koenig equation also be a better straight line? Yes. Can be. It can be. It can be. Just a second. Pair of straight lines. We are here. Okay. So let's have a very, very simple question over here. Okay. I hope you can read this question. The question says find the condition that the pair of straight lines joining origin to the intersection of this line and the circle may be at right angles. Okay. So the situation goes like this. I'm just drawing it roughly for people to follow. So this is a circle. Okay. Having a center at origin. Okay. And there is a line y equal to mx plus c. So let me show that with a blue line over here. Okay. So this line is y is equal to mx plus c. And this is your circle, which is actually a Koenig. Now the condition they're asking is what is the condition that a pair of straight line joining origin. This is your origin with these two points. Okay. So let me just draw them. What is the condition that this becomes a right angle over here? That's the diameter. I'm sorry. No, sir. Try this out. Try to see how you can fit in the homogenization concept. Very easy. Let's have two minutes for this case, not more than that. Let me know when you're ready with the condition m is equal to plus minus one. Okay. So any line with a slope of either one or plus one will make a 90 degree at the center. Are you sure, Shamik? Any line I made that will separate the 90 degree. They can be infinitely many lines with slope of plus minus one. Yeah. That seems to be correct. Okay. Let's see what others have to say. Okay. That seems to be correct. I'm not commenting whether it's actually correct or not. Y is equal to plus minus one. Why would a condition involved... Okay. And I was also saying m is equal to plus minus one. Okay. Again, the same mistake which I think Shamik is also making. Can we discuss this quickly? Anybody else who wants to respond? Trippan, Ajay, Ananya, Anjali, Arimaan. Anybody? One sometime. Okay. I'll give you one more minute. Okay. So Trippan also is ready with the condition. Okay, Trippan. Okay. Okay. Time to discuss. Fine. Ananya. All of you, please. Okay. Now those of you who said m is equal to plus minus one. Can you unmute yourself and tell me how did you get that condition? Sorry. I made a mistake, sir. Same, sir. I forgot. People who have said m equal to plus minus one. Okay. They take back their answers. Okay. Fine. Now, coming to the answer of Shristi and Trippan. Okay. Good. Let me discuss it. One of you is correct, by the way. So now what is happening here is it is the concept of homogenization. Isn't it? So first of all, if I say, get me the pair of straight lines L1, L2 equation. So what I'm going to do is the second degree terms I'm not going to disturb. Okay. Whereas this a square, I will write it as a square one square. Remember, a square is a constant. It is like your C of that general second degree equation of a corner. What is m over here? All of you see here, m will be obtained by doing this step. Okay. First you take the mx on this side and then divide by a C. So this is your one. Okay. This one you have to replace over here, my dear. Getting my point. So this one y minus mx by C, you have to replace over here. And when you do that, let's see what comes out from this exercise. So it gives you y minus mx by C the whole square. Getting my point. Right. Let's try to simplify this even further and express it as a homogeneous second degree equation. So please allow me to take the C square on the other side. Okay. Something like this. Right. Now let me bring x square to one side. If I bring x square to one side, I get a C square from here and I would get a minus a square m square from here. Okay. Now let me take y square on the other side and there's already a C square and from here I would get a minus a square. Okay. Now there would be an xy term also. Okay. With some coefficient, but I don't care about it. Can somebody tell me why? Okay. Suddenly I'm saying I don't care about this. I don't care about this. Why? Can you tell me why? Sir, because for the perpendicular condition only a plus b should be equal to C. Absolutely. Very, very good. See for the perpendicular, now what I'm claiming here is that these two pair of straight lines represents perpendicular lines. And when can that happen? When can a pair of equation, let me write it like this. When a pair of equation ax square, ax square plus by square plus 2hxy equal to 0 represent a pair of straight lines when a plus b is 0. Remember this. This is the condition for perpendicularity of the two component lines. So what I'm going to do here, I just need to do this and I'm done. This should be 0. That is the coefficient of x square, coefficient of y square will add up to give me 0. So that will give me 2c square is equal to a square 1 plus m square and this is the condition that I desire. So this is going to be my answer. Any question, boys and girls? Please stop me and ask. So I think some silly mistake had happened. Yeah, yeah. She has corrected. Ashish, now it is clear where you went wrong. Shankin. Oh yes, sir. Shankin, I understand your question. See, my aim is to make a second degree term everywhere. Now there are so many ways you can write one. I understand it. But does this expression of yours help me to make a second degree term? Tell me. No, sir. No. That is why this expression of one is desired. This expression of one is desired. Okay, sir. Any more? Okay. Let's have one more question. Let's be confident about it because we are going to use this concept. I think in one of the doubts, Shomik asked about homogenization. Okay, so you will get the answers over here, Shomik. Don't worry. So let's have another question. Let's take this one. Now, try to solve this by homogenization. Okay, there may be other ways also to solve it. But see whether you can apply homogenization to solve it. Read the question. If you know that the pair of lines joining origin to the intersection of the curve, this curve is an ellipse curve. We all know that. And this line are coincident. Okay. So the pair of lines are coincident if this condition is met. If this condition is met. Now, if you read this question, it is actually like saying that this line is a tangent to this ellipse. Isn't it? So you can solve it by condition of tangency also. You can't deny it. But let's use our homogenization, which we have learned just now. Let's have two minutes for this. I'm sure you can do it in two minutes. Just type done once you're done. Okay. So let me also pitch in. I can see some people writing done. Okay, Shristi is done. Very good. So see what is happening is that there is an ellipse. Okay. And there is a straight line. Okay. Of course, this is an ellipse whose center is at origin. Now, what is the question saying? The question is saying that if you connect this point over here. Okay. This angle over here should become zero because they are coincident. Isn't it? Okay. Now, we all know that when will this happen? When this will become a tangent at this point? Okay. So you could solve this question very easily by using the condition of tangency. Okay. So approach one. So I'll write it down. Method number one. Those who were with us last year would know what is the condition of y equal to mx. I'll write it down. Cot condition of tangency for the line y equal to mx plus C to be tangent to an ellipse. What is this condition? Can anybody unmute and tell me what is the condition? C square is equal to a square and square minus plus B square. Very good. This is the condition of tangency. If you use this, you can solve this question in one shot. Okay. So C here, what will be the C here? C will be minus n by m. Am I right? Yes or no? Okay. So C square is equal to a square. M will remain minus l by m. Okay. So the m of this line. So basically you have to convert this to y is equal to mx plus C form. So it will look like this. Y is equal to minus lx by m minus n by m. Okay. So this is playing the role of my m. This is playing the role of my C. Right? So I need this m and this C. Okay. Plus B square. Now multiply throughout with m square and your answer is ready. And your answer is ready. But this is not how I wanted to solve this question. So I will use another method that is method number two, which is the concept of homogenization. I mean, while any question with respect to how I solve this by using condition of tangency. No, right? So those who have not done this chapter, you need to look into my videos on Sentom channel. There are almost 42 lectures on phonics. So please have a look at them. Okay. We'll be also doing some concepts after our theory of equations and limits chapter. Now step, method number two, how do we solve this with homogenization? Now in this equation, will I disturb this term? No. Will I disturb this term? No. It's already second degree. Will I disturb one? Yes. I'll write one as one into one square. Correct? And what is this one going to be? This one is going to be your Lx plus My by minus n. Okay. So this one here, I would replace it with this. Okay. So I am homogenizing it. Okay. Very good. So that. So now if I do that, see what happens. It becomes x square by a square, y square by b square is equal to Lx plus My by minus n the whole square. So let me write it like this. This by n square n square. I'll take it to the other side. Okay. First, let me write n square here. Okay. Let's take the n square on the other side. So you have something like this. Okay. And on this side, you have L square, x square, m square, y square, and you'll have plus two L, M, x, y. Is that fine? Now let me convert it to a proper homogeneous form that we have known it like before. So this minus L square. Okay. And here y square n by b square minus m square and minus two L, M, x, y minus two L, M, x, y. Okay. Everybody is fine till this step. Now this is your a, this is your b and this term that is two L, M is basically acting as your two h term. Okay. Now what is the condition that the angle between the lines here is zero square minus a, b is equal to zero. Correct. So that means head square should be equal to a, b. Correct. So head square will be what? Head square will be L square, M square. Correct. a, b will be the product of this and this. Correct. Let me open the brackets. Let me open the brackets. I get L square, M square. Here I will get n four a square, b square. I'll also get minus M square, M square by a square minus L square, N square by b square plus L square, M square. Okay. So I'm just opening the brackets on the right-hand side. So this term and this term will get cancelled. Zero will be left off over here. Now what I'll do, I'll also drop off N square. So drop N square. If you drop N square, you'll end up getting one, sorry, zero is equal to n square by a square b square minus M square by a square minus L square by b square. Okay. Now let me simplify this further. Let me just take it to one side and multiply with a square b square. So L square b square plus M square by a square multiplied with a square b square will be your n square. So that gives you a square L square and M square b square or you can say b square m square equal to n square. Is this the condition we desired? Yes. This is the condition we desired. I understand this is a long route to do it, but I just wanted you to apply the homogenization concept. Okay. So multiple way of looking at the same thing. Are you confident guys? Type confident on your chat box if you're confident. Okay. Now there are certain cases where it doesn't need a line and a second degree curve to homogenize it. Okay. Homogenization can occur between two second degree terms also. Okay. How it happens, let me show you in a different type of a question. So let me just pick up a different question altogether. Right. Let's have this question. Everybody read this question carefully. Show that the straight lines joining origin to the point of intersection of these two curves are at right angles if this condition is met. Now see here for a change both are second degree equations. Isn't it? Okay. How will you solve this kind of a question? Let us try to solve this. First of all I would give you around one minute to try. Let's see if anybody is able to get this and write done once you're done. Only two people were confident. Okay. Never mind. It'll take me a few more questions. Just give it a try. I'm not expecting you to solve this question guys. Just give it a try. If you have to solve this question anyhow whether homogenization or any other process how will you solve this question? Anybody any success? Now see here. Ultimately I have to get the equation of the lines joining the intersection of these two curves with the origin. Okay Aditya. For Aditya I'm going to give one minute. Okay Aditya. Oh Anjali also wants one minute. Fine. I'll wait for one minute. Yeah. Are we good to go guys? Anjali. Aditya. Okay. Thank you dear. Okay. So let's see. See what is my aim? My aim again I'm reiterating I need a homogeneous second degree equation boss. That's it. This is my requirement. Now by hook or by crook whatever I want that equation to come from these two equations. Okay. Now what I'll do is they already have second degree. So what I'll do I'll linearly connect them. Okay. I'll just like you form a family of curves. Right. I'll connect them as one plus lambda times the other. Okay. All of you please see what is lambda. I will tell you what is lambda in a while. Don't worry. Okay. Now I'm claiming that the second degree term that I would get would be a family member of these two second degree curves. Now what lambda I need. I need such a lambda. I require such a lambda that makes the first degree terms vanish. Remember this should be a homogeneous second degree equation. That means there should be no term containing X. There should be no term containing X. That is my aim. Very good. Ayush. Okay. Excellent. So now what I'll do is the second degree terms. You can just collect it like this a lambda a dash X square. Okay. A lambda a dash X square. You'll have a XY in fact to XY H plus lambda H dash and you will have B plus lambda B dash Y square. Okay. So that from that you'll have 2 X G minus lambda sorry plus lambda G dash. Correct. So if you want to see a homogeneous second degree term, tell me boys and girls should this term be present? Should this term be present? No, right? This term should not have been present. Right? Only these three should have been present. So if you don't want these to be present and it can only happen in the way where you have this term as zero. So a lambda switch be should be actually such a value which will make this as zero which indirectly gives you lambda value as minus G by G dash. Are you getting my point? If you choose your lambda like this, you'll end up getting a homogeneous equation which will be a pair of straight lines passing through the origin which will be a homogeneous second degree equation. Are you getting my point? This is a very, very crucial step. Feel free to stop me if you have not understood why did I do this? All right. So let's complete the question. So now if you complete the question, you have already eliminated the last term. So you can put back your value of lambda. So A and lambda is minus G by G dash A dash X square and two. Okay. By the way, I don't care about this term. Why don't I care about this term once again? I need the condition for them to be at right angles. So whatever is this X, Y term. Okay. I don't care. Okay. Don't waste time doing unnecessary things in comparative exams. You don't have that, you know, leeway of time to spend. Now Y square will be B minus G G dash into B dash into Y square equal to zero. Is that fine? So the condition of tangency, sir, can we equate both the initial equations? If you equate both the initial equations, you'll end up getting the two points where they intersect. But however we needed the equation of the pair of straight lines connecting those two points with the origin. Okay. All right. Now all of you, we have reached the final step. If this has to represent perpendicular lines, what can I comment about this plus this zero? Yes, I know. So A minus G by G dash A dash plus B minus G by G dash B dash should be equal to zero. Should be equal to zero. Correct. Which clearly gives me the final blow to this problem. A plus B equal to A dash A dash plus B dash, which is nothing but G dash A plus B is G times A dash B dash. Done. Proved. This is what we wanted. Any question? Boys and girls. Okay. I can see a few questions. Correct. Shashank A plus B should be zero. Okay. Okay. Before we close this topic, I want you to be still more confident. And one more question I'll give you before I end this and then start with the new chapter of the theory of equations. So this is the question for you all prove that I would have given this as an objective question also, but that would be prove that the angle between the lines joining the origin to the intersection of this straight line with this curve is this this type of question is very common in CT. Okay. Aditya has a question. Can we divide the two equations? Well done. See, just give me a second. Just give me a second. There is some question coming from Aditya. Guys, you could have also done the same problem. Let's say if you don't want to get into this Lambda business, you could do one more thing. You could multiply all of you PC here on the top, multiply this with a G dash, multiply this with a G and subtract it. You would automatically get rid of the first of first degree term. Getting the point. So this is also another way to do it. Actually, you're doing the same thing indirectly. Okay. So if you want your this term to go off, these terms to go off, another root would have been multiply the first entire equation with G dash multiply the second equation with a G, subtract it off. So these two terms will get cancelled and you'll be left with a pure second degree homogeneous equation. Okay. Aditya, does it solve your purpose? Let's do this. Can we have three minutes to do this? Oh, it was there in the homework question. Okay. Just type done once you're done. Yes, three minutes is over. Anybody with any success? So the process is well defined, you guys. You just have to follow it and avoid any silly mistakes. Okay. Let me do it along with you. Okay. Done in the homework. A little more time. Okay. Let's have one more minute. Aditya has done. Very good, Aditya. Okay. Very good. Many of you have done. Let's discuss this quickly without much waste and then we can start a new chapter theory of equations. Okay. So all of you please pay attention. What is the agenda over here? The agenda is there was a curve. Okay. And there is a line. I need to find out the angle between these two lines which you're making from the origin to P and Q. So obviously homogenization will help you out. Okay. So first of all, what I'll do, I'll write y is equal to 3x plus 2 as y minus 3x by 2 equal to 1. Okay. And then the second degree term, I'm going to write it like this. Okay. Instead of 4x, I'll write 4x into 1 and 1 will be replaced by y minus 3x by 2. Okay. Similarly, 8y into y minus 3x by 2 minus 11 into y minus 3x by 2, the whole square. Okay. Let's do a simplification of this. Let's do a simplification of this. So I end up getting x square, 2xy, 3y square. Okay. I'll end up getting from here 2xy minus 6x square. From here, I'll end up getting 4y square minus 12xy. From here, I'll end up getting minus 11 by 4y minus 3x the whole square equal to 0. Okay. Let's do one thing. Let's multiply, throw it with a 4 and then I'll open this bracket. So let me make those changes directly 4x square, 8xy. Okay. Let me group of terms first without before doing that. Let me group of whatever terms we have outside. So x square will be obtained from this guy and this guy, which is minus 5x square. Okay. I'm taking care of these two. Don't think that I'm cancelling out. I'm just grouping up terms. 2xy, 2xy into xy will be like minus 8xy. Taking care of these. Okay. And I'll have a 7y square. I'll have a 7y square minus 11 by 4y minus 3x the whole square. Okay. Now let me multiply with the 4 first. So minus 20x square minus 32xy plus 28y square minus 11. And let me open the bracket. Y square minus 6xy plus 9x square equal to 0. I've been hoping the bracket will give me minus 20x square minus 13. Now here 66 was there and minus 32 is here. So I'll end up getting how much? Plus 34. Plus 34xy. Okay. And 28 was there and this will give me 17y square. Correct. Is there anything which I missed out? Anybody please point out if I missed out on anything. So even the x square There was an x square term. Oh, I forgot that. This will change. This will become minus 190. Plus 190. Yeah, minus 190. This gives very ugly figure. Okay. Now let me divide toward with minus 17. That will give you 7x square minus 2xy minus y square. I divided by I divided toward with minus of 17. Okay. Now what is the angle between the two lines which comprises that pair of straight lines to under root of x square minus a b by modulus of a plus b. So that is to under root of x square h here is one x square is my one minus a b is going to be plus 7 by mod of a plus b. a plus b is 7 minus 1. Okay. So 2 root 8 by 6. That's nothing but 4 root 2 by 6. That's nothing but 2 root 2 by 3. So theta is going to be tan inverse of 2 root 2 by 3. Okay. Slightly calculation intensive but the approach was pretty much the same. Correct. So I'm sure after this you are very very confident about how to use homogenization. So thank you very much to this chapter and now I'm going to begin with a new one. Okay. And the chapter is theory. Let me show you my artistic skills theory of equations. Okay. Let me put some decoration also. Okay. Yeah. It's a completely new chapter. So welcome to the new chapter now. The chapter's name is theory of equations. Now actually this chapter is not a chapter at all. It is actually an ingredient. It's actually a tool which everybody must know because in maths you are solving problems everywhere. Correct. In fact in real life also you are solving problems. Corona right now is a problem. And getting the vaccination is like getting the root or getting the solution of that equation. So in maths you'll come across different types of equations. It may be trigonometric equation. It may be algebraic equation or polynomial equation. It may be equations involving modulus functions. It may be equations involving gif function, fractional part function. It may be differential equations. Right. So so many equations are there in maths. So we need to learn how to solve at least the basic ones. Trignometry you have already done guys. Correct. In the trigonometry chapter. So in this chapter I'll give you an overview what we are going to learn in this theory of equations chapter. This chapter actually is not only about equations. It's actually about solving. Sorry. Let me change the pen killer. It's about solving equations and in equations. So not only we'll be learning how to solve equations we'll be also learning how to solve in equations. In equation means inequality involving the variable. Okay. So the structure of this particular chapter is we are going to cover up how to solve first of all a polynomial equation. How to solve a polynomial equation as well as any equation. Okay. Under that a lot of emphasis will be given to quadratic. Okay. Because we never did quadratic officially in class 11. So now the time has come that we go we take a deep plunge into this topic quadratic equation and also we'll see some glimpses of how to solve higher degree polynomial equations. Okay. How to solve higher degree polynomial equations. Okay. Second thing that we're going to talk about how to solve equations which involve special functions. Equations which involve special functions. What are these special functions? Special functions are like your modulus function. Okay. They're like your greatest integer function. Yes. We are going to introduce something called least integer function also. The fraction part function. The fraction part function. Okay. So equations which involve them we are going to talk about those equations as well as any equations. Okay. And finally we are going to spend some time solving exponential and logarithmic equations and any equations. Okay. Now it seems as if there are only these topics to be dealt with but let me tell you this will minimum take two, two and a half class for us. So here one hour is already gone. So don't consider this class to be complete. So probably this class, next class and next to next class will also be required. Okay. It's a huge topic. Well, let's start waste time and move forward. The first thing that I would like you all to answer is what is the difference between what are the difference between a equation and an identity. I know this is a very silly question to ask but let me tell you people do make mistakes when questions related to them are asked in the comparative exams. Equation has a solution that means identity doesn't have a solution. Okay. Shashank is saying identity is true for any case. Okay. Okay in general. Equation is fixed value. Okay. Any other opinion? Identity is variable. Okay. Any other response? Identity has condition. Okay. Oh my god. Okay. Okay. Okay. Thank you guys for your response. Okay. Some of you are absolutely correct. So let me clear up this misunderstanding. Identity identity is something. It's a kind of something equated to something which is actually true for all values of the variable. For all values of the variable. Okay. That means any value provided the functions involved in that identity are defined for those values will satisfy an identity. A simple example of it is your Pythagorean identity cos square x plus sin square x. Okay. Now any value of x for which cos x and sin x is defined, we know it's defined everywhere, will actually satisfy this equation. So this will be satisfied for all values of x. This will be true for all real numbers. Okay. So when I say sin square x is 1 plus tan square x. Okay. This is going to be true for all values of x provided it doesn't make your tan x and sin square to become undefined. So basically you say excluding odd multiples of 5 by 2. Okay. And being some integer. Okay. So this is also an identity which is defined, which is true for all values of x provided they are defined. Okay. Whereas a equation whereas a equation is true only for certain values of x. Is true for certain values of x. Certain values of the variable. Okay. And these values basically are called the roots of that equation. Okay. Since I mentioned a trigonometric example over here I will continue with the same. Let's say cos x is equal to half. Okay. Is this equation is this an identity? No. Because it may not work for all angles. 0 will not work. Okay. 30 degree will not work. It is only satisfied for values like pi by 3. 2 pi 2 pi minus pi by 3. 2 pi plus pi by 3. So in general what do we write for this? 2 n pi plus minus pi by 3. Okay. So only these values of x will satisfy it. Okay. So these are called the roots of this equation. So they will be called as the roots of this equation. Okay. Now you must be wondering why is sir speaking about such a simple concept. We all know this but if you know this then I have a question directly for you and let me tell you many people have not solved this question which I am going to give you next correctly. Sorry. Where is that? Theory of equations. Okay. So let's have this question. Okay. Now what I will do is I will give you options. Okay. Option a. So the parameter can be so lambda can be option a. Option a. Minus 2. Option b. 2. Option c. 3. Option d. All the above. Okay. Read this question carefully. I am running a poll for this. Once you are done I want you people to respond by pressing a poll button. Let's have minutes. Time starts now. If this equation is satisfied by more than 2 values of x find the parameter lambda and these are the options. Minus 2. 2.3. All the above. Okay. Very good. Guys last 10 seconds to vote. 5. 4. 3. 2. 1. Go. Okay. I will stop the poll now. Look at the results. Look at the results. 36% Janta says d. Okay. Then 32% Janta says b. Okay. So Janta is going with option d. Okay. Let's check it out. Now if this is... If you look at this equation very carefully it actually is a quadratic. Right. Nobody can deny that it actually is a quadratic expression. Isn't it? This expression is a quadratic polynomial. Okay. And we all know something called fundamental theorem of fundamental theorem of algebra. Okay. What does it state? It states that any degree n polynomial any degree n polynomial equation polynomial equation can have or has exactly n roots. Okay. So if it is a polynomial equation then it will have exactly n roots. Okay. Now these roots may be real. These roots may be non real. But altogether if you add the roots they will always be as many number of roots as the degree of that polynomial equation. This is something which we call as the fundamental theorem of algebra. It's a very very basic theorem which you cannot violate. Okay. Now this is a quadratic equation and under no means I will have more than two roots for this. In fact I should have exactly two roots. Not less not more. But the question gives us a surprise that it has more than two values of x satisfying it. Which clearly indicates that this is actually not an equation it is actually an identity. This is something which you need to understand and appreciate. It is not an equation because had it been an equation it cannot violate the fundamental theorem of algebra and it will not have more or lesser number of roots than two. Okay. Now here comes the approach. If it is an identity then it is not only true for more than two values but it is true for any value of x. Okay. So if you want this to be true for any value of x it can only happen when these coefficients that you see. Okay. I am underlining these coefficients with black. Sorry with pink. Okay. These coefficients that you see they should all be individually be zero each. Because then only outcome will become independent of x that means left hand side and right hand side both will be zero every time. So you need to see for what value of lambda will these three equations simultaneously be satisfied. So you have to see they are simultaneously satisfied. Simultaneously satisfied. Okay. Now these are very easy to solve because they are factorizable. So the first one is factorizable as lambda minus 2, lambda minus 3 that gives you lambda value as 2 or 3. The second one gets factorized as lambda minus 2, lambda minus 1 so that gives you lambda value as 2 and 1. The third value again the third equation that is lambda minus 4 again gets factorized like this. So lambda is 2 or lambda is minus 2. Now which value is simultaneously satisfying all of them? 2. So your answer is lambda is 2. This is your answer. So janta didn't I tell you many people would make mistake janta was wrong. Option b is correct. So those 32% people who replied b they were correct. Guys you understood your mistake? Okay. Is that fine? Because 1, 3 etc will make few of them 0 not all of them 0. Is that fine? Any questions? Sorry Sorry Sir in the in the in the nth degree can have Sir if in complex equations we solve equations then they add infinitely. No if it is complex equations they also have the same number of equations. But sometimes then it was satisfied for like supposing some 2 degree equation is satisfied for more than 2. See it may have repeated roots. It doesn't mean that it has lesser roots. It may have repeated roots. Because j is minus 2x plus 1 equal to 0. 1 and 1. But the conditions will be satisfied. Yes sir not less. It had more than 2. I can't remember exactly. I don't know. No, were they polynomial equations or were they special functions equations? Sir Maybe they had modules. They must have special functions there. They are not polynomial equations. Fundamental theorem of algebra works for polynomial equations that's why I categorically wrote there. If there is a degree n polynomial so any polynomial you make out of complex numbers or whatever see complex numbers are just names they are just variables. It will have exactly the same number of fruits as a degree. But if there are special functions involved let's say you are taking modulus or you are taking mod or argument then probably the number of equations will become more or less. I understand your concern you are coming from one of the questions that we had solved in the class. They were not polynomial equations my dear. Okay. Now this is something which I would actually ask every year just to surprise people there is a quadratic actually that I have. Some of you would already be knowing this. Okay. This is the quadratic. Okay. All of you please look at the expressions carefully which I am writing so this is a quadratic equation that I am actually writing. Okay. This equal to what? So basically prime of AC it's a quadratic equation kind of thing. I am saying prime of AC because it appears. It's a quadratic equation. And of course A and B and C are not equal to each other because if they are the denominator will become 0 and the expression will be undefined. Okay. Now there is something very surprising about this equation. If you put X equal to A everywhere you would realize that this equation is satisfied by this. Check it out. If you put X as A this will become 0 this will become 0 and this will become a 1. So 1 is equal to 1 so A is satisfying so A is the root of this equation. Correct. Anybody trying to contest that? No? Okay. In the same way even B will satisfy it. Check it out. Even B will satisfy this. So B will make this as 0 this as 0 and this will become 1. So 1 is equal to 1. Okay. So that is getting satisfied. So B is also a root of this equation. Correct. You will see that it is going to satisfy it. Check it out. So C will make this as 0 this as 0 while this will become a 1. Correct. So 1 is equal to 1. Now isn't this surprising that this equation is having more than 2 roots. Right. It is violating the fundamental theorem of algebra. So actually speaking this is not a quadratic equation. What is it? Identity. It is an identity. I would request all of you to actually check this out at home. If you expand this you will realize I have written something like 0x square plus 0x plus 1 equal to 1. That is actually always true. Okay. So don't be you know fooled up by such kind of expressions which are claiming to be a quadratic but having more than 2 roots. If it is there straight away you have to declare it is an identity. Not only 3 it will have more than 3 it will have any value of X satisfying it. Is that fine? Okay. So as we discussed in our overview of this chapter we are going to talk about quadratic equations. So this is a giant chapter actually. It looks like a simple you know chapter which you have already done in class 10 to a certain extent. But this chapter is huge. Okay. So what is the quadratic equation? So any equation of this nature ax square plus bx plus c okay. Where your abc belongs to complex numbers okay. It can be complex also. But remember a should not be 0. It actually is a quadratic equation. Okay. But primarily we will be dealing with those quadratic equations where abc will be real. Okay. In some cases we will also talk about where abc are complex also or non real also. Okay. There are actually few types or categorizations that I have divided quadratic equation. There is something called pure quadratic. Pure quadratic equation. So what is a pure quadratic equation? It is an equation where b is 0. Okay. Obviously a should not be 0. But b is 0. C may or may not be 0. Okay. So example of it can be something like x square minus 6 equal to 0. This is an example of pure quadratic equation. When we are doing integration we will be learning some integration which involves pure quadratic. Okay. Then the second categorization is something which we call as affected. Adfected. This word means varying power. Okay. Adfected quadratic equation. In this equation basically your b should not be 0. And of course a should not be 0. C may or may not be 0. So a typical example would be something like x square plus 2x is equal to okay. This is an affected quadratic. Okay. Then there is a standard quadratic. Okay. Standard quadratic. Which we are going to talk about. In that basically we are not going to have your a as 0. Yes. Everything can be 0. So a typical example is like x square plus 3x plus 1. Okay. This is a standard quadratic equation. Okay. Now this is not important just for your know-how. No question will be asked which kind of quadratic is it. Okay. So don't get scared. Next important thing when we talk about quadratic is the roots of the quadratic equation. So I am going to spend quick amount of time on this because in 10th you have done it and think in the school also you would have done it to a certain extent. And you have been using it also again and again in different verticals. Roots of quadratic equation. How to find roots of quadratic equation? Basically we will be talking about the Shridhar Acharya Shridhar Acharya Formula. Or the quadratic equation formula. Okay. So before that I like to first start with a quadratic expression. Yes. Let me tell you this is a quadratic expression. It is not an equation. Many people say it's an equation because equal to is involved. This equal to was put only to show that this is plotted against the y-axis. See it's because you are plotting the values of ax square plus bx plus c on the y-axis that's why it was called as y. But ideally it is just an expression or you can say a polynomial in 2 degrees or degree 2. Okay. It is equation only when it is equated to some number. Okay. So let's say if I call it as an equation like this what does it mean? It means that you have equated this with y equal to 0. So if you equate these two graphs you end up getting this quadratic equation. So don't have a wrong notion about equation. Okay. And there is another wrong notion that I have seen. When I ask people what is the graph of quadratic equation? What say they will make a parabola? Let me tell you this is the graph of quadratic expression. It is this is the graph of y is equal to ax square plus bx plus c. This is not the graph of equation. The graph of equation doesn't need a y-axis. It is a single variable. It just needs the number line and the two roots of that. Okay. There are many people asking this question. So if somebody says draw the graph of x square minus 3x plus 2 equal to 0. Okay. Don't start drawing a parabola and all. It just means one and two values. So take a number line put a 1, put a 2. This is the graph. Okay. Anyways, my purpose was not to tell you things which is not required right now. But please remember most of you are attending KBPY again. By the way congratulations to Trippan. In the interview you may be tricked by asking these things. Okay. Now with this quadratic expression, I will be deriving the quadratic equation roots. How will I do that? All of you please pay attention. So first of all I will work with y is equal to ax square plus bx plus c. So I will take an a common okay. Just follow the procedure. Right. So basically I have taken a common from the first two terms. Then I will complete the square over here. Please note the Sridharacharya formula is actually an art of completing square. He did not get this formula by doing Kapasya and all. He actually just did completing the square. So it has not come from sky. It has come from a very simple act. And that is the act of completing the square. Okay. Now if you want to get the roots of this quadratic equation, please pay attention over here. This arrow. If you want to get root of this quadratic equation, you are trying to solve it or you are trying to see the intersection with y equal to 0. Intersection with y equal to 0 right. y equal to 0 is what? The x axis. Okay. So wherever this particular quadratic polynomial will need the x axis. Those are actually the roots of the quadratic equation which we normally refer as alpha and beta. Okay. Now see so y is 0. I'll put a 0 over here. Okay. So I'll put 0 is equal to a. Okay. And these terms that is the constant terms over here I will take it to the left hand side. So B square by 4. Let me take LCM also minus 4 AC is equal to a. I'll drop over here. A square. So it will become something like this. Okay. Now take the square root. When you take the square root it becomes x plus B by 2a is equal to plus or minus under root of B square minus 4 AC by 4 a square. Now normally in a quadratic equation, you can always choose your a to be positive. You can multiply it throughout with a minus 1. So I'm doing that. I'm assuming that I'm making my a positive. So I'll take it out of the root sign and make it 2a. Okay. Else it would have been 2 mod a. Okay. Anyways. So now I'll take the B by 2a to the other side and it becomes minus B plus minus under root B square minus 4 AC by 2a. Okay. This is your Shridhar Acharya formula that you had derived and see since a is in the denominator a should not be 0. That's why that condition which I had quoted in the beginning of the discussion that means a should not be 0. I think it is in the previous page. That is important. Okay. Now a couple of things here. So this will correspond to your 2 roots alpha and beta. So if you take a plus 1 root will come if you take a minus another root will come. Now there is a relationship between the roots and the coefficients which you already know. So if you add alpha and beta you'll get minus B by a and if you multiply them you'll get B by AC by a. I'm sure you already know this. Okay. There is another way to get to this. If you treat your quadratic to be factorized like this then you can expand and compare the coefficient. Now many people ask me sir why do you put an a over a? See if I don't put an a let me drop an a. Something wrong will occur. So if I drop an a what will happen see. So if you're saying this quadratic has 2 roots alpha and beta that means I could equate this to product of these 2 factors. Now when you expand it you end up getting something like this. Okay. Now you can't say that these 2 expressions are equal because this has an x square and this has an ax square. That's why the a over here is important. Okay. So don't forget that a. So now I'm going to get the same relation here by comparing the coefficients. So if you compare the coefficients let me write here compare the coefficients. See guys you already know this so I'm slightly fast here. Okay. But in case you want you know further discussion I can always stop and discuss more about it. So this gives you alpha plus beta as minus b by and if you compare the constant c is equal to the alpha beta so alpha beta will be c by okay. Now something important yes any question anybody anybody any question yeah please mute yourself. Yeah. Okay. Now guys I'm going to get the same thing that is the condition that you have sorry. Now I'm going to talk about nature of these roots. Nature of these roots. This is also something which is known to you so I'll be not spending a lot of time. So if you see the Shridharacharya formula the formula was minus b plus minus under root b square minus 4ac by 2a right. Okay. Now this term over here which you see under the root this term is very important. What is this term actually called this term is actually called the discriminant who said that it is not determinant my dear friend it is discriminant. Okay. Determinant is something else now the name discriminant itself means it discriminates discriminates between what it discriminates between the roots or discriminates between the nature of the roots. Okay. So this term b square minus 4ac is actually called the discriminant which we also write as d. Okay. Now from this formula it is very clear that if d is positive you will end up getting two values of alpha and beta which are real and distinct. So here alpha and beta would be two separate values we say real and distinct so alpha and beta would be real and distinct. Okay. If d is equal to 0 then alpha and beta would be real and equal and if d is less than 0 we say alpha beta would be non real in nature. Okay. Many books call it as imaginary but it is just written as non real in nature. Right. Yes. So now the same thing I will try to get it from the graph of the parabola. Okay. See how it comes. Now before that I would like all of you to answer this question. What happens to the graph when you vary your A? Anybody? What is the change in the shape of the parabola when you vary the A in the given quadratic expression? So it becomes thinner or fatter. Very good. Okay. Let me take you to GeoGebra students who have not installed GeoGebra yet my sincere request to you would be please do so after today's class. Dhyan says it widens or narrows. Okay. Now that's the very crude way of explaining it. I'll tell you actually what happens. All of you please focus over here. So Y is equal to AX square plus BX plus C. Okay. Now since I have not specified ABC GeoGebra will actually assign some random values to it. That means it will treat it as some parameter, something like this. You can see it. So automatically it has made ABC 111 but nevertheless it has given me an opportunity to vary A from minus 5 to 5, B also from minus 5 to 5, C also from minus 5 to 5. You can always change it. You can always change it. You can make it minus 10 to 10 also. That's your call. Okay. With a step off you can say 0. 0.4. Okay. Something like this. It's your call. Okay. Now guys, all of you have your eyes glued on this parabola. I'm varying my A. So from A equal to 1, so right now you can see my A value is 1. Okay. Check it out. This A value right now is 1. Okay. Now I'm going to change this value. I'm going to increase it. Okay. What do you think will happen? Graph will become thinner or fatter? Thinner or fatter? Thinner or fatter? Thinner or fatter? Thinner. Okay. Let's check it out. Correct. Thinner. So I'm increasing the value. See it's becoming thinner. Okay. As you go lesser it becomes fatter or it starts expanding. Okay. The moment you become 0, you can see it becomes a straight line. That means the quadratic loses its parabolic structure and it becomes a straight line. Okay. Now what do you think will happen when I take A as negative? It will better the other way round. See that. Okay. So from 0 if I go negative, see it has started bending the other way round. Okay. Now this is something which we call as the change in the concavity of the quadratic expression. So A decides the concavity of the parabola. So those who are saying fatter and thinner, this is the technical word. What is the technical word? Concavity. So if it is up, let's say like this. If it is up, it is said to be concave upwards. This is called concave upwards. Okay. Later on we'll learn this in calculus that if the graph of a function, let's say there's a function like this. So if a double derivative of this function is always positive then this graph will be concave upwards. Okay. I don't worry about it right now. We'll talk about this when we deal with the chapters, application of derivatives. Okay. So as you can see when you double differentiate when you double differentiate this quadratic expression, what do you end up getting? See, single differentiation would give you single differentiation will give you 2ax plus b. Double derivative will give you 2a, correct? And this will always be positive if your a is positive, right? So if your a is positive for all values of x belonging to the domain of this function that means the graph will open upwards. In a similar way a function will concave downwards. Okay. Concave downwards if the double derivative of the function is always negative for the entire value of the domain, correct? So if this a becomes negative then this value will be less than 0 and that is what makes the graph become Ulta. Is that fine? Is that fine? Okay. How does b change this graph? Anybody? Or let me ask this question very specifically. You know this point? What is this point called? Vertex. Okay. Vertex. Okay. Now how would this vertex change if I change the b value? Okay. I want everybody to give their opinion by typing it up, down, left, right or it will do some kind of a particular motion. Anybody? Let's see what is your option. Up, down by Ananya. Aditya also says up, down. Okay. Aditya also says up, down, left, right. Up, down, left, right. Left, right. Left, right. Okay. Alright. So guys see what is going to happen. First of all I will locate this point. By the way there is a special function available in GeoGebra which tells you the vertex of the conic. Okay. So I can always locate this vertex. Okay. Now I am going to trace the locus of this point. So I am going to run short trace. Okay. And I am going to vary b. What do you see? People are saying up, down. You are partly correct. People are saying right, left. You are also partly correct. It actually traces the parabola itself. See this. If you want I can increase this interval and show you further. 15 to 15. Okay. Let's say 0.2. Okay. Let me vary it for the more. Okay. So it will again become a parabola only. Okay. Now this is a question that I would like to give you. If there is a parabola let's say this parabola. I take this parabola 1.6. Okay. Let me take some nice value. Okay. 2x square. 2x square plus bx okay. Plus 3. Okay. This is a quadratic. Fine. If I vary this b if I vary this b that means b is the parameter. Okay. What is the locus of sorry I am making a locus question here. Okay. What is the locus of the vertex? Okay. This is a question for you which is a homework. Note this down. You will send this respond to me. Send the answer to me before you sleep tonight. I think class will end up at 8 o'clock. So before you sleep you have to send me for this parabola if I vary the b value what will be the locus of the vertex. Okay. I will not answer this till I have received response from you guys. You have done so many locus questions. Okay. Good. Next question to you all is how would this vertex change or how would this parabola change if I vary the c value? Now please put your response. How will it change when you vary the c value? Up down or left right or again a parabolic motion? Up down. This time up down. Okay. You can see it happening now. See the point a. You see that? Okay. So now vertex is changing a straight line. Okay. Now add one more question. If you see is vary right? What is the path traced by what is the equation of that straight line that you see? Okay. Now anyways coming back to my previous discussion. So I want all of you to tell me if this is the parabola or this is the graph of this quadratic polynomial expression. Okay. What is the coordinate of the vertex? What is the coordinate of the vertex? So let me draw this. Please note I have assumed it is concave upwards. That means I have assumed a is 0. But let me tell you let me assure you that it will not change the coordinates of the vertex. If you don't know the coordinates, I'll help you out. But if you know, can you please type it out? Could you repeat the question? That's correct Shankin. Our question is in terms of a, b and c tell me the coordinates of the vertex of this parabola. Siddhartha, that's not correct dear. Anybody else? Shankin, is it a y is equal to this or x equal to this? It's a line parallel to the y axis. So it should be x equal to something. Yeah. Anybody who can tell me the coordinates of the vertex? One minute. Then it is as good as me deriving it. All of you please pay attention. All of you please pay attention. I would like you all to go to this expression. Sorry, I'm going to the previous board for one second. I would like you all to go to this expression this expression that you see over here. Sorry, what I'm making I have to make a box this expression. I would invite you all to have a look at this expression because I'm going to use this to get the vertex coordinates. So all of you please copy this down on your notebook. I'm going to the next board. Copy it. Now see how I'm going to use that. So the expression was y is equal to a times x plus b by 2a the whole square minus b square by 4a plus c. This was the expression. Now this is the point where my y is minimum. Do you all agree with this? Vertex is a point where the y is at its least value. There is no other point on the parabola where y is lesser than the y that you have on the vertex in this present scenario. Agreed. So now all of you listen to this since your parabola is opening upwards a is a positive quantity and since this is a square this quantity is always greater than equal to 0 and this quantity here is fixed. Now boys and girls think about it carefully and then respond if you want your y to be minimum what should happen to so a cannot change this fixed quantity cannot change what should happen to this fellow if you want your y to be minimum then what should be the value of the square thingy 0 very good Suda very good Arpita very good very good. So if that is 0 my dear friend you would agree that x has to take this value only under this condition my y will become minimum. So your x coordinate of this point should be minus b by 2a do you agree this term becomes 0 then your y minimum would be let me write it over here your y minimum would be negative b square by 4a plus c which is actually negative of b square minus 4ac by 4a which is actually minus d by 4a so in future please remember this expression I would not derive this again and again so please remember this okay sorry I am asking you to memorize something but it is only for your good now sir partial derivative gives you the center pratham it doesn't give you the vertex partial derivative gives you the center of the conic it doesn't give you the vertex of the conic got it okay I will go through it once again remember in the previous board I had come from this expression to this expression by completing square now my aim here is my problem statement is I want to get the coordinates of this point I want to get the coordinates of the vertex okay vertex is a point where your y is minimum correct when can this expression become a minimum let me tell you the ingredients of this expression it has a value in it which is always positive but how we are fixed for that particular expression there is a variable quantity but since it is subjected to a square it is always greater than equal to 0 and there is another fixed quantity that is b square by 4a plus c minus b square by 4a plus c now if you want your y to become least as less as possible right can I say this positive thing should be as less as possible and for that this quantity which is greater than equal to 0 should actually become a 0 correct or no Anitha correct so if it actually becomes a 0 it can only happen when x becomes minus b by 2a right then only your y will take this value which is minus d by 4a so the point where your y is minimum will actually have this as your coordinate now all of you please pay attention here now by using this vertex by using this vertex I will get the condition for the nature of the roots all of you please see this again I am sketching the graph of the parabola okay so this is your parabola this coordinate was minus b by 2a comma minus d by 4a okay now let me assume a scenario that a is positive okay now if your roots are real and distinct if your alpha and beta are real and distinct think carefully and answer what should be the y coordinate of the vertex should it be negative or should it be positive or should it be 0 I am repeating my question if you want your roots to be real and distinct like what I have shown in the diagram should the y coordinate of the vertex be positive negative or 0 negative very obvious from the diagram so if it is negative my dear you would agree that this y coordinate which is minus d by 4a should have negative value now let me take the 4a on top which means minus d should be negative value remember inequality will not change because 4a is positive that means d should be positive correct multiply throughout with minus 1 inequality will flip isn't this the condition that we had derived in the earlier page also by using the Shridharacharya formula now see I am bringing the same concept to you on the table but from a different angle from a different perspective are you getting my point are you getting this point if you want your alpha and beta to be real and equal ok now watch the diagram how will it change if you want your alpha and beta to become the same value can I say it would be a condition where the graph is just touching the x axis ok now the alpha and beta will come to the same point ok this will happen when the y coordinate exactly becomes 0 that means d is equal to 0 the same condition for real and equal roots non deriving it from a different angle if you want your alpha and beta to be non real or imaginary what will happen to this graph it should be hanging in the air like this something like this correct if it is hanging there will be no real roots being created ok in that case minus d by 4a should be positive that means minus d should be positive that means d should be negative the same condition getting my point now many people ask sir will the same condition hold true if your parabola was opening downwards yes it will not change so I will just do one of it and leave it leave the rest up to you to realize for homework if your parabola was having concave downwards position like this and let's say this was your alpha beta ok now tell me what is the vertex coordinate now again I have leave this as an exercise it will remain the same hint is now you should search for see I am so patriotic or hint of it ok so for the vertex now you should be looking out for y max ok trust me the vertex coordinate doesn't change now if you want your roots to be real and distinct if you want this to be real and distinct if you want it to be real and distinct what should be your y coordinate sorry sorry minus b by 2a it doesn't change sorry by mistake I wrote thank you Sana yeah if you want your roots to be real and distinct what can you comment about the y coordinate of the vertex shouldn't it be positive now since a is negative over here guys let me tell you it's a downward opening parabola so if a is negative and if I take the a on the other side won't this switch the inequality sign correct and if you drop your minus sign again the inequality sign will be switched that means the condition doesn't change condition of the nature doesn't change irrespective of whether your parabola is opening upward or opening downward is that clear is that clear everybody ok now this can also help us to know the sign of a quadratic expression so I'll take this as the next concept sign of a quadratic expression normally lot of questions are based on this so a quadratic expression like this ok this is an expression not an equation ok when will it always be positive can somebody tell me when will this be always be positive that means it is greater than 0 for all values of x can somebody tell me from the previous exercise that we had what are the conditions that I need to honor such that this quadratic will always be positive no matter whatever x you feed in a should be positive ok d should be less than 0 awesome very good guys this is the condition which we need to honor for this quadratic expression to be always positive now question is why ok now why a should be positive sorry sorry a should be positive yeah why a should be positive see if a is negative your quadratic will be facing downwards ok so I'm just drawing a rough figure so if you have a quadratic like this let me tell you these are infinitely extending arms they will both go to minus infinity ok you cannot stop the parabola from acquiring negative values so a must never be negative right now if a is positive why do we want our d to be negative because if a is positive and it is like this then what will happen for these values for these values of x your graph here is negative isn't it so it will not be always be positive right so if you want this to be always positive the ideal situation is it should be hanging in the air like this that means your y let me call this as y even though I don't want to write it so I'm just putting brackets let's say you're calling it as y y should always be positive that means your graph should be entirely above the x axis for that you should have imaginary roots or you should not have any real roots so that's why this condition that is your second condition must be true ok now carrying on with the same discussion if you want your quadratic expression to be always negative for all values of x then what are the conditions we need to honour think and tell me carefully I'm putting a box for you to tell me the answer you can type it out if you don't want to speak you can type it out a less than 0 and a less than 0 a less than 0 d also less than 0 very good so it that means it should be hanging down like this ok so this is the scenario where it will have always a negative value very good we have aditya, d should be negative, not positive ok fine now before we solve any question I would like you to know these properties properties of the roots of a quadratic of a quadratic equation ok so let's say I have a quadratic equation like this ok now these are the properties which you must note very important number one if your abc that is the coefficients that you see over here if they are all real if abc are all real and one of the roots and one of the roots is imaginary that is p plus iq ok then this property says that the other root that the other root will be p minus iq that means it would be its complex conjugate ok please remember this property will only hold good if your abc are all real if let's say any one of them is non real that means if let's say a is an imaginary number let's say 2 plus 3 i then will you apply this property no it is only applied when your coefficients are all real then you can say if one of the root is imaginary like p plus iq the other must be of the type p minus iq ok also note my dear students that this is applicable to this is applicable to polynomial equation of any degree polynomial equation of any degree that means if you have a polynomial equation whose coefficients are all real and if let's say any one of the root becomes p plus iq p minus iq must also be present that means they will always come in pairs they will always come in pairs are you getting my point ok next property next property yeah any question anybody yes ashore I will repeat it again see if any polynomial equation no matter whatever is the degree if let's say one of the root is imaginary in nature then it's conjugate root must also be a root of the same equation ok at a simple example is let's say xq minus one equal to zero ok so this has three roots one minus half plus i root three by two and minus half minus i root three by two so this is a cubic equation this is a cubic polynomial equation isn't it this is a cubic polynomial equation ok as you can see one of the roots is of the nature p plus iq then the other must be of the nature p minus iq correct so they will always come in pairs said it's not possible to have one root real and other root imaginary if abc are see imaginary roots when your coefficients are real it will always come in pairs it will not come in isolation so it can never happen that you have two real root and one imaginary root if you have abc as real please note that I am repeating this again and again ok so exactly if there is a quadratic we cannot have one real and one imaginary never it will happen if your abc are real ok by the way what is this called guys omega omega and this is omega ok good next property that is property number two property number two property number two says if abc are rational ok now I am changing the coefficient to rational ok and if one of the roots of the quadratic one of the roots of the quadratic is p plus root q that means it is irrational that means let's say it belongs to q bar bar means not a rational ok then the other root then the other root must be p minus root q that means it is going to be its rational conjugate is going to be its rational conjugate ok so if one root is let's say 2 plus root 3 other will be 2 minus root 3 if one root is let's say 5 minus root 6 the other will be 5 plus root 6 ok but guys unfortunately this property is not applicable for all polynomial equations for some it works for some it does not ok now this is a shocking thing for many of you because in rs agarwal polynomial equation it was stated that if one root is 2 plus root 3 other has to be 2 minus root 3 irrespective of whether it is quadratic or cubic or bi-quadratic or pentic let me tell you this is not true this is not true I will give you a classic example for this because in many books also take this equation x cube minus 3x plus 1 equal to 0 x cube minus 3x plus 1 equal to 0 ok how many roots it should have 3 roots ok now my claim is that the irrational roots will not exist in pairs so there is no pairing just like we have it for complex but in case of rational irrational roots this pairing is not a must it is not compulsory now later on when I am doing the solving of cubic polynomial I will be discussing a method called cardens method with you cardens method ok I will be telling you how to solve cubic equation just like you know how to solve quadratic equation ok if you apply cardens method you will get the roots of this as the roots of this as 2 cos 40 degree 2 cos 80 degree and 2 cos 160 degree ok those who have calculators or phones in front of you can you quickly tell me what are these values ok please tell me these values anybody 2 cos 40 is 9.53 ok 2 cos 80 is 0.34 2 cos 80 is 0.347 right minus 1.879 879 so let's say 1.88 ok even though they look like rational numbers they are not actually ok now sorry my bottle just fell ok sorry now these quantities that you see here these are all irrational values now many people even question me on this how do you know that these are irrational values how do you know that you cannot express them as p by q ok now to support this there is a theorem which we call as niven's theorem niven's theorem ok you can do a quick study on this after the class niven's theorem says that if theta is rational and belongs to 0 to 90 degrees ok then only the following angles theta will give you rational values of sin theta so this will be rational only for only for these 3 angles 0 degree 30 degree 90 degree and similarly cos theta would be rational only for theta equal to 0 degree 60 degree and 90 degree that means other than 0, half and 1 there is no angle which is rational for which sin is going to give you a rational output this is niven's theorem ok now cos 40 doesn't belong to this right 40 doesn't belong to this angle 80 also doesn't belong to this angle in fact cos 60 you can directly write it as negative 2 cos 20 ok so even cos 20 doesn't belong to this angle so cos theta for these angles which is 40, 80 and 160 will be irrational ok 37 is 37 not a rational value shawmic are you sure it gives exactly the same value no it is an approximation 37 is an approximation getting the point so now these 3 are irrational and it is actually surprising because if they are irrational they should have existed in pairs but all the 3 are irrational all the 3 roots are irrational so where is the question of a pair getting a pair so even if you form a pair between the 2 of them the 3rd one is without a pair are you getting my point so it is not necessary that your irrational roots will exist in pairs it is true for quadratic I agree with that it is true for a quadratic but it is not true for any polynomial ok and those of you who are wondering how are these the roots I will quickly show you on the GeoGebra I don't want you to sit in confusion here I will just delete it delete it ok so the equation was y is equal to x cube minus 3x plus 1 ok this is already some let me open a new one don't save yeah it is equal to x cube minus 3x plus 1 ok I can show you the roots over here by the way you can also solve there is a function in GeoGebra that says solve solve and you press the function name ok yeah sorry as you can see here it is showing you the 3 roots minus 1.88 minus 0.35 1.53 is that understood so there is a condition where I had a cubic polynomial equation which doesn't give me rational roots or irrational roots in pairs yeah you can extend it to tan also you can extend it to tan also ok so I would suggest you to go through Neven's theorem after the class ok yeah tell me sir if one of the roots we can't have just one irrational root no all the 3 are irrational over here yes sir but is it possible to have just one irrational root and other to a rational root now with this I am sure it should be possible also sir there is one irrational I have seen a case where there was a pentic equation pentic means degree 5 where there was 2 pairs of complex and one was irrational like that sir but then like if one is irrational the other to a rational then if we add the coefficients it won't give a rational value but then the second coefficient the coefficient of x square should be it should be irrational right so one of the roots is coefficient of x square see in this case it is happening that the 3 of irrational roots but you are claiming that only one irrational root yes sir I am asking if that is possible I have to think of a condition right now it may not strike me but prima facie it appears that there may be a situation where there is just one irrational root and the other 2 you are saying have to be real yes sir both should be irrational rational one is irrational other 2 rational yes sir in that case it cannot happen that is not possible now let us take the next property so I am switching the board ok this is a property which has come up from lot of questions which have come in j in the previous years the property says if there are 2 quadratics let's say this and this ok and let's say the discriminant of this is d1 and discriminant of this is d2 ok so this property says if your d1 plus d2 is positive or greater than equal to 0 then at least at least one of the quadratic at least one of the quadratic will have real roots now this is a very trivial thing because if the sum of 2 things is positive then at least one of them should be positive correct so both positive means both will have real roots so this will still be holding true but if let's say one is negative then the other has to be definitely positive to make the overall sum positive ok so this is something which is a very trivial kind of observation and lot of questions have come on this in the past that's why I am revealing I am putting this as a property if the sum of the discriminant is negative then remember at least one of them will have non real roots at least one of the quadratic has non real roots or non real roots yes any questions are we good to go are we good to start solving questions ok so here is the first question for all of you I will start with a very basic question that's one solve for x so the answer is that the complete solution or they can be still further values I want the exhaustive solution ok Shusti ok Aditya Sudha guys last one minute ok Ayush alright guys time up let's discuss most of you have given incomplete answers incomplete answer means there are more solutions possible now all of you please focus over here we know that 5 plus 2 root 6 and 5 minus 2 root 6 will actually give you a 1 right 25 minus 24 that means 5 minus 2 root 6 5 minus 2 root 6 is actually reciprocal of 5 plus 2 root 6 or you can say 5 plus 2 root 6 to the power of minus 1 right so if I do this activity that means if I raise the left hand side to this here I will realize I am actually doing 1 by 5 plus 2 root 6 to the power of x square minus 3 correct now let us call let us call 5 plus 2 root 6 to the power of x square minus 3 as t ok this is called convertible to quadratic so it will become t plus 1 by t remember this term is actually 1 by this expression so I am calling it as 1 by t right so here you have a quadratic in front of you which you can easily solve t square minus 10 t plus 1 equal to 0 let us use our Shridharacharya formula for this b square minus 4ac by 2a that is nothing but 10 plus minus root 96 by 2 root 96 is 4 root 6 so 2 factor will get cancelled so it will be this that means 5 plus 2 root 6 to the power of x square minus 3 is either 5 plus 2 root 6 or it is 5 minus 2 root 6 correct now this implies that x square minus 3 has to be 1 and this implies x square minus 3 has to be minus 1 now most of you gave me the response 2 and minus 2 because you only ended up solving this you did not care about the other one so if you solve this one you get x square is equal to 2 that means x will be plus or minus root 2 so there are 4 solutions possible that is 2 minus 2 plus root 2 minus root 2 is that fine did you realize your mistake where you left off one of the equations so guys something to make you happy let us have a break sorry 2 hours of class I am sure you must be hungry will resume at 6 21pm okay so this is a multi-correct question I would request you all to type out the answer individually privately to me it is a multi-correct question read the question carefully it says the roots of this equation are non real complex and a plus c is less than b then which of the following options are correct let us have 3 minutes for this time starts now yes any success anybody aditi says okay a lot disclosure answer it is multi-correct aditi so there can be more than one options correct also so just look at all the options and be doubly sure before answering his last one minute I can give you already 3 minutes have passed okay keerthana good I am nothing right or wrong to your answer we will discuss it in some time okay aditya okay dhyan okay arpita his last 20 seconds okay krishna okay siddharth let's discuss this and then we will come back who has given the right answers okay so your answers are in front of me so first of all this is a quadratic expression which has got non real roots right non real root means I am sorry non real roots means the graph is either hanging in the air above the x axis or below the x axis so it could be either be a situation like this okay or it could be a situation like this so I am just drawing two possibilities correct so white one is a situation where your a is positive and yellow one is a situation where your a is negative now all of you please look at this expression a plus c less than b okay it is actually saying a minus b plus c is less than 0 correct now let me call this expression as f of x so let's say I call f of x as ax square plus bx plus c right if I want to get this tell me which value should I put as x minus 1 absolutely so if I put a minus 1 this is what I will get now what is the question saying this is less than 0 that means the value of the function or the y coordinate of the point whose x coordinate is minus 1 is negative that means let's say this x coordinate is minus 1 so my y coordinate for the same point is a negative value okay it can only happen when your graph is this one and not this one because for this at minus 1 value should have been a positive value isn't it but this is clearly saying it is negative negative means it is a downward opening parabola and not only that it doesn't have any roots okay so if a is negative and d is negative that means remember the sign of the quadratic expression that we were discussing this is always going to be negative correct now look at the options if you see this option I am going to use a different colour if you see this option this is like saying 4a minus 2b plus c is greater than 0 okay that means indirectly it is trying to say f of minus 2 is greater than 0 do you think it is correct see the property says the quadratic must always be less than 0 because these two conditions are satisfied so can this ever happen no it will never happen so option a is wrong so sorry to those people who have marked option a in their answer that is wrong but can option b happen yes because it is saying f of minus 2 is less than 0 that is possible so b is possible okay now look at this okay it is as good as saying sorry it is as good as saying a by 4 minus b by 2 a by 4 minus b by 2 plus c is greater than 0 right if you just divide by 4 throughout and bring the b to the left hand side this is what it is trying to say isn't it like f of minus half they are saying f of minus half is greater than 0 but please note this condition must be holding for all values of x so even c is wrong and by the same reason d is correct so the right options here are b and d these are your correct options let's see who all give the right answer okay arpita is asking how did you get this as a download opening parabola arpita this is a condition which says f of minus 1 is negative right f of minus 1 means the value of the function minus 1 should be a negative value so at minus 1 if you go up you will get a positive value right that means this graph is not possible only the downward graph will give me a negative value that's why a download opening parabola would be there okay so ajay incomplete answer araman one correct one wrong angeli wrong shomik wrong dharth wrong arpita wrong aditya wrong kirtana that's correct kirtana is the only one person who gave this correctly well done next question so this is a single option correct question and I would run a poll for you all let's have 3 minutes to do this okay I got one response last 30 seconds make sure everybody should poll or go for the poll okay the countdown has begun last 10 seconds 5 4 3 2 1 press on the poll button guys everyone okay let's stop oh my god if such a response were given in KBC my god equal vote for BCD and just by marginal A is more okay let's check whether janta is correct or not okay shomik let's check okay first of all if you see the given situation you would realize A is negative right for sure because it's a downward opening parabola on KVT is downward correct second thing you would realize that this value is actually your 0,C let me use a 0,C so C must be positive so product of AC should be negative for sure that means C cannot be correct I don't know why you people voted for C even a 10th grader can see that C is not correct okay next thing is if you see alpha value is lesser than minus 1 and beta value is greater than 1 you can see the position of 1 and minus 1 okay so the product of the roots that is alpha beta should be definitely be less than minus 1 you can assume some value let's say if you take this value as minus 2 and if you take this value as 2 then product is minus 4 minus 4 is definitely less than minus 1 which means C by A should be less than minus 1 so that means D must be also be wrong I don't know why you voted for D next thing let's see this one when X is greater than beta your function is acquiring positive value which is clearly not correct because if you take a beta value higher beta value higher than beta value let's say I take gamma your function is acquiring a negative value so this is also not correct it only leaves you with A as correct answer but still I will justify this all of you please pay attention what is mod of beta minus alpha indicate here mod of beta minus alpha actually indicates the distance between this position to this position so this distance here is mod of beta minus alpha correct can I say that is definitely going to be greater than 2 because this distance itself is 2 and C by A C by A is a value which is less than equal to 1 sorry less than minus 1 so if you multiply it you will end up getting something which is less than minus 2 which is actually what option A says okay guys I can see people are making mistakes in obvious things another question this is a very interesting one again multiple options may be correct time starts now single option or multiple option multiple option I already told you it's a multi option multiple options may be correct can you do one thing instead of giving you a straight forward 4 minutes to do this can we have one minute for each one of them let's have the first one is this option correct or wrong just type let's say A is correct and B is for incorrect I am running a poll all of you please comment about the first one wait let me run the poll it will be interesting to see how you guys are responding to that let's have a minute for that is press A if you think A is correct press B if you think A is wrong just one minute I will give you not more than that press on the poll button also those who are responding to me privately press on the poll button also okay time up time up see the poll result most of you said A is correct 58% of you have said A is correct let's check everyone now if you see this expression it is actually trying to state this expression is always less than equal to 0 the probability of it is happening is 0.3 okay and your X can only take first 100 natural numbers so your X should only come from 1 to 100 okay let's check I am sure this is factorizable right yeah 30 and 5 so X minus 30 so X plus 5 30 so we all know these inequalities that it will be true only when X lies between minus 5 to 30 okay now minus 5 to 30 the favorable conditions would be only when it takes 1 to 30 because it cannot go to the negative side because we are only allowed to choose natural numbers okay so my favorable condition will only be 30 which is 1 to 30 and my total sample space is 100 so I can say my probability is going to be 30 by 100 which is 0.3 so yes you guys are correct A is right okay now I am running a poll for B option and likewise you have to choose A and B A for correct B for wrong okay one minute for that A for correct B for wrong okay Shomit press on the poll button A for correct B for wrong yeah yes Krishna A for correct B for wrong alright end of poll maximum people have said 88% of you have said B option is wrong let's check it out let's check it out now you want this expression to be greater than equal to 0 right let's write it over here let me put a demarcation here so x square minus 17x plus 30 is greater than equal to 0 that means I am sure it is factorisable minus 15x minus 2x plus 30 greater than equal to 0 that means x minus 15 x minus 2 greater than equal to 0 that means x should be either be less than equal to 2 or x should be greater than equal to 15 let's count the favorable events how many national numbers are less than equal to 2 1 only 2 2 itself is included so 1 and 2 how many numbers from 15 onwards still 100 so how many national numbers are from 15 to 100 86 including 15 and 100 86 so total will be 88 so probability would be 88 by 100 that's 0.88 so option B also is correct my dear so janta was wrong guys I don't see a point to make a mistake here what happened people went wrong for this on a massive scale 288% people here were wrong answer yeah sure why not is that fine dhyan why people went wrong in this see follow this A is correct B is wrong now let's have the response for C again the same thing you have to press A if it is correct and press B if it is not poll starts 100 minus 15 is 75 but you have to add a 1 also because corner positions are also included guys you can take a little longer time for this probably you can take one more minute no worries last 10 seconds please press on the poll button 5 4 3 2 1 and a poll look at the results maximum of you have said C option is wrong 63% of you have said C option is wrong who is pressing C you have to only press A okay let us discuss it let us discuss this x2 minus 30x-x2 is a perfect square the probability of it is 0.07 okay let's do this thing again let me draw a line over here 30x-x2 let's say it's a perfect square which is let's say A2 now if this is a perfect square by the way this can be completed to be 15 square minus 15-x2 isn't it am I right I have tried to complete a perfect square from the left hand side like this am I correct okay so I can write 15-x2 to be 152-A2 or you can write it as 15-A into 15 plus A now please note since there is a 15-A I can only go from 1 to 15 so A can go from 1 to 15 okay we have to do this manually now now put A as 1 if you put A as 1 you get 15-x the whole square as okay by the way can you put a 0 first yeah 0 is also permissible right yeah if you put a 0 I'll end up getting 15-square right that gives me 2 values of x one is 0 and the other is 30 okay so one value I have over here 30 is going to give me a perfect square you can check this is going to become a 0 for that 0 is itself a perfect square so one solution I get from it now if I put A as 1 so this is when your A is 0 if I put A as 1 what do I get what do I get I get 15-x square as 14 into 16 no I don't think so this is square of something okay so I'll get no solutions from here okay let's try out the other values I'm writing it over here once again 15-x the whole square 15-x the whole square is equal to 15 okay does A equal to 3 work check 12 into 18 is it a perfect square 12 into 18 is it a perfect square 216 no it is not a perfect square okay A equal to 4 will it work check 11 into 19 that is also not a perfect square it will also not work A equal to 5 10 into 20 it will also not work A equal to 6 6 gives you 9 into 21 that is 189 that will also not work A equal to 7 8 into 22 that will also not work A equal to 8 that will give you 7 into 23 that will also not work A equal to 9 that will give you 6 into 24 that is 144 yes it is 12 square so A equal to 9 will work and for that I'll get 2 values 1 is plus 12 so 15-x is plus 12 and 15-x is minus 12 and both of them will give me 2 values so 1 will be 3 other will be other will be 27 okay now let's carry on 10 10 will give me 25 into 5 25 into 5 125 125 is not a perfect square A equal to 11 it will give me 29 sorry 4 into 26 again it's a perfect square 4 into 26 is no 4 into 26 is 8 oh sorry sorry this is not a perfect square it is 114 this will also not work A equal to 12 will give me 3 into 27 81 yes it will work so 15-x the whole square could be plus minus 9 so 1 of the values could be 24 and other value could be 6 okay A equal to 13 no it will not work 2 into 28 A equal to 14 no it will not work A equal to 15 I think 15 we have already seen when we had put 15 will give you 0 so x can be 15 correct yes or no so what are the values that we have got 30 3 27 24 6 15 anything else that I am missing out on I cannot go higher than 15 for A right because 1 of them will become negative so what about A equal to 2 A equal to 2 gives you 13 into 18 13 into 17 is that a perfect square no didn't I take that up no I didn't I took this up no A equal to 2 will give me this has 13 into 17 this is not a perfect square anything else that I am missing out so guys the answer is 1 2 3 4 5 6 by 100 0.06 which unfortunately is not there because it is 0.07 so C is wrong ok so can we say that if it should be a perfect square then it should have equal roots see if you say it has equal roots x may be non real sorry x may be not between 1 to 100 also so many other values will come into picture ok and if it is equal roots that will definitely be satisfied see this cannot have equal roots because if you say this actually 30 minus x into x the roots are clear the roots are 0 or 30 they cannot be equal I understand where you are coming from you are coming from the straight lines concept right that was expression which was supposed to be a perfect square here it is a value that we are talking about last one let's run the poll for the last one again sorry stop sharing again A for correct B for wrong 1 minute for this last 10 second guys time starts now 5 4 3 2 1 please press on the poll button ok I don't know why people are voting for D only A and B have to be voted for ok now maximum people say it is correct ok let's check so you are saying 30 minus x square is less than 0 the probability for that to happen is 0.7 so I will use this space 30 minus 30x minus x square is less than 0 means you are trying to say x square minus 30x is greater than 0 that means x could either be less than 0 or x could be greater than 30 greater than 30 means from 31 to 100 we have to count 31 to 100 there are 70 values including the corner value so the answer will be 70 by 100 which is 0.7 so yes you are correct Janta was correct D is correct so A C and A, B and D are the right options very good guys interesting question isn't it so a bit of probability also getting tested ok let's have another question ok let's move on to another property question I will just take it up from so it's a show that question show that if PQRS are real and PR is equal to twice of Q plus S then at least one of the equation has real roots we have already done a property based on this so try to apply that I will give you 2 minutes for this not more than that done Pratham we were ok good ok let's discuss guys this is a very easy question see if you look at the first quadratic let's say this one ok so for this quadratic let's say your discriminant is D1 and D1 would be P2 minus 4Q ok let's look into the other quadratic for this the discriminant is R2 minus 4S now let's add these 2 discriminants if you add this it actually becomes P2 plus Q2 minus 4Q plus S ok now time to utilize this condition that we have PR is equal to twice of Q plus S so can I say 4Q plus S could be written as 2PR ok so from the condition given to us that is PR is equal to 2Q plus S I can write this as D1 plus D2 equal to P2 plus R2 minus 2PR isn't this P minus R the whole square which means it is definitely greater than equal to 0 correct now remember the property which I told you if D1 plus D2 is greater than equal to 0 that implies at least one of D1 or D2 at least one of D1 and D2 must be greater than equal to 0 that means at least one equation must have real roots is that fine easy everybody could do this any questions? good guys the next concept which I am going to talk about is very important I would like all of you to pay attention here that concept is called the transformation of quadratic equations now even though I am writing quadratic equations here it could be applied to other polynomial equations also ok so let's talk about this concept ok I have a question for you in general let's say we have ax squared plus bx plus c equal to 0 as a quadratic whose roots are alpha and beta ok now we all know that alpha plus beta is minus b by a and alpha into beta is c by a right everybody knows this ok I will talk about this also in detail this is something which we call as beta's relation beta's relation by after the name of the Italian mathematician now I don't know what you are thinking when you hear of Italy ok anyways let's not talk about corona now enough messages I am getting man on corona and all those things aren't you all getting messages on corona too much of discussion going on right ok so let's say the roots of this quadratic equation are alpha and beta I have a question here for everybody tell me a quadratic whose roots are let's say 3 alpha by alpha plus 2 and 3 beta by beta plus 2 how do you solve such questions normally what is the approach when you solve these questions anybody can unmute himself or herself and say what approach normally do you take when you are solving these questions Aditya my question is tell me a quadratic whose roots are these two if this quadratic has roots alpha beta tell me a quadratic whose roots are these two I don't multiply and put in the place of the answer right very good so people do this let's say I call this gamma and delta they get this equation by doing this activity x square minus gamma plus delta x plus gamma delta right this is what you do ok for that you have to find the sum and you have to find the product ok now I am going to tell you an approach correct Krishna correct correct now I am going to tell you an approach which will not require you to find the sum and the product and that approach is the transformation of equation so all of you please listen to this very very carefully transformation of equation is applied when you already know a root of a certain equation and you want to find out an equation whose roots are a symmetrical change in both these roots so you can see there is a symmetrical change happening in both these roots that means both of them are undergoing a similar fate in fact the same fate you realize here 3 alpha alpha plus 2 3 beta beta plus 2 so wherever you realize that your roots are getting transformed under the same function then this method which I am going to tell you next is going to be very very handy ok but however it has the limitation it only will work for those situations where both the roots are undergoing the same fate ok so this method is useful only when only when what happened to this color man are you able to see this no sir ok just let me do one thing here it is working I don't know why it is not working on the previous board let me choose some different color ok let me just scroll this up or down ok I think there is it's getting hanged ok so let me do one thing let me save this document because we will lose out the details over here so let me write it as theory of equations equations for nps r and r I am just going to reopen this again because there is some issue with the this thing ok let me go to the next page ok so now let's say you want 3 alpha and alpha plus 2 and 3 beta beta plus 2 as your new roots of a equation correct now what I do here all of you please watch carefully since alpha is a root of since alpha is a root of ax square plus bx plus c equal to 0 can I say alpha will satisfy this equation correct let me call this as 1 ok now you want to construct a equation whose roots are these 2 fellows so let me do one thing let me call this as x let me call this as x ok and now what I will do is from this expression I will make alpha the subject of the formula I will make alpha the subject of the formula so let me cross multiply and let me do this so alpha will become 2x by 3 minus x ok let me call this as 2 ok now this 2 you substitute in 1 that means in place of alpha that you have written over here you substitute this guy 2x by 3 minus x so it becomes something like this ok and now you simplify this expression to bring it to another quadratic that quadratic will be your answer ok so this is how we work on those questions where there is a transformation of equation required because the roots are undergoing the same fate so if alpha become this and beta become this you can apply transformation to solve this question are you getting my point let me tell you this method is very very fast ok but it has a limitation also it will not work when let's say this becomes 3 alpha by alpha plus 2 and this remains let's say beta only then this method will not work right is this approach clear to you because approach is important so please make a note of this approach why it works I'll tell you why it works because see you are actually transforming the roots and this transformation is basically applied to both the roots that means the entire equation is undergoing the same transformation ok so instead of separately finding out the sum and the product and then creating an equation out of it you can directly apply this transformation let me give you a simple example let's say if your roots were alpha beta and I change them to alpha minus 2 beta minus 2 what is happening actually look at the graph if let's say this was alpha and this was beta and you want the root to be alpha minus 2 and beta minus 2 what is happening to this graph says that these two will become the root can anybody tell me move on to the left move to the left very good so if I move it to the left basically then I will achieve basically I'm moving this by two units correct so when the roots are undergoing the same fate ok you can actually make a change in the equation itself are you getting my point by applying the same transformation or you can say reciprocal of the transformation now I am saying reciprocal of the transformation or inverse of the transformation is because if I want let's say alpha beta was the root of this equation ok if I want the equation with this two as the roots ok remember the process which I have discussed over here what do I do first of all I put alpha in place of x correct first of all I put alpha in place of x secondly I call this as x and make alpha the subject of the formula so then I replace my alpha with 2 plus x or x plus 2 whatever you want to call it ok do you see that by doing this you have taken this graph two units to the left remember changing x with x plus 2 shift the graph two units to the left ok so that is why Aditya this method works getting my point ok now I will do a quick exercise with you let us say alpha and beta are the roots of are the roots of this equation ok quickly tell me a equation whose roots will be alpha minus k and beta minus k first just tell me what would you replace your x with very good this is something which should come very naturally to you after today's session very good ok next question what if your roots become alpha plus k and beta plus k what would you replace your x with x minus k very good do you realize what is happening ok let us take few more cases then I think it will be clear to you what if you replace your alpha and beta with k beta by k x by k very good this is what I wanted to you to appreciate ok by the way let me write as a equation sorry I miss 0 ok tell me this one what about if I put alpha by k and beta by k k x k very good very good tell me what will happen if I do 1 by alpha 1 by beta 1 by x absolutely absolutely but this is not exactly in a quadratic shape you have to bring it to a quadratic shape by multiplying with x squared very good one take you and also I will ask you what if I change it with alpha square and beta square root x very good but when you write root x you have to be careful because the moment you write this as the moment you take this fellow as x alpha will become plus minus root x correct so it will actually look like this alpha sorry ax plus minus b root x plus c ok then you have to actually bring it to one side let's say minus plus b under root x and then square both the sides so it actually becomes a square x square to acx but c square is equal to b square x and then you club the like terms ok this will be your final answer now I am sure you would have got a hang of what is going on here so one final question ok I will give time for this what if I replace my x with px minus sorry I replace my alpha with p alpha let me read it I messed it up p alpha minus q and p beta minus q what will I do so this is something like inverse function yes it's working like that's what I was trying to say you are actually applying the inverse kind of transformation very good x plus q by p very good very good ok I am not simplifying it I am leaving up to you to simplify this my main agenda was to make the things clear to you ok if this is clear and if you are happy with this may I ask you a question sorry sir could you go back let me just dump the question then we can go back let me dump the question you want me to scroll up just could you show the question for a second guys I can't do two things at the same time somebody asked me to come to this page is that person done there is no need to copy by the way if there is something to ask you can ask because anyway the notes will be shared with you if alpha and beta are the roots of this equation find the value of 1 minus alpha by 1 plus alpha whole cube and 1 minus beta by 1 plus beta the whole cube let's see how many of you get this correct let me put an option also so that I can check who all are correct option A 10 option B minus 10 option C 8 option D minus 8 let me run a poll let's have 3 minutes for this not more than that guys last 30 seconds last 5 how do you read this word 5 4 3 2 1 go for it okay end of poll look at the answer maximum people have gone for C 63% people say 8 26% people say B and 2% sorry 11% people say D let's see what is the answer now if you see this question the moment I see this 1 minus alpha by 1 plus alpha and the same with this the transformation of the equation comes in my mind so let's say I call this expression as A and I call this expression as inside one not the overall let's say this inside one I call it as A so let me write it let's say A is 1 minus alpha by 1 plus alpha and this inside expression and let me call it as B okay now what are we looking for we are looking for A cube right okay and everybody knows the formula for this A plus B the whole cube minus 3 AB A plus B right now looking at this what runs in my mind that somehow if I am able to form an equation let's say I am able to form an equation with A and B as roots then from that equation I can know the sum I can know the product by looking at the coefficient right that's the V taz relation correct and then I will be able to apply it to this formula and my job is done okay so now my effort is to find out that equation whose root will be A and B that means these guys okay again let's take the normal root itself so if alpha is the root of this that means this condition will be true okay then what I will do is I will call 1 minus alpha by 1 plus alpha to be X okay and I will make X the subject of the formula guys can you suggest me a quick way to get alpha in terms of X very good component no and dividend no so that will give me 1 minus X by 1 plus X okay so let's place over here 1 minus X by 1 plus X whole square to 1 minus X 1 plus X plus 1 equal to 0 okay let us multiply with 1 plus X square whole throughout so this will become this this will become 1 minus X square and this will become 1 plus X whole square okay if you expand it correct me you will have 2 X square okay and you will get minus 6 X minus 4 X and there will be a 6 that means you get X square minus 2 X plus 3 equal to 0 as your equation okay now if a and b are roots of this can I say a plus b will be 2 that is minus b by a this is your b and a b will be 3 by vitas relation now what is this vitas relation I will take care of it little later on after this problem itself okay so I can say this value over here can be easily be found out by putting these values that is 2 cube minus 3 into a b is 3 a plus b is 2 so answer is 8 minus 15 which is minus 10 sorry janta you are wrong 63% of you said c but the answer is b okay now as a homework I would like you to solve the very same question by now assuming that the roots are not 1 minus alpha 1 plus alpha and this assume that the roots are 1 minus alpha by 1 plus alpha whole cube and 1 let's say delta delta 1 minus beta and 1 plus beta the whole cube please try to solve the same question by creating an equation whose roots are these 2 and then find the sum of the roots okay so this is an alternative approach which I am giving you as a homework okay any question with respect to the approach guys please ask no question now officially I am going to talk about officially I am going to talk about beta's relation that I have been telling you from beginning beta's relation beta's relation is basically the relationship between the roots and the coefficient of a polynomial equation so it's basically a relation between between the roots and the coefficients of any polynomial equation any polynomial equation okay so in quadratic you already saw that relation what was that alpha plus beta is minus b by and alpha into beta is c by so basically you are connecting the coefficients with the roots okay now I will be generalizing this so let's say I have an nth degree polynomial like this I have a nth degree polynomial like this now let's say the roots of this relation are sorry the roots of this polynomial equation are alpha 1 alpha 2 alpha 3 and so on till alpha n okay by fundamental theorem of algebra it must have n roots okay the relation beta relation goes like this if you add all the roots that is what I am writing in short form as summation of alpha 1 please read this as I am adding up alpha 1 alpha 2 till alpha n it is actually minus 1 a1 by a0 that is what you used to call it as minus b by okay if you take the product 2 at a time that means if you keep on taking every 2 terms or every 2 root and multiplying it by the way how many such values will come up if you take any minus 1 no nc2 nc2 will come okay write it as alpha 1 into alpha 2 as a short form the beta relation says it will be minus 1 square a2 by a0 okay in a similar way if I take 3 at a time that means sum of product of 3 at a time by the way how many such terms will come up nc3 nc3 terms okay so this will become minus 1 to the power 3 a3 by a0 do you see a pattern in this so if I keep doing this if I take summation of product of k at a time of course k should be less than equal to n then this sum would be what okay this is called the beta relation so many people ask me how do you prove it see the proof is very easy if you take your polynomial equation let's say this is your polynomial equation and you know your alpha 1 alpha 2 are the roots can't you break it up like this yes okay if you expand this and compare the coefficients of like powers compare the coefficients of like powers of x you will end up getting all these relations that you have seen okay so you end up getting the beta relation is that fine any question here okay so not only in a quadratic you can apply it to a cubic you can apply it to a bi quadratic you can apply it to any degree polynomial equation okay I'll give you a simple example for example if you have a cubic equation let's say ax cubed plus bx square plus cx plus d equal to 0 and let's say roots are alpha beta and gamma okay so beta relation says that is summation alpha this is what we are I'm writing it as a shortcut for this it's actually minus b by a summation alpha beta which actually is a short form of saying alpha beta beta gamma gamma alpha that is going to be c by a and your alpha beta gamma that is nothing but minus d by a so this relation will hold to for cubic as well is that fine now on this I have a very very interesting question okay in fact don't solve it I'll be solving it okay all of you just read the question first okay done with the reading part yes sir so the question says a1 is greater than a2 greater than a3 and so on till a6 p is that expression q is the other expression r is another expression and they have given you a cubic equation here and they're asking you which of the following options are correct so it's a multi correct okay let's say it's a multi correct okay let me solve this question but before I go into this question I have an important theorem to discuss with all of you okay so please be attentive the last 20 minutes of the class I think we should be able to wrap this up the concept is intermediate value theorem IVT the intermediate value theorem is actually a very obvious theorem which probably you will realize once I explain it to you it says that if you have a polynomial equation let's say f of x equal to 0 okay so let's say it's a polynomial equation okay even though it can be applied to other equations as well but right now since I'm talking about polynomial equations let's say I talk about this as a polynomial equation polynomial equation okay such that there are two values alpha and beta for which the polynomial function over here has opposite sign are you able to understand this so let's say there's a polynomial equation and there's a polynomial function over here and there are two values alpha and beta which you feed into this function they will give you two answers which are of opposite sign okay then this theorem says that odd number of roots of f of x equal to 0 will lie between alpha and beta now let me explain why it happens see let us say I make the graph of this function a rough graph okay so let's say the graph is like this okay alpha is here let's say beta is here now you yourself will realize that the value of f of alpha which is this value is positive okay and the value of the function at beta which is here f of beta is negative that means they are of opposite sign then you would realize that one root will definitely lie between alpha and beta now I have written odd roots why I have written odd number of roots because let's say your beta was here then still they would have opposite sign because now f of beta will still be negative in that case you will have three roots between them are you getting my point so your number of roots that will lie and beta will always be odd number will always be a odd number of roots okay so either it can be one root or it can be three roots or it can be five roots or it can be seven roots that depends upon what type of polynomial you have or what is the degree of polynomial you have is this understood or not any question with respect to this okay if there is no question then there is another follow up if f of alpha f of beta has the same sign okay then it implies even number of roots even number of roots of this function of this polynomial equation will lie between alpha and beta will lie between alpha and beta I will repeat it once again aditya let me finish this off now why even number of roots so again see look at the same graph I am redrawing it again so something like this okay let us say alpha is here and beta is here then only the value of f of alpha and f of beta will be having the same sign correct let's say yeah if you see if that were to happen two roots will lie between alpha and beta now many people say alpha let's say beta is here also then same condition will hold true in that case zero roots will lie but zero still is an even number okay now many people say what if it touches and goes so let's say alpha is here beta is here still both are positive never mind you still have even number of repeated roots so this is a repeated root okay let's say this value is a then a is a repeated root so a and a are two roots still the condition will hold good is it understood I'll repeat once again if let's say f of x equal to zero is a polynomial equation given to you right and there are two numbers alpha and beta for which the function here which is on the left hand side gives you opposite signs then it implies odd number of roots of this equation lie between alpha and beta if they give you the same sign that means even number of roots of f of x equal to zero lie between alpha and beta is that clear aditya manjunath okay now with this understanding let us solve this question okay I'll do one thing I'll copy this question on the next page also so that there is not hodgepodge over here as there would be a hodgepodge this was the question sir it doesn't matter whether roots are real or not right no no no I'm talking about real roots only sir but even if it's not then there will be zero roots between alpha and beta so one bit like even the root oh yeah zero real roots I'm talking about I'm talking about the roots here being real okay okay now let me solve this question for all of you all of you please see here okay in the interest of time I'm solving this so let us first write down this equation 2x cube minus p what is p a1, a2, a3, a4, a5, a6 okay x2 plus qx q is this guy a1, a3, a3, a5 a5, a1 a2, a4 a4, a6 a6, a2 minus r minus r so minus let me write it slightly more to the left minus a1, a3, a5 a2, a4, a6 now this is the pattern in which these numbers are written you can see here all the odd ones have been paid up all the odd ones have been paid up all the even ones have been paid up all the odd ones have been paid up all the even ones have been paid up okay so now what I'll do is taking clue from that I will break up this equation into two equations see how one of the x cube I'll take from here and all the odd numbers I'll pick up from here a1, a3, a5 okay all the odd ones I'll pick up from here a1, a3 a3, a5 a5, a1 okay and from here also all the odd ones I'll pick up okay plus plus then the remaining x cube and now this time the remaining even ones a2, a4, a6 a2, a4 a4, a6 a6, a6, a2 and from here minus a2, a4, a6 okay now everybody has understood till this step so what I did I broke this up as two different equations are two different you know expressions under the same equation is that clear everybody no doubt okay now recall beta's relation if you recall beta's relation this expression that you see which I am underlining with yellow it's actually nothing but x minus a1, x minus a3 and x minus a5 if you expand this up you'll end up getting this yes or no plus now this expression which I am underlining with white this is again from beta's relation x minus a2 x minus a4, x minus a6 do you agree with me so tactfully what I did this equation that was given to me I broke it up as two such expressions everything is fine till this stage please let me know could you scroll up a bit yeah sure clear please type clear on your screen so that I know it's clear to you clear okay good now listen to the music let's say I call this as my f of x so basically there is a polynomial equation here that you are handling f of x equal to 0 isn't it now look at the options option says one root lie between this two root lie between this blah blah blah so they are all talking about how many roots are lying between these numbers okay so for the first option I have to see what is the sign of this polynomial f of x for a1 and a2 so can I do this act can you first find out what is the sign of f a1 now let me put forget about 0 as of now this you forget as of now we will talk about it let us just check f of a1 so in this function if I put x as a1 what will I get first part will be 0 second part will be a1-a2 a1-a4 and a1-a6 all of them are positive positive because a1 is greater than everything a1-a2 is also positive a1-a4 is also positive and a1-a6 is also positive so it is positive value okay now what about f a2 f a2 can anyone tell me f a2 positive or negative let's erase this again thank you a2 will make this fellow as 0 and this will become a2-a1 a2-a3 and a2-a5 negative positive positive so yes negative okay now if there is a sign change happening what does it mean odd number of roots of f of x equal to 0 lie between a1-a2 so let me make a dummy graph let me make a dummy graph since this is a polynomial since it is a polynomial with the leading coefficient as 2 so as you can see the leading coefficient here is 2 so it will come from down like this see all like this okay and a1 has to be on the right so a1 is somewhere over here okay so a1 is positive now you are saying f a2 is negative that means a2 could either be at this position that means somewhere between these two roots so these are the three roots or it could be here also right so either one root can lie or three roots can lie I can only say odd number of roots can lie I am not very sure whether one root will lie or three roots will lie so I will as of now park this because I don't have an idea whether one root will lie or three roots will lie okay so I am not in a position right now to say right or wrong for option a okay so now I will move on to option b what it says this says two roots lie between a1 and a3 so a1 already we have seen the sign sorry let's see the sign of f a3 can you help me with the sign of f a3 again let me erase this a3 will make this as zero because there is an x-a3 over here but this will become a3-a2 a3-a4 and a3-a6 so this is negative positive positive correct so this is a negative quantity okay now if it is a negative quantity either a3 can lie here or a3 can lie here there are two positions also for a3 correct but if a3 lies here then there will be either one root between a1 and a2 or there would be three roots between a1 and a3 sorry if a3 lies here there will be one root between a1 and a3 not a2 by mistake I said a2 or if a3 lies here there would be no root between a1 and a3 so under no condition we will have two roots so this option is definitely wrong this option is definitely wrong does it make sense okay but unfortunately still I have not got an answer for a so let me move further let's check a1 and a4 so a1 is already known positive let's say a4 can you help me with the sign of f a4 positive okay this will become 0 okay this will become 0 a4-a1 a4-a3 and a4-a5 a4-a1 is negative a4-a3 is also negative a4-a5 is positive yes correct it's positive it's positive it's positive means if a2 lies here a4 has to be to the left of a2 and it is positive so it is somewhere over here because it can't come over here then it will be negative if a4 is here that means my a3 cannot be at this position so these two are wrong positions so my a2-a3 must be here and if my a2-a3 is here now I am in a position to answer my option 1 that means this is correct sorry this is correct okay is that clear why my a2 has to be in this position not on this position because a4 is greater than a2 right so it cannot lie here it cannot lie to the right of a2 okay it can go on this side okay a4 is lesser so it has to go to the left okay now two roots lie between a1 and a4 two roots lie between a1 and a4 yes that is also correct so option c is also correct let's look into d two roots lie between a3 and a5 so I need to know a5 sign also can you guys help me with a5 sign I think it's positive Sana is saying again positive are you sure Sana yeah I can this is negative positive yes that's again positive it has to be somewhere over here okay it can't come in this branch so between a3 and a5 only one root lies that means even this option is wrong let me write it in back even this option is wrong so can you explain a5 again okay a5 if I put the values what do I get positive yeah positive so positive means the function should show positive nature so I can't have a5 here because a5 will become negative then so it has to be in this loop only somewhere between this point and this point it has to be between this point and this point okay so I just chose an arbitrary position it's in this position so between a3 which is here and between a5 which is here there is exactly one root which I am showing with this white dot okay that's why option d cannot be correct is that fine guys this is a typical je advance question this type of analysis is what je wants you to be clear with okay I know it has been a very tiring session I will not teach any further but I will just stop after telling you what is the things which I left for quadratic equation sorry sir why did you start the graph going from down to up okay yeah it's a very good question whenever you have a qubit polynomial let me go back whenever you have a qubit polynomial with leading coefficient as positive this is called the leading coefficient this is called the leading coefficient if leading coefficient is positive your cubic equation graph will always start from down and end upwards okay I will explain you if ax cube plus bx square plus cx plus d is to be plotted and if a is positive the graph is always like this it may not cut everywhere I am just giving you a rough example it could be like this also it could be like this also okay but the left hand side will always go to minus infinity and the right hand side will always go to plus infinity why is that so is because if this is positive and your x becomes very very negative in value then the entire function will move towards negative infinity because this is the dominant term similarly if your x becomes very very large then the entire quantity will move become very very large irrespective of whatever is the sign of a b c and d that's why when you go towards plus infinity this graph will also go towards plus infinity and if you go towards minus infinity the graph will also go towards minus infinity okay on the other hand if you have a as negative your graph will look like this ulta so in the left hand side when you are going towards minus infinity it will tend towards plus infinity and on the right hand side when you are going towards infinity this will tend towards minus infinity getting the point okay i have a question if i have a cubic in y like this okay tell me if a is positive how would the graph look like okay i am giving you two options like this or like this option a or option b b b b it's option b which is correct because if a is positive and your y is becoming very very large even x will be very very large both are going towards infinity here so x is becoming infinity y is also becoming infinity so b is the answer if your a were negative then a would have been the answer okay there is something related to the inverse function which i will do the inverse yeah meanwhile the agenda for the next class i will talk about common roots first okay i will talk about common roots this will be the first thing that i will discuss then i will talk about location of roots and then i will talk about polynomial equations in general what is rule rose theorem cardens method of solving i am not sure i will be able to do more than this so this will be the agenda for the next class so thank you so much i am going to stop here