 So quadratic equations, before I talk about quadratic equations, I would like to comment upon, I would like to comment upon the general difference between equation and identity. Okay, so what are the difference between an identity and an equation? So identity versus equation, who will tell me the difference? What are the difference between an identity and an equation? I think this was something which I talked about during the bridge course also. Yes, Ashindya, I think offline, online, whatever is the issue, if it is offline, we'll be coming to your school campus. Okay, 315, normally we normally reach the school campus and we run the class till 7, 7, 15 in the premises only. Okay, if it is online, because the online offline, this thing is actually slightly shaky right now. Is your school started offline? I mean, are you having your semester exams offline? No, I think you told me it's happening through online, exam.net. Okay, so depending upon whatever is the situation, if it is offline, we will be there in your school campus. If it is online, then how it is continuing. But mostly 90% chance is that it should convert to offline. Okay, yes, exactly. So identity is what? Identity is basically a equality relation between two expressions. Equality relation between two expressions, between two expressions, two expressions that holds, that holds true for almost all values of the variables. Let me write all values of the variable or variables. Okay, under the subject to the validity of the expression, subject to the validity, valley, what happened to my spelling today? So today I'm sad that last class of 11th and I'm not going to see you for the next one month. Oh sorry, not see you, have classes with you. Subject to the validity of the expressions. Okay, now see, I have used a very, very interesting word. I have not used the word equation. I have used some kind of an equality relation between two expressions, between two expressions, that hold true for almost all values. Almost all values means it might not be all values. But let's say it can be true for all the values. And why almost not all values? Because it is subjected to the validity of the expression. See, example is, let's say if I say sine square x plus cos square x is equal to 1. This is a identity. Why? Because it is going to work for all the values of x and in fact, all real values of x. Now, in international books, they normally write it with triple dash over s, showing that these two are actually equivalent relations. That means both of them are stating the same things. Okay, left hand side is also one, right hand side is also one. But if I talk about something like this, tan square x plus 1 is secant square x. I'm putting the three dashes here. Don't think that your books are also going to do the same. This is, I'm just trying to copy the symbols that is used by international books. This is true for all values of x provided it should not be an odd multiple of 5 by 2. So identity doesn't mean it's going to be true for all the values of the variable. They are true for almost all the values, but subject to the fact that the functions given to you or the expressions given to you should be real and defined. So don't start putting it, putting into that expression anything that you want under the name that it's an identity. All right, so what's an equation on the other hand? So this is an identity, my dear. This is an identity. So what's the equation on the other hand? Equation on the other hand is basically the same concept is just that they are true. I might not write everything down, but they will be true. Let me write it in white. Yeah, they will be true for certain values of the variable, certain values of the variable or variables, which are called the roots. So these are called the roots of that equation. Of course, again, subject to the fact that it makes the expression of valid one. That means the expression should be real and defined. Now, long, long back ago, see today we are doing quadratic equation. Quadratic equation is the type of polynomial equation, isn't it? Cubic equation, bi-quadratic equation, pentic equation. These are all examples of polynomial equation. So long, long time ago, there was a German mathematician called Karl Friedrich Goss. Okay, Goss, you would have heard. He gave a very important theorem in mathematics, which is actually called the fundamental theorem of algebra. And Karl Friedrich Goss in his fundamental theorem of algebra stated that any polynomial equation of degree n will have exactly n roots, whether real or non-real. That's a different thing, but it cannot have more or it cannot have less. If it has less means there is some repetition in the roots happening. For example, if a quadratic equation, if I talk about, it's a degree two polynomial equation, so it'll have two roots. Now, both the roots can be same also, like one one, let's say x minus one whole square equal to zero. There's only one root, which is one, but he said it is a repeated root. So there will be two roots, which will be one one. Okay, if let's say if I talk about x square plus x plus one equal to zero, it will have two complex roots, omega and omega square, but two only will be having. It will not be having more than that, it will not be having less than that. Okay, and this particular theorem given by Gauss is not violated. That is why this is called the fundamental theorem. It's not going to get violated. So when I talk about an equation, let's say I give you a very surprising equation. Please consider this interesting example. Please consider this interesting example. Okay, so let me give you an equation to you x minus a x minus b by c minus a c minus b x minus a x minus c by b minus a b minus c and x minus b x minus c by a minus b a minus c equal to let's say one. Let's say minus one equal to zero. Okay, now if you look at this equation prima facie and of course your a b and c should be all distinct. Why? Because if any two of them are equal, your denominators are going to become zero, which you don't want to happen because it'll make the expression undefined. So here, if you see prima facie, this looks like a quadratic equation. So people who will look at it will say this seems to be a quadratic equation, isn't it? Doesn't it look like a quadratic equation? This is a quadratic term. This is a quadratic term. This is a quadratic term. Okay, so overall it's a quadratic equation. So as per fundamental theorem of algebra, it should have two roots, not more than that. Okay, so now I'll give you a surprise here. Try to check by putting x equal to a. Does it satisfy the equation? Does it satisfy the equation? Let me call this as E equation. Does it satisfy E? Yes, it satisfies E. Correct? Okay, now try putting x as b. Does it satisfy the equation? Put b in place of x and check, is it satisfying the equation? Yes, it'll satisfy. Very good. Cello. Put c. Does it satisfy the equation? And to our surprise, it does. Now, how is this happening? Because it is supposed to be prime of SE, a quadratic equation, and it is showing me three roots. Ayurama, why? How is it possible? Now, one part of the definition I did not tell you, he added that if let's say a polynomial equation of degree n shows more than n roots, then that polynomial equation actually becomes an identity. In the other words, that particular equation will not have only two or three or four roots. It will have infinitely many roots possible. So that will become an identity. So right now, many of you got deceived because actually I wrote an identity in a very camouflage way, in a very discrete way that nobody was able to identify. You know what I wrote actually? So you'll be surprised to know I wrote something like this, 0x square, 0x plus 1 minus 1 equal to 0. So this equation is actually this. And it could work, it would work for any value effects that you want to put. It will not only work for ABC, you put anything inside it, it'll work. Okay. So what I gave you was actually an identity. So this becomes, this is actually was an identity in disguise. So when we're trying to look at any equation, let's say a polynomial equation, many times the question setter would have actually given us an identity in the shape of a quadratic. So how do we identify whether a given equation, a polynomial equation is an identity or not? So in that case, please understand, you just look at the coefficients present to the left and the right. Ideally, all of them should be 0, 0 each. If the right side is 0, ideally all the coefficients should be 0. Okay. That is why these equations will work for any input you provide to them. So whether you put A here or a B here, or a C here, or any other value, it's going to become a 0 because there is a 0 sitting in front of it. And of course, this will ultimately also become a 0. Okay. That's why 0 is equal to 0 is happening, which is a universal truth. Okay. So now I'll give you an example. I'm sorry, not example, a question based on the identity to crack. Okay. So let me give you a simple question. Question is this particular equation, this particular equation has or is satisfied by, is satisfied by, by more than two values of X, more than two values of X. Okay. Find the value or values of lambda. Find the value or values of lambda. Okay. Simple question based on the theory that we covered in the previous slide. I'm sure everybody can answer this. So this has more than two, satisfied by more than two values of X. Yes, done. Excellent. Satya, very good. Harshita. Okay. Now, see everybody please pay attention. So as per the prima facie, you know, equation as it appears to me in the first glance. So for my prima facie experience of this equation, I can see that it's a quadratic. So as per Karl Federich-Goss fundamental theorem of algebra, it should have exactly two values of X, which shouldn't be satisfying it. But the question setter has given it a satisfied by more than two values, which means this actually equation must be an identity. And if it is an identity, no matter whatever X you put on it, it should always be giving you a zero on the right side. And this can always only happen when the coefficients are becoming zero together. That means the coefficient of X square X and the constants, they are simultaneously zero. So what we have to do is we have to take the intersection of the values of lambda, which we get by solving these three, that value is going to be the value that will make this an identity, which will make it you can say immune to any value of X you put, it's just going to give you zero every time. Okay. So everybody knows that this is factorizable as this, which is nothing but two and three. Okay. This is satisfied for one and two, or you can say two and one. And this is also satisfied for two and minus two. So out of these, you'll see that two is a value which is common to all of them. Yeah. So if you choose a lambda value as two, it will make everything zero simultaneously. So the value of lambda is actually a two here. So this is your answer for this question. Is it fine? Okay. If you put a one, see one will make only the second one zero. The first one still will be dependent on your, the first one will still give you a value which is dependent on X. Okay. And obviously the constant term will also become a minus C. So it did not be zero for all values of X. So if it is an identity, it is going to be true for all values of X, not only three or four or five or six, it will be true for so many values. Okay. So if it is not an identity, if it is an equation, it should have exactly n roots. And if it is having more than n roots, it definitely becomes an identity. There is nothing in between. Right? It cannot happen like, sir, is it possible that it'll have seven roots exactly being a quadratic equation? No. Either two or infinitely many, nothing in between. Is it fine? Okay. Now coming to a equation and in fact, a quadratic equation, everybody is aware what's a quadratic equation. I had a brief discussion about it in the bridge course as well. So equation of this nature AX square plus BX plus C equal to zero, where A, B, C could be complex numbers. Now, when I say complex numbers, it will also include real numbers into it. Okay. Proof for, I'll definitely send you one if I find it. Okay. But mostly they state the theorem like this. Even in Olympiad books, if you see, they'll just state the theorem. But I will check for that. I'll just check the higher algebra books and find the proof if at all is there. Okay. Now, in a quadratic equation, please understand A cannot be zero because if A is zero, it will actually become a linear equation. Now, we normally categorize quadratic equation as two types. One which is called a pure quadratic equation. Okay. What's a pure quadratic equation? A pure quadratic equation is where your B is zero. So A should not be zero. B is zero. And C could be anything. Okay. C could be any complex number. Second type is called an adfected quadratic equation. Adfected. A name given to it. There is nothing very important about it. It's not like the J people will ask you what are the types of quadratic equation. But just for your general, you know, knowledge. So here B and A and B are not zero. C could be any, any complex number. Okay. So we'll talk about pure quadratic equation in our upcoming academic year where we'll be discussing a few types of quadratic equations. Sorry, a few types of integration problems involving pure quadratic expressions. Okay. Anyways, so let's talk about roots and nature of roots of a quadratic equation. Okay. Now, I know many of you would be saying I already know this. I already know the Sridharacharya formula, but I would be trying to put in some geometrical, you know, analysis also into this, which actually I covered in the bridge course. So I'll be slightly going at a faster rate. If I talk about a quadratic equation, sorry, if I talk about a quadratic equation, a quadratic equation basically gives you some values of x, right? We satisfy this particular quadratic equation. Okay. So this gives you some roots, right? Now these roots are basically nothing, but they are points of intersection. They are points of intersection of these two curves, y equal to ax square plus bx plus c and y equal to zero, which happens to be your x axis. Okay. So when you're trying to solve this equation, it is like asking yourself, where are these two curves actually intersecting? Okay. The points of intersection is what we call as the roots of the quadratic equation. Now, this particular expression is a familiar term to you because you have already done parabola chapter in quite depth. So this is basically going to be a parabolic curve. Okay. So this is a parabolic curve or a parabola, you can call it not parabolic. Let's call it as a parabola. Okay. Now this parabola has, okay, has its axis parallel to the y axis, correct? You're already aware of this. So such kind of a parabola will have their axis parallel to the y axis. If it was x equal to a quadratic in y, it would have been a parabola whose axis was parallel to the x axis. Okay. I hope this is not a news to you. You're already aware from the parabola chapter. Now a is a very important, you know, term over here, which we call as the leading coefficient. So a basically decides the concavity of this parabola. So if a is positive, then your parabola will be opening upwards. Okay. Having its axis parallel to the y axis, but opening upwards. Okay. You're already aware of this. If a is negative, it will be a parabola which will be opening downwards. Okay. So what I'm going to do here is, okay, I'm going to, I'm going to start with the parabola equation and y equal to zero, I will substitute it at certain stage. So let me just begin with the equation of this parabola. I want to show you something very interesting. Okay. So let's start with y equal to ax square plus bx plus c. Okay. So here what I'm going to do, I'm going to just complete a bit of square over here like this. So a times x plus b by 2a, the whole square, minus b square by 4a square plus c. Okay. Or opening the brackets, you'll end up getting something like this. Correct. Or you can write it in a much refined way as this minus b square minus 4ac by 4a. Okay. All of you please look at this expression. Now here in this expression, I will now substitute the y as zero. Okay. Just to make, just to find out the points where it is intersecting the x-axis. So when you put y as zero, you will end up getting a situation like this. Okay. By the way, if you permit me, I can write this term as a D. Okay. Where D is your expression, b square minus 4ac. Now, why does given a special treatment? Why are we calling it as by a different name? It is basically the discriminant of the quadratic equation. And I will talk about the discriminant in some time, how it is useful. So when you take the D to the other side, in fact, let me rewrite it like this. It becomes something like this. So if you bring this a down, it becomes D by 4a square. Take the under root on both the sides, it will give you plus minus root D by 2 mod a. Now remember, mod a becomes irrelevant for us because there is a plus and minus both sitting over here. So we don't write actually a mod a, we just write it as an a, right? Because even if a is positive or negative, it has to undergo both the plus minus scenarios, isn't it? So from here, you get x value as minus b by 2a plus minus root D by 2a. Okay. Yes or no? And this gives us the two roots, the two roots as minus b by 2. In fact, minus b plus, let me start with the minus. Yeah, this is one root and this is another root minus b plus root D by 2a. Okay. So two roots come out from this, but this is something which we are, you were already aware of. Okay. And this formula that you have seen here, this formula is actually called as the quadratic equation formula or the Sridharacharya formula, right? Quadratic equation or Sridharacharya formula. Okay. Now I would show you something related to the graphical aspects of it also. Yeah, sorry, a lot of drilling noise was coming here. Now here, I would like to connect this thing with the, the vertex of the parabola. So how is the vertex of the parabola very important? But before I proceed towards that, let's analyze this formula. Let's analyze the quadratic equation formula, analysis of Sridharacharya formula, Acharya formula. So when we talk about this equation, minus b plus minus root D, which is b square minus 4ac by 2a. Okay. When we talk about this formula, this term b square minus 4ac, which we call as the discriminant. By the way, many people call it as determinant. It is actually discriminant. Discriminant is something which helps you to discriminate between the nature of the roots. So three situations arise. If your discriminant is positive, we say that the roots, these are your roots. The roots are real and distinct. The roots are real and distinct. Okay. Now, how do we infer this from our graph? Now, all of you, please focus on these two graphs, which I'm going to show you on the screen right now. Okay. So I'm going to make the graph of ax square plus bx plus c equal to y for two cases. One case is where your a is positive. So if a is positive, your graph will be opening upwards like this. Correct. And when your a is negative, your graph will be opening downwards like this. Correct. Now, we know that the points where it is meeting the x-axis, those are called the roots of the quadratic equation. Correct. So these are the roots of the quadratic equation. Okay. Now, this condition that you have on your screen, when D is greater than 0, you get real and distinct roots. How is it associated with this graph? Now, let us understand that this vertex, this vertex is basically a point whose coordinates is minus D by two a comma minus D by four a. Now, how, how is it that? And you'll be surprised that even for this also, the expression is the same. So it doesn't depend on a whether, whether a is positive or negative, your vertex is still at minus b by two a comma minus D by four a always. Now, how do I get this? How did I get this coordinate? See, to get this coordinate, I'll go back to the previous slide that we had discussed here at this stage, if you realize, at this stage, if you realize, let me write a D for this. Yeah. Now, all of you please pay attention. When your parabola is opening downwards, opening upwards, your vertex position is where your y is minimum, isn't it? So if you want this expression to be minimum, minimum least, what are you going to do? So here, if you see this guy is the only guy which is changing, this is fixed, this is fixed, and this is the variable guy. Okay. And that to this variable is always greater than equal to zero. So if you want your y to be the least, what would you do? You will say, I will put this variable to be the least value so that, you know, I whatever I get is the least value of the entire expression, because the second guy is anyways fixed. It's not going to change. So you're going to put your x value as minus b by two a in order to make it the least, which is zero. So for that, you have to make your x value as minus b by two a. And when you do that, this entire term will go for a toss and you'll be left with minus d by four a as your y value. And this actually becomes your coordinates of the vertex of the parabola. Okay. This is already known to you. Right. Now, what happens if your parabola was opening downwards? Why didn't the vertex coordinates change in this case? In this case, vertex is a point where your y coordinate is the maximum possible value. Right. Now here, the situation will slightly change. Here, this is variable, but this variable will now be lesser than equal to zero. Why? Because a is negative. If a is negative and this is a perfect square, the whole thing is going to be a negative quantity. Right. So this fellow, the first guy is reducing the value of the entire expression. Isn't it? So if I ask you a simple question, what would you do to that variable so that the value is maximum? What will you do? You'll say, sir, the guy which is reducing the value, I will make it zero so that it doesn't further reduce my value. So for that to happen, x will be again minus b by two a and y will be again minus d by four a. So that's the reason why the vertex coordinate irrespective of whether the parabola is opening upwards or opening downwards remains the same. Okay. So that is what I have written over here. Okay. Now see, if you want your roots to be real and distinct, that means you want the parabola to cut the x-axis. So when your a is positive, should your vertex be below the x-axis or above the x-axis or on the x-axis? What will you do? What will you say? You'll say, sir, it has to be below the x-axis. Then only the branches will go and cut the x-axis. So if it is hanging like this, it will not get the x-axis. So no real roots will be generated or if it is touching only and going, again, you'll say no real roots will be generated. So the diagram which I've shown you, that is the exact scenario that should happen if you want your roots to be real and distinct. So for this, your y coordinate of the discriminant should be negative. That means minus d by four a should be less than zero. That means minus d should be less than zero because four a, you can take it to the other side without affecting the inequality. Correct? So that means your d should be greater than zero and this is what I had written over here. So when your d is greater than zero, your roots will be real and distinct. So that is further supported by your vertex, I mean, the graph of the parabola using that vertex coordinate. Are you getting my point what I'm trying to say? So if somebody asks you, can you justify why discriminant greater than zero is required for you to have real and distinct roots graphically? Graphically. Then this is the justification for it. Getting the point. Similarly, we also know from the Shridharacharya formula that if d is zero, the roots will be real and equal. The roots are real and equal. Roots are real and equal is basically a scenario where your graph is going to, many people ask me this question. Sir, what about this case? What about this case? So if a is negative, does the same situation be greater than zero hold for real and distinct roots? Yes, the same condition will hold true. See how in this case, your minus d by 4a should be positive. Why? Because if you want your graph to cut the x-axis at two places, your vertex should be above the x-axis if your a is negative, right? Which means minus d should be less than zero. Why? Because 4a will be negative quantity. 4a is a negative quantity. Correct? Yes or no? So if you take the 4a to the other side, your inequality will stop or will flip. So that means d should be again greater than zero. So it doesn't matter. It actually doesn't matter whether your a is positive or negative. Your discriminant should be greater than zero to get real and distinct roots irrespective of whatever is your sine of a. Okay. So this a d greater than zero is a condition which is going to be universally valid for both the situations. Okay. Now when you talk about real and equal roots, your discriminant should be equal to zero. Does the graph justify it? Yes, it does. Why? See, if you want your equation to have real and equal roots, your vertex should just touch the x-axis. So if your vertex just touches the x-axis, please note, your minus d by 4a should be zero, which means d should be zero. Correct? And from here we can actually know what's your roots also because this point is your minus b by 2a. So it will have real and equal roots and each one will be equal to minus b by 2a. Okay. So please make a note of this. And this situation doesn't change even if the parabola was opening downwards. The same situation will be valid here as well. Okay. So we'll go and just touch the x-axis like this. Thereby, the vertex will have a y-coordinate which is equal to zero. Okay. So this will be zero and hence d will be zero. Is it fine? Any questions? Any concerns here? The third situation is where your discriminant is less than zero. Discriminant is less than zero means your roots are, roots are non-real in nature. In fact, I would say roots are complex in nature. Roots are complex in nature. Complex in nature means it will not be appearing to touch the x-axis. So let's say I take these two scenarios. So when your roots are complex, it may be hanging up in the air like this. Okay. Or hanging down like this. So you will not see a visual cutting happening of the x-axis with the parabola. Okay. So in such situations, what will you say? You will say that the vertex of the parabola should be positive in the case where a is greater than zero. That means minus d by 4a should be positive, which means minus d should be positive, which means d is negative. In our case where a is negative, this discriminant, sorry, this vertex should be having a y-coordinate, which is negative. So minus d by 4a should be negative in this case, which means minus d should be positive because 4a is negative. Isn't it? If 4a is negative and if you multiply throughout with 4a, your inequality sign will switch, which means d should be again less than zero. So same condition will arise here also. Okay. So your discriminant should be negative if you want your roots to be complex. And please note, please note that these roots will be complex conjugates of each other. So if one root is let's say p plus iq, then the other root, if one is p plus iq, then the other root will be p minus iq for sure. If your coefficients of the quadratic equation, they are all real in nature. I will talk about this little later on. Okay. First of all, this concept is clear to you that how vertex plays a very critical role in understanding the nature of the roots. Is it fine? Any questions, any concerns here? Anybody has? So graphically also we have understood it. Mathematically, we already knew since our class 10 days. Okay. Any questions, any concerns? All right. If there's no question, no concerns, we'll talk about, we'll talk about some important points. The first thing that I would like to discuss here is that if a, b, c are all real number, when I say real means purely real. Okay. No non real part should be there with it. So purely real. Okay. If one of the roots, roots of the quadratic equation is p plus iq. Okay. Then the other root must be, a root must be p minus iq. Okay. Now, why it is basically happening like this? See, very simple. Let's say if one of the roots is, one root is p plus iq. Okay. And let's say for the sake of generality, I take the other root to be, maybe, what should I take? Give me some two variables, x plus i y. No, x I don't want to use. Alpha plus i beta also I don't want to use. R, oh, very good. R plus i s. Thank you. So let's say I use R plus i s. Okay. Now, there's a very important relation that you would have already learned in class 10. That is called the Vita's relation. So by, from Vita's relation, I'll talk about this also from Vita's relation. Vita v capital. From Vita's relation, we know that the sum of the roots, the sum of the roots should be minus b by a. Okay. Now remember, if a, b, c are all real number, this is a completely real number. That means I can safely say it is minus b by a plus 0 i. Now, if I group up the terms here, as per our complex number, algebra, what I realize is that q plus s should be 0, which means your s should be minus of q. Okay. By the way, here, when I'm assuming that the root is p plus i q, I'm assuming it under the fact that q is not 0. That means I'm saying one of the root is actually having a non-real part with it or imaginary part with it. Okay. So your s is actually minus q in this case, right, which makes our root. So the other root is actually now becomes r minus i q, isn't it? Okay. Now we also know from Vita's relation that the product of the roots, the product of the roots is c by a, isn't it? Okay. So we also know this. Now here also c and a both are real numbers. So I can say it doesn't have an imaginary part. So I'll put a zero i next to it. So now again, pull out the imaginary part and equate it to zero because right hand side we have the imaginary part as zero. So there'll be some real part. I'm not worried about it. But what will be the imaginary part on the left side? Okay. Let's try to equate it to zero. So this will be minus p q plus q r. Right. And there is something over here, which I'm not interested in. So can I say from here that p q should be equal to q r? Correct. And since q is not equal to zero, can I say r is equal to p? That means the same root will now become, so this will now be transformed to p minus i q. Right. Thereby signifying that the other root should be the complex conjugate of the given root. Okay. But this is only to be used, my dear, if your coefficients of the quadratic equation are all real, this is a very, very important criteria. If any one of the coefficient is non real in nature, you cannot claim this. You cannot claim that if one root is 2 plus 3i, other will be 2 minus 3i or if one root is 3 minus 4i, other will be 3 plus 4i unless until the coefficients of the quadratic equation are purely real quantities. Is it fine? Any questions here? Okay. Anybody wants to write this down? Please do so. Else I will move on to the next important point to be noted. Second point to be noted is that if a, b, c are all rational quantities and one of the roots of the quadratic equation is irrational, is irrational. Okay. Then the other root, then the other root must be the rationalizing factor of the given root. Okay. That means to say if one root is p plus root q, okay, where root q is a, this is a third, okay, then the other root will be p minus root q. Are you getting my point? Now, why is this happened? Because again, you want the sum to be rational because sum will be, sum will be minus b by a and b and a are all rational numbers. So sum will be rational, product will be rational. So if sum and product both are rational, it can only happen when the two roots themselves are rationalizing factors of each other because on adding, the third will get canceled and on multiplying, the third will get squared up. Are you getting my point? Okay. So very, very important. So if you realize that your coefficients are all rational, this is rational. Okay. And one of the root is let's say two minus root 3, the other will become 2 plus root 3 by default. Is this fine? Okay. The third thing that I would like is actually the Vita's relation, which you are already aware, but I will like to talk about it a little bit more. See, Vita's relation is basically a relation which relates the coefficients of any polynomial equation with its roots. Again, I'll repeat, it's a relation which relates coefficients of any polynomial equation with the roots of the polynomial equation. So let's say if you have a polynomial equation of this nature. Okay. Any polynomial equation. And when I say polynomial equation, your a0, a1, etc., they can all be complex numbers. Okay. So don't be like they have to be real, purely real. No, not necessarily. They can be complex also. So Vita's relation will work even for those polynomial equations where the coefficients are complex in nature, need not be real. Okay. So Vita, Italian mathematician, he figured out that if this particular equation has roots alpha 1, alpha 2, till alpha n, let's say, then the sum of the roots, which I will write it as summation alpha 1 will be minus a1 by a0. Okay. So if this equation has got these many roots, alpha 1, alpha 2. Now again, alpha 1, alpha 2, all of them need not be real. Some of them may be complex also. And product of 2 root at a time will be plus a2 by a0. Product 3 at a time will be minus a3 by a0. Okay. So in general, he said, so in general, he said that if you take the sum of product k at a time. Okay. This sum will be minus 1 to the power k ak by a0. Now this actually comes from a very simple fact that many people ask me, sir, is there any proof for this? The proof is just you writing this expression like this, a0 x minus alpha 1, x minus alpha 2, x minus alpha n. And just compare, compare coefficients of coefficients of coefficients of x to the power n minus 1 on both the sides. Okay. You will automatically get the first relation. You will automatically get the first relation. Okay. Similarly, if you compare the coefficients of x minus 2 on both the sides, you'll get the second relation and so on and so forth. Is it fine? Any questions? So please note this down. So in class 10s, you would have already used this relation for cubic polynomial equation, quadratic equation, but most of you would not be knowing that this is true in general, actually. So it could be used for any polynomial equation in general. Is it fine? Any questions, any concerns? Okay. So for our case, for a quadratic equation, you know it's going to be if let's say this equation has two roots, alpha and beta, then remember alpha plus beta is minus b by a, alpha beta is c by a, that is already known to you since class 10 days. Okay. Now time for some questions. Let's take some questions. All right. Let's take this question just to give you up, you know, just to make everybody recall their class 10 ways of solving equations. So a very simple question is here in front of you. Don't get scared. Solve for x and give me a response on the chat box. Okay. Are you sure Vishal, Satyam, that this is a complete answer? Okay. So we all know that these two are harmonic, sorry, these two are rationalizing factors of each other, isn't it? Because this gives you 25 minus 24, which is one, right? Clearly meaning that this guy, this guy is reciprocal of this guy. Correct. Now let's take five plus two root six to the power of x square minus three as a t. So can I say five minus two root six to the power of x square minus three will be one by t? Right. It's like, you know, you're raising x to the power two minus three on both the sides here. Okay. So what will happen to this equation? This equation will become t plus one by t equal to 10, which means t square minus 10 t plus one will be zero. Let's try to solve it using our Shridharacharya formula. So it's minus b plus minus b square minus four ac by two a, which is nothing but plus minus root 96 by two a, which is going to give you five plus minus two root six. Yes or no? So now if you relate five plus two root six to the power x square minus three with five plus two root six, you'll end up getting, you'll end up getting x square minus three as a one, which means x square is four, which means x is plus minus two. Many of you have given me this answer. Well, accepted. No problem with that. But most of you forgot that this could actually be five minus two root six also, which makes x square minus three as a minus one, which means x square could be two as well, which means x could be plus minus root two also. Okay. So there are four values to this question. X could be plus minus two or plus minus a root two. Is it fine? Any questions? Any concerns here? Any questions? Any concerns? Okay. All set. Great. So let's take the next question. It's a show that question, but again, it can always be objectified. So the question says, show that if p q r s r real numbers and t r is equal to twice of q plus s, then at least one of these two equation has real roots. Okay. So after solving it, just say I'm done. No need to give me any justification. Okay. I'll give you around two minutes for this or one and a half minutes for this. So you have to show at least one of them has real roots, at least one of them. Okay. Nobody? All right. We'll start with this quadratic. So let's say I call the discriminant as d one. So can I say discriminant here will be b square minus four ac, which gives you p square minus four q. And let's say this quadratic x square plus rx plus s equal to zero. Let's say discriminant is d two, which is given by r square minus four s. Now let's add these two discriminants, d one and d two. So let's add them. We'll end up getting p square plus r square minus four q plus s. Okay. Now we have been given that two q plus s is equal to PR. So this is given to us. Okay. So what I'm planning to do here is that I'm going to replace this minus four q plus s with minus two PR. Isn't it? Can I do that? And it actually becomes p minus r the whole square. And if p and r are real numbers, so it's only given that all of them are real numbers. Can I say p minus r the whole square will be a quantity greater than equal to zero? Which means the sum of the discriminant is greater than equal to zero. And this can only happen when at least, when at least one of d one, d two is positive. Okay. If both are negative, their sum can never be positive. Right. So one has to be positive for sure. If both of them are positive, very good. Okay. The sum will be definitely be positive. But even if one is negative and other is more positive, means having a magnitude more than the guy which is negative, then the total sum will still be positive. So for that, at least one of them should be positive, which means at least one of the equations, at least one of the equations, equations must have, must have a real roots. Hence, sure. Is it fine? Any questions, any concerns on this? Do let me know. Is it fine? Any questions? Should we move on to the next problem then? Okay. Let's take this question. If alpha beta are the roots of this equation, find the roots of the second equation. Of course, your answer will be only in terms of what is known to us. Okay. A, B, C, maybe alpha, beta like that. So please give me your answer for the roots of the second equation on the chat box, on the chat box. Yes. Anybody, I thought this was an easy question. Okay. See, when you say this equation has roots, alpha and beta. Right. Right. Can I say in that situation that this is as good as saying X minus alpha into X minus beta by factor theorem, which means I'm trying to say X minus A, X minus B minus C is as good as saying X minus alpha, X minus beta. Correct. No. Correct. Right. Absolutely. Right. Vishal. So can I say if this is true, that means this expression is going to hold true. Correct. So if somebody is asking you what are the roots of what are the roots of this equation, then it will be as good as asking you the roots of this equation. And that is very obvious that it is going to be A and B. Correct. So the roots of this equation will be nothing but A and B. Problem is done. Is it fine? Any questions? Any concerns? Any questions? Any concerns? Okay. All right. Let's take another one. Yeah. Find all roots of this equation. If one of the root is 2 plus root 3, come on. This is actually a class 10th question. But still, I would like to ask this question to you all in case anybody has forgotten that basic important point that I have given. So please solve this question. Find all the roots of this bi-quadratic equation. So this is a bi-quadratic. So it'll have four roots. Now, when I say roots, it will include real complex, all the type of roots. So one of them is already given to you. You have to find the remaining three. Let's do it and give me a response again on the chat box. Okay. So as I already told you, all the coefficients here, if you check all the coefficients, 1, 2, minus 16, minus 22, 7, they are all rational numbers. They are all rational numbers, isn't it? And if I have one of the roots which happens to be an irrational number, this is an irrational number, q bar. Bar means not q, complement. Then it tells us that other root I can actually guess. So other root has to be or one of the other roots, one of the other roots should be 2 minus root 3. Okay. It is for sure. So though two root have already found out, right? So first of all, I'll try to frame a quadratic with these two roots. Okay. And everybody knows how to frame a quadratic with two given roots. So it's nothing but x square minus sum of the roots times x plus product of the roots. Okay. In fact, I just need a quadratic expression, not a quadratic equation from here. So I can say that, that given bi-quadratic polynomial basically has this quadratic as its factor. Again, I'll repeat the word. This bi-quadrate, the one which I've shown with the curly braces, that is a bi-quadratic polynomial. See, guys, don't use the word equation and polynomial interchangeably. I've seen many people doing it. This is a polynomial equation. I agree. But just this part is a polynomial. It is not an equation. That expression is a polynomial. So what I'm trying to say is that this quadratic is a factor of the given bi-quadratic polynomial, x square, what was that? x4, sorry, x4 to x cube, to x cube, minus 16x square, minus, minus, minus, minus, minus, 22x plus 7. Correct. That means if you divide by this, you should be able to get another factor without leaving any remainder. So doing it in a slightly matureish way, rather than doing it like a class 10th grade. So this is going to be one quadratic multiplied to another quadratic. Now let me guess few of the terms of this quadratic. Since there is x4 here and there's already an x square sitting here, it has to be x square, for sure. Since there's a plus 7 here, there has to be, and there is a 1 over here, there has to be a plus 7 here also. And in between, there will be something like kx. So all I need to do is find my k. Many of you would have literally gone for a long division by this particular quadratic polynomial, but it was not required. You could have done it like that also. So for getting my k, what I will do, I will compare the coefficient of x cube. I'll compare the coefficient of x cube on both sides. Okay. So when I do that, left side will give me 2, but the right side will give me minus 4, minus 4, plus, plus k. Correct. That means k value is 6, k value is 6, which means I could factorize this guy as x square minus 4x plus 1 times x square plus 6x plus 7. Okay. So when you're equating this to 0, it means you're equating this to 0. And this is already equated to 0 and we got 2 plus root 3 and 2 minus root 3. So now I have to equate this guy to 0. So now equate this guy to 0 and get the other two roots from there. So you can use again your Shridharacharya formula minus b plus minus b square minus 4ac by 2a, which is minus 6 plus minus 2 root 2 by 2, which is nothing but minus 3 plus root 2 and minus 3 minus root 2. Okay. So four roots, four roots are now as follows. 2 plus root 3, 2 minus root 3, minus 3 plus root 2 and minus 3 minus root 2. And as you can see here, they're occurring in rationalization factors. Okay. So these two are rationalizing each other and these two are rationalizing each other. Is it fine? Yeah. Of course, Satyam. But why are you getting a wrong answer? Why is your answer different from Achintya got it right? Why are you getting minus 7 and 1? Maybe you factorize it. Ah, you factorize it incorrectly. Okay. Is it fine? Any questions? Any questions? Any concerns? Okay. So with this, we are going to a very interesting concept called transformation of quadratic equation. Transformation of quadratic equations. By the way, this concept is going to be true for any polynomial equation. Many times it works for all the type of equations. So what is this concept of, what is this concept of transformation of quadratic equation? Okay. Let me give you a brief introduction to this through an example. Okay. Very simple example I'll give you. Let's say I have a equation like this. Okay. Whose roots are some value alpha and beta. In fact, you can find it out, but I don't want you to find it out. Okay. Now I have a question to you. Okay. Question is, find a quadratic equation with roots, with roots alpha plus 3 and beta plus 3. Okay. How will you solve this question? Now, the two ways that students take to solve this question is, I'll tell you both the ways. One way is you find out the roots. Correct. You find out the roots of this equation. So you find the alpha value and beta value from the given quadratic, which is actually factorizable. Correct. So let's say I call this as alpha, call this as a beta. Right. And then find your alpha plus 3, which is going to be 4 and beta plus 3, which is going to be 5 and then frame a quadratic equation by using this. Okay. And you get your answer from there. Okay. No denying that this is not going to give your answer. Obviously, this is going to give you an answer. Very correct. No problem with that. Another way of doing it is you don't find alpha beta directly. Okay. Rather, you know that alpha plus beta will be minus B by which in our question should be 3 and alpha beta will be 2. Correct. C by, isn't it? And since I have asked you to formulate a quadratic equation with roots alpha plus 3 and beta plus 3, this is what you will be doing. I mean, essentially the same step that I did in the first approach also towards the last. Okay. And here what you will be doing, you will be literally adding these things up. So alpha plus beta plus 6 and here you will end up getting alpha beta plus 3 alpha plus beta plus 9. Correct. And now you will be using these two values, alpha plus beta and alpha beta over here. Okay. So when you use that, you'll get 9x. I'm so sorry. I forgot an x here. Yeah. So you'll get a 9x and here this will be 2 plus 3 into 3 plus 9, which is actually a 20. And you get your answer. Okay. 90% Janta will use this method. I know that. Okay. But what if I tell you a method which will not require you to use any of the two, but still will do the work for you in maybe slightly time efficient way, right? Then that method is going to be your transformation of quadratic equation method. Now what is this method? This method basically works when you want to find out the equation, a quadratic equation whose roots are, whose roots are transformed in the same fashion. Okay. So let's say alpha and beta are the roots, right? And what you see is that this, this root is obtained by adding a 3 to it. And so is this also. So there is a same destiny which is given to both alpha and beta. So whatever you're doing with alpha, the same thing you're doing with beta. So in such situations, there is a shorter way to solve this question. And that is by the use of transformation of quadratic equation. So transformation of quadratic equation, I would like to repeat works when you are trying to find out a quadratic equation having such roots, which is obtained from the roots of one of the quadratic equation, giving same treatment to them. Are you getting my point? Is it understood? So how does this method actually work? So we know that alpha and beta are the roots of, so this is the third method for you, alpha and beta are the roots of this equation, correct? Okay. And since both the roots are given same treatment, can I use one of them and say alpha will satisfy this equation, which means alpha square minus 3 alpha plus 2 is 0, correct? Now you want a equation whose roots are alpha plus 3 and beta plus 3, correct? So what I'll do here is I'll put my alpha plus 3 as x. Why I'm putting alpha plus 3 as x because I have used alpha here. So my alpha will become x minus 3, correct? Put it back over here. Put it back over here. You'll end up getting something like this. x minus 3 the whole square minus 3 x minus 3 plus 2. And there you go. This is the answer to your question because if you simplify it, it becomes x square minus 9x plus 20 equal to 0. This is your required answer. Okay. So please note it is only working because both the roots were given the same treatment. So your new roots were obtained from the old roots by giving the same treatment to both of them. If let's say one was alpha plus 5 and other was alpha plus 2, sorry beta plus 2, then this method is not going to work, my dear. This method is not going to work, right? Are you getting this point? So of course, it has got a limited use, but it really saves your time. If you realize that in your question, the question setter is asking you for a quadratic equation by using certain transformation on the roots of a particular equation and they have the same set of transformations. Then it is going to work. Are you getting my point? So I'll give you more such examples. Don't worry. I'll give you more such examples here. So first, everybody understand this, then we'll be taking in more such scenarios and you will really enjoy using this method. It's very, very time efficient in certain scenarios. Okay, done. All right. So let me ask you another question, another example. Let's say I have an equation. Now, at this time, I will not give you the equation. I'll just keep it in terms of unknowns. Okay, let's say these equation, this equation has roots alpha and beta. Can you get me an equation? Okay, so find a quadratic equation. Find a quadratic equation. Quad quadratic equation whose roots are, whose roots are alpha plus one by, let's say five and beta plus one by five. Can you find a quadratic equation whose roots are these two? Just give it a try once. See, anybody's able to crack this. Or you can just tell me what should I do with my x over here? How will this get transformed? I'll just put some brackets here for you to tell me what to do. In one shot, you can solve this by just telling me what transformations the x will undergo. You want to go from basics. So if you want to go from basics, take any one of them. Let's say this time I take a beta. Okay, so beta satisfies this quadratic equation. Correct. Right. Now your beta plus one by five, take it as x because you want this to be the root. So which means beta is five x minus one. So just put this back over here. So your answer will become a five x minus one, the whole square. Okay, b five x minus one plus c equal to zero. Plain and simple. Done. Of course, if you want to simplify it further, you can do it. So you can further write it as 25a x square. You will get, if I'm not mistaken, minus 10a from here and 5b from there. Okay. And your constants will be a minus b plus c equal to zero. Oh, sorry. I'm already putting x. Okay. So this is going to be your answer. So it is so efficient that you don't need to get alpha beta directly. You don't have to strike any kind of a beta relation on it. You don't have to make alpha plus beta is minus b by alpha beta c by etc. etc. All those things are already taken care of. Is this fine? Okay. So now I'll give you a quick, you know, small exercise on this. Let's say if a x square plus bx plus c equal to zero has roots, has roots alpha and beta. Okay. Then the quadratic equations with roots. Let's say I take this example. So if you are adding any number to both the roots, then what will happen? Your quadratic equation will become a x minus k the whole square bx minus k plus c equal to zero. You don't have to do anything else to get the quadratic equation. If you're subtracting any number from k, in fact, it is the same approach. You just have to add that k to x. Are you getting my point? If you are multiplying any non-zero number with k, also non-zero number k with alpha and beta, what will happen? Your x will become x by k. I'm doing it quickly, but you can always do it at your own pace. If you divide a non-zero number k to both the roots, then it'll become xk square bxk plus c. Yes or no? Okay. And even you can do a non-linear change also. For example, if you have root alpha and root beta, then your x will become x square square like this. Okay. Isn't it? If you have something like p alpha plus q by m and p beta plus q by m, then what will happen to your x? It'll become mx minus q by p. Okay. This is a more generic version. This is actually a generalized form of a, b, c and d. Are you getting my point? So this list can go on and on. I don't want to write all the scenarios possible. But what I want you to give an idea is that if you are asked to find out an equation whose roots is a transformed version of the roots of a given equation and the transformation is applied same to both the roots, then you can always try out this method which will be actually time efficient. Of course, one time we'll go in simplifying it. But once you have written it, simplifying is just a clerical work. Is it fine? Any questions? Any concerns? Anybody? Okay. Maybe when we solve a question, things will become clearer for us. Let's take this question. If alpha beta are the roots of, oh sorry, if alpha beta are the roots of x square minus px plus q equal to zero, then find the equation whose roots are these two. And you can see here that both the roots are given the same treatment, right? So you can definitely try out for transformation of equation concept. If done, you can give me a response in the chat box. Okay, Satyam. So what did I tell you with this approach? Since alpha and beta are roots of these two equations, we can say this is going to be satisfied. Okay. All you need to do is take one of them which contains alpha equated to x, correct? Which means p minus alpha is q by x, which means alpha is nothing but p minus q by x. Or you can say px minus q, but let it be p minus q by x. Okay. Just replace it over here. Whatever answer you get, that is your answer. Okay. So it will be p minus q by x the whole square minus p, p minus q by x plus q equal to zero. Okay. Let's expand it. This will be p square, q square by x square minus two pq by x. Again, minus p square plus pq by x plus q equal to zero. This and this gets cancelled off. And I think this and this will get adjusted also. So q square by x square minus pq by x plus q equal to zero. Okay. So if you multiply with x square, you'll end up getting something like this. Okay. Yes or no? Drop a factor of q if you want. So this is a very surprising question because despite transforming the roots, your equation did not change. Okay. So basically, this is such a relation where you would realize that probably this guy became a beta and this guy became an alpha back. Interesting question though. So despite making the change to the roots, the equation did not change, which means the equation, the two transformed roots were actually the same roots. Okay. Anyways, good. Let's take another one. Another question. Is it clear everybody any doubt related to this? See, a lot of questions have come based on this and people who have not understood this concept took a longer time to solve the question. Let's try this one. If alpha beta are the roots of this equation, find the roots of this guy. Okay. So this question has been framed in a slightly reverse way. Okay. So look at this equation and look at this equation. You will see the presence of A, B, C in both of them. But there is a small change done to the root over here, to the x value over here. Okay. So let me bring your attention to something which is very interesting. Divide by x minus 1 the whole square. Okay. If you divide by x minus 1 the whole square, you end up getting something like this. Now I'll do a small activity here. I will observe this sign and make it 1 minus x. Here also I will make it 1 minus x because being subjected to square power, it will have no meaning. It will have no issues actually. Okay. Now see, treat this term to be like your y. Right. Okay. So let's say I call this term as a y. Okay. So this term is a y. Okay. Now please understand that if this is satisfying it, which means if you compare it with this equation A alpha square B alpha plus C equal to zero, your y is actually like either your alpha or your beta. Okay. Alpha or is like your beta. Correct. Right. Because the same quadratic equation cannot have three roots. So either that y will be alpha or y will be beta. So if y is alpha means you're trying to say x by 1 minus x is alpha. Correct. And if you're trying that, if you're trying to say it is beta, that means you're trying to say x by 1 minus x is beta. So from here, if you try to solve for alpha x, what will you get? This is nothing but alpha minus alpha x. So which is alpha plus 1x equal to alpha. So x is alpha by alpha plus 1. Similarly here, I will not do everything here. Similarly here, your x will be beta by beta plus 1. Right. In short, the other quadratic equation, the roots will be alpha by alpha plus 1 and beta by beta plus 1. Right. So this is just a reverse of what we were actually learning. So we were learning that if the roots are transformed in the same fashion, what will be the new equation? So here the question setter has given a new equation to you and asking you what is the transformed root. Okay. So both the directions, the question can be framed. Is it fine? Any questions, any concerns here? Okay. All done, copied, understood more importantly. I understand. This is a slightly new concept for you because in tenths you were not taught about transformation of equations. All right. Let's take one last question on this. If alpha and beta are roots of, if alpha and beta are roots of 3x square plus 2x plus 1 equal to 0. Okay. Then find the value of, find the value, find the value of 1 minus alpha, 1 plus alpha cube plus 1 minus beta by 1 plus beta cube. Give me a response on the chat box. Yes. Anybody, nobody. Okay. So let's solve this question. Let's call this expression as a p and let's call this expression as a q. Okay. So can I ask you a simple question? Give me an equation, give me a quadratic equation whose roots are p and q. Of course, many people want an answer. Fine. Sorry. Many people want some time. One minute. Okay. Fine. I'll wait for one minute, not an issue. Last class. So as you say, last class for the year, by the way, for the academic year, not the actual year. Okay. See, if you're taking this much time, means your approach is not right. That means you will find that you're doing a lot of unnecessary things. Okay. See, let me ask you this question. What will be the equation? Can we get an equation with roots, with roots p and q? Okay. What is this equation? Can we find that out? Very simple. Here we can see that p and q are actually the same transformations of alpha and beta. Right. So I can use the transformation skills over here. I will just take this as an x. Okay. Apply componento and dividendo. If you apply componento and dividendo, alpha will be 1 minus x by 1 plus x. Correct? No. Yeah. Right. Componento dividendo is a very good tool for such cases. Okay. Now put this in your original equation. So the original equation, that is 3x square, that will be satisfied by alpha. Right. So put it over here. So you end up getting 3, 1 minus x, 1 plus x whole square, 2, 1 minus x, 1 plus x, plus 1. Okay. And put it to zero. So multiply throughout with 1 plus x, the whole square, that will give something like this. Correct? Collect your x-squares, x-square terms together. So if I'm not mistaken, this will give you plus 3x square minus 2x square plus x square. That's nothing but 2x square. Okay. Collect your x terms together. This will give you minus 6x. Correct? Correct? Minus 6x and plus 2x. So that's 4x. Oh, sorry. Minus 4x. Okay. And let's collect your constants together. This will be 3 plus 2, 5, 5 plus 1, 6. Right. In other words, x-square minus 2x plus 3 equal to zero will have roots P and Q. Correct? Which means you know P plus Q, which means you know P into Q also by Vita relation. Correct? By Vita's relation, you know these values also. Now, what are the questions it is asking you? Questions it is asking you PQ plus QQ. Okay. So you are supposed to find out PQ plus QQ, which is nothing but P plus Q the whole Q minus 3 PQ P plus Q. Isn't it? So which is 2Q minus 3 into 3 into 2, which is nothing but minus 10. Problem is solved. This is your answer. Is it fine? Any questions? Any concerns? Please have a good look at it and tell me if you want me to explain any part again. Okay. All right. Convinced? Any questions? All right. So I'm convinced that you all have understood this. The next concept, which is the second last concept in this topic that we are going to talk about is the conditions for common root. Conditions for two quadratic equations to have common roots. Quadratic equations to have common root or roots. Okay. So guys, we all know that a quadratic equation can have two roots. Right? So let's say there are two quadratic equations Q1 and Q2. Okay. One is A1x square B1x plus C1 equal to 0 and another is A2x square B2x plus C2 equal to 0. Okay. Now, both of them can have two roots. Right? So what is the condition that should be satisfied between your A1, B1, C1, A2, B2, C2? That means these coefficients of these two quadratic equations so that they have both the roots common. Let's start with this scenario. So let's discuss if you want two quadratic equations to have, sorry, both the roots common, what condition must be satisfied? Can anybody tell me that? Two quadratic equations, same roots. Both of the roots are same. What condition must be satisfied between the coefficients of those quadratic equations? That means what relation must exist between A1, B1, C1, A2, B2, C2? Right, Satyam. So it's very obvious that the quadratic equation 1 and 2 must be proportional to each other in the sense that one is the other equation multiplied with some k. In other words, I can say the coefficients of x square, x and constants must be the same. Isn't it? Yes or no? Now here a very important add-on I want to put here. If these two quadratic equations are giving you the same roots, that doesn't mean this expression is same as this expression. You can't say A1 is equal to A2, B1 is equal to B2, C1 is equal to C2. You cannot say that but you can definitely say that one quadratic equation or you can say one quadratic expression. The word is expression, not equation. One quadratic expression is a multiple of or you can say it's proportional to the other one. So you can say A1 is lambda times A2, B1 is also lambda times B2 and C1 is lambda times B2. This is what you can comment. So there was one student last year I asked in this question that if these two equations have same root common, does it mean that if I draw a graph like this for both the let's say quadratic polynomials, will I get the same graph? Will they have the same graph? Means this parabola and this parabola will be exactly the same? Yes, not necessarily. So the graphs need not be same. But yes, what is same is the points where they will be cutting the x-axis. So let's say one is this white one. Let's say I'm just making a dummy one. The other maybe something like this. But still cutting the x-axis at the very same position. So these two graphs need not be the same. So please understand these. These are small small things where you might get stumped. Okay. Is it fine? All right. Let's talk about the second case here when at least one root is common. Let's talk about that. Can I move on to the next one? Yeah. So what are the conditions that must be satisfied if two quadratic equations have at least one root common? At least one root common means it can have both the roots common also. Okay. By the way, we will say that both the roots common we already talked about. Okay. So this situation, the second situation is going to be a superset of the first situation. I will tell you why once we start discussing it. Okay. So as of now, we'll say let's say this quadratic equation Q1, this quadratic equation Q1 and this quadratic equation Q2, this quadratic equation Q2. Okay. Has one known common root. Let's say has one common root alpha. Okay. There may be more roots also. Let's say, but I know for sure that one common root is there. And let's say that root is alpha. Okay. Now what we are doing, we're trying to find out the condition. Okay. So that is our main purpose. Now our main agenda is to find the condition that must be satisfied if these two quadratic equation has at least one root common. Okay. So let's say I call that root, which is common to be alpha. Okay. One of them is sufficient. So that means alpha should satisfy both the equations. That means alpha should satisfy this also. Alpha should satisfy this also. Correct. Now here for the time being, for the time being, let's call alpha square as capital X and alpha as capital Y, just for the time being. Okay. So let alpha square be capital X and alpha be capital Y. So can I say the same set of equations will now be something like A1 capital X, B1 capital Y plus C1 equal to 0 and A2 capital X, B2 capital Y plus C2 equal to 0. Can we do, can you all do me a favor? Can you solve for capital X and capital Y? Can somebody give me capital X and capital Y? Okay. One interesting approach would be to use cross multiplication method. How many of you know cross multiplication method? Huh? Everybody knows here. Yes. No, sir. I knew but I forgot. Oh, sorry. This is A1, A1, B2 minus A2. My bad. A1, B2 minus A2. Yes, tell me. Learn but never got to use. Now you'll get to use it. And you'll, you'll realize that this actually is called the Cramer's rule of solving a system of linear equations. Okay. You learn that again in your matrices and determinants chapter, which is going to come like four, five, three, four months down the line. Okay. By the way, you are, you should, you should start considering yourself to be in class 12 now. Unofficially. Okay. Yeah, till the final exam is not over. But normally, you know, I mean, NPS students, super bright students. Okay. I should not say any further. Okay. So X will be B1 C2 minus B2 C1 by A1 B2 minus A2 B1. And your Y will be just note that there's a minus sign sitting over here. So I will just swap the positions of the denominator term while taking it to the next side. Okay. Is it fine? Now, the reason I found out capital X and capital Y is because now I want to say alpha square is this and alpha is this, right? Which means Y squared is your X. So if you square this guy up, if you square this guy up, you should get the X guy, which is your square of this, which is square of the alpha. Indirectly, I'm trying to say this, right? Because if X is this and Y is this, then X is actually the square of Y because of our assumption of X and Y, right? Which further means, which further means I'll be just writing it on the top, which further means B1 C2 minus B2 C1 times A1 B2 minus A2 B1 is equal to A2 C1 minus A1 C2 whole square. Correct. Now, this is the condition that we were actually looking for. And you would be thinking, sir, such an ugly condition. How do I, how do we remember this? It's so ugly to remember. The previous one was much better, right? A1 by A2 is B1 by B2 is C1 by C2. So how do you expect us or how do the examiner expect us to remember this condition? Okay, not to worry. I'll give you some memory aid for this. By the way, I'm expecting everybody to be aware of determinant expression, right? Everybody knows determinant. Have you seen determinant before? Something like this, let's say X, Y, A, B. Can you, can you write it as BX minus AY? You're all aware of this, everybody? Everybody knows this determinant? You would have used it in your cross-product concept. Okay, now I'll be using my determinant idea over here. Look at this term. Can I write it as determinant A1 A2 B1 B2? Correct me if I'm wrong. Right? Look at this term. Can you write it as B1 B2 C1 C2? Correct? Look at this term. Can you write it as C1 C2 A1 A2 the whole square? Right? By the way, squaring doesn't make a difference. You're still getting the same expression. Okay, now this is much easy to remember. Why it is easy to remember? Because it is like A, B, B, C, C, A. Okay, you just have to remember the square over here. Right? Now it is simple to remember. Okay, so please note this down. This is your condition to have at least one root common. Now, many people ask me, sir, why do you say at least one root common? Isn't it only exactly one root common? See, if you look at the derivation in the derivation, I did not talk about the other root. I've just taken one root to be common and I started with it. Even if other root is common, the same procedure will be followed. No. Right? And not only that, not only that, if both roots are common, then you would realize that let's say both roots are common. If both roots are common, as per our previous discussion, A1 will be equal to lambda A2. Correct? Or in other words, I can say A1 by A2 is B1 by B2. Okay. Which means A1, B2 minus A2, B1 is zero. Right? Which will definitely make this guy as a zero. Correct? Similarly, if roots were both roots were common, we know that B1 by B2 is C1 by C2, which means B1 C2 minus B2 C1 would have been zero, which means this guy would have been zero. Correct? And not only that, we also know that A1 by A2 is C1 by C2, which means A1 C2 minus A2 C1 would also have been zero, which means this guy will also have been zero. So your zero and zero will be equal to each other. Are you getting my point? So this is also addressing this concern also that both roots may be common. Are you getting my point? So you can say this condition to be more robust, more generic. Getting it? Is it fine? Any questions, any concerns here for the situation of at least one root common? So both root common, the situation, the condition is very simple. Straightforward A1 by A2 is equal to B1 by B2 is equal to C1 by C2. If you have at least one root common, it becomes determinant of A1 A2 B1 B2 times determinant of B1 B2 C1 C2 is equal to square of the determinant C1 C2 A1 A2. And this is cyclic in nature. You can remember it. A, B, C, C, A kind of a thing. Is it fine? Any questions, any concerns? Let's take a question. Let's take a question. Question is if this equation and this equation have a common root? Show that A is to B is to C is 1 is to 2 is to 3. Just write it down on the chat box if you're done. Yes, this was actually an easy question. See, all of you please pay attention here. Let's say this has got two roots. Okay, alpha and beta. And this equation has got one of the roots common. I don't know about the other one. Okay, one of the roots I know it's for sure common. Okay. But what I also realize is that this quadratic discriminant is actually negative. That means B square minus 4 AC is a negative term. Which means if this is P plus IQ, then this should have been P minus IQ. That means if this is P plus IQ, then can I do something about the other root because A, B, C are real numbers here. Look at the question. A, B, C are real numbers. So if the coefficient of a quadratic equation are real and if one root is P plus IQ, what is the other root? It has to be B minus IQ. What does this eventually mean? Eventually means both roots are common. Both roots are common. Yes or no. That means A by 1 is equal to B by 2 is equal to C by 3. And 2. Is it fine? Any questions? Any questions? Okay, let's take one more question. Yes. Any response? Did you get some lambda values? Okay, Harshitha, can you type them out on the chat box? Okay. Let's try to write down that situation which we have written. A1, A2, B1, B2 into B1, B2, C1, C2 is equal to C1, C2, A1, A2 determinant whole square. Okay. So now let's write down what is A1, A2. A1, A2 will be 1 lambda. B1, B2 will be minus 1, 10. Okay. Minus 1, 10. C1, C2 will be minus 12, 3. This will be minus 12, 3, 1 lambda square. Okay. On simplification, this gives you 10 plus lambda, the 10 plus lambda. And this is going to be minus 310 is minus 117. Oh, sorry. Plus 117. Plus 117. This will be 12 lambda plus 3 the whole square. Let's simplify it. So let's collect the lambda square coefficients. Lambda square coefficients will be 144 minus 117, which is going to be 16. Am I missing out anything? Any important thing here? Oh, sorry. What am I doing? Yeah. Okay. And lambda coefficients will be 117 minus 12 into 3 into this. So 117 minus 117 minus 144. What am I doing? 117 minus 72. Sorry. That's going to be 45. That's going to be 45. So plus 45 lambda and constant terms would be 1170 minus 9, which will be minus 1161. Okay. I think we can drop a factor of 9 throughout. So that will give you 16 lambda square. This will be minus 5 lambda. Sorry. Plus. This was minus I believe. And this will be minus 129. Correct. Now this is factorizable as 16 lambda square minus 48 lambda plus 43 lambda minus 129 equal to zero. Take 16 lambda common. Take 43 common. And this is factorizable as this. Correct. So you should all be getting the lambda value as 3 and minus 43 by 16. Are you getting this? Is it fine? Okay. So if lambda value is 3, then what will be the common roots? First of all, if I just simplify this, okay. Is this going to be factorizable? Yes. It is factorizable as x minus 4 x plus 3. Okay. So this has a root 4 and a minus 3. Okay. So if you take lambda value as lambda value as 3, okay. Then what will happen? Then the other equation will become 3 x square, sorry, 3 x square plus 10x plus 3 equal to zero. And this is easily factorizable like this. That means the common root will be minus 3. Okay. So the common root will be minus 3 in this case. Okay. So common root will be minus 3 in this case. But if you take lambda value as minus 43 by 16, minus 43 by 16, without much hiccup, I can say the common root will be 4 in that case, right? Because any one of the root has to be common, right? So if minus 3 is addressed when lambda was 3, then with lambda is minus 43 by 16, your common root should be 4. Done. Is it fine? Any questions? Any concerns? All right. So with this, we'll take a quick break of 15 minutes. Right now the time, sorry, time is 6, 10 p.m. We'll meet exactly at 6, 25 p.m. On the other side of the break, I'll be doing the concept of location of roots. Okay. I owe it 6, 11. So let's meet at 6, 26. I'll be fair. Okay. So let's meet at 6, 26 sharp. See you on the other side of the break. Oh, the last leg of the chapter is where we are going to talk about location of roots of a quadratic equation. Okay. Of a quadratic equation. Now, many times you will be given a quadratic equation with certain parameters, right? For example, they would mention the value of A, they will mention the value of C, but they will not mention you the value of B. Maybe they will write a, you know, lambda for B. Okay. Or it could happen with all of them also. A, B, C themselves could be written in terms of parameters. Parameters means something which is not known actually. Okay. You will not call it as a variable because, you know, it's, it's something which is unknown to you. Okay. And they will say that this particular quadratic equation has both the roots positive. What should be the range of value of that lambda? Right. Or they will say that particular quadratic equation has both the roots lesser than a given number. Right. So these type of questions are going to be, you know, addressed under the present discussion that we are going to have that if you give, if they give you a quadratic equation with some unknown, right? And it could be like A, B, C could all be unknown to you in terms of some lambda or M or whatever. And they give you that this quadratic equation should have, you know, the two roots satisfying so and so conditions. What should be the interval in which lambda should lie? Okay. So let's talk about it. Let's talk about it through cases. So I'll be addressing six cases over here. Okay. So what are the six cases? Let's talk about it. So my first case is where I will say both the roots, both the roots are lesser than a given number K. Okay. So what should be the condition between A, B, C, or what should be the number of conditions that must be satisfied so that your both the roots are less than a given number K, a given real number K. Okay. So let's discuss it one by one. See, when you talk about a quadratic equation, all depends upon your A value, right? So whether it opens upwards or opens downwards, it depends on your A value. Okay. So irrespective of whether A is positive, that means your quadratic is opening upwards or your quadratic is opening downwards, we will try to devise a set of conditions which will be universally applicable to both of them. Okay. So what conditions you must honor so that both the roots of this quadratic equation are less than a given number K, both the roots of this quadratic equation are less than a given number K. Okay. Now as per you, what should be the conditions that must be satisfied? For me, the first condition that must be satisfied is your discriminant must be greater than equal to zero. Your discriminant must be greater than equal to zero and I'm sure all of you would agree with me on that. Why? Why discriminant must be greater than equal to zero? If you're saying your roots are lesser than a given real number, they must exist first of all. Without existing, how can you say that a number is lesser or greater than the other number? So your roots must exist first of all. So this is number one condition. Number two condition which I say should be fulfilled if you want your both the roots to be lesser than K is that f of K into a should be positive. Yes or no? And if you know down, I've written this condition by taking into consideration both the scenarios. In this scenario, f of K is positive. But in this scenario, f of K is negative. So if you're wondering, why did I put this A here? Because in either of the two cases, if you see A into f of K, A is positive, f of K is also positive. So your product is positive. In the second scenario, in the downward opening parabola, your A is negative, f of K is also negative. So the product again becomes positive. So can I say this condition is the universal condition for both the scenarios where your parabola is either opening upwards or downwards? It doesn't matter. Yes or no? Now tell me, are these two conditions sufficient enough or are they necessary? Are they only necessary or are they sufficient enough? That means if a quadratic equation discriminant is greater than equal to 0. By the way, some of you would be wondering, what is this f? I'm actually calling this guy as f. This guy as f of x. So are these two conditions sufficient enough or are they necessary? What do you think? If these two conditions are met, will my quadratic equation roots will always be lesser than K or there is something more that I need? So sattam is saying it is necessary. So sattam, tell me the condition that is required to make it sufficient. So you were right in saying that it is necessary but not sufficient because why is it not sufficient? Because even if your K was here, isn't it? Or even if your K was here, can I say both these conditions would still have been satisfied? Yes or no? So how do I ensure that my roots are actually lesser than K but not greater than K? So here I got the third and final condition so that it becomes sufficient or they all become sufficient together. Can I say the midpoint of this which is minus b by 2a? That will always be lesser than K. So together they now become sufficient conditions. Isn't it sufficient conditions? In other words, if I take the overlap of these three conditions, intersection, whatever answer I get, that would be my acceptable answer. Let me show you a question like this so that you get a better idea of what I am trying to discuss here. Let's say this question. So there is a quadratic given to you and you can see that this quadratic is not completely expressed. It has got some m's and all in it. So certain values are not given to you. So m is a parameter you can say here. Now read the question. The question says find the value of m for which let's target the first part of the question here only. Subsequent parts I will come with after certain discussions. Both the roots are smaller than 2. So what should be the value or values of m such that both the roots are smaller than 2? Now how will I solve this question? Let me address this question so that you get an idea here. First of all, thankfully my a is 1. So it's an upward opening parabola. No need to worry about the second case. So only one case is what you are required to address. Now you want your both roots to be smaller than 2. So let's say this is 2. So can I say first of all, discriminant should be greater than or equal to 0. That means b square minus 4ac should be greater than or equal to 0. Let's try to solve this. So this will give you b m square minus 10m and 9. Yeah, greater than or equal to 0. So this is factorizable. So if you look at a wavy curve here, your m should be either lesser than 1 or greater than 9. So this is only the first condition. So let me move on to the second condition. Let me just box it first. Second condition is your f of 2 here would be positive f being the polynomial involved in that quadratic equation. So this guy should be positive. So f of 2 should be positive. Let me write it in white. So f of 2 should be positive. Positive, not greater than or equal to 0. Exactly greater than 0. So f of 2 is what, in this expression, you have to put your x as a 2. So that means 4 minus 2 times m minus 3, if I am not mistaken, 2 times m minus 3. Yeah, plus m should be greater than 0. Okay, which means 10 minus m should be greater than 0, which means m should be less than 10. Let me write it in white so that we rhyme with our second condition. But as I told you, there is one more condition required. And what is that condition? Minus b by 2a should be lesser than 2 in this case. So what is minus b by 2a? Minus b is m minus 3 by 2. So m minus 3 by 2 is less than 2. That means m should be less than 7. Yes or no? m should be less than 7. Now, these three conditions must simultaneously be satisfied. In order to give you your roots, both less than 2. So I will try to take an overlap or intersection of 1 and 2 and 3. So let's take an intersection of 1, 2 and 3 over here. Okay, so I'll use my number line method to do it because I find that method easier to use. So first condition says, I want to be less than 1 or greater than 9. Second condition says, I want to be less than 10. And third condition says, I want to be lesser than 7. Less than 7. Now, which part of the graph do you see all the three lines overlapping? Obviously, this part. So your overlapping condition will be m should be less than equal to 1 or you can say m should belong to minus infinity to 1. This is your final answer to the question. So if you choose any m value, which is minus infinity to 1, you would realize that the both roots of this quadratic equation will come out to be smaller than 2. Is it clear? I hope this example has given you an idea about what type of questions we will be getting with respect to this concept. Yes or no? Anything that you want to copy, ask, please do so. Okay, so from here on, you will be driving the story. Okay, you will be driving the theory and story also. So next case, which I'm going to ask you only, okay, if your both roots are greater than a given number k, both roots are greater than a given real number k. Then tell me what conditions must be satisfied? For your purpose, for your convenience, I'm going to draw the two graphs which you will require. One is an upward opening parabola, other is a downward opening parabola. Okay, you want both your roots to be greater than k. That means k should lie somewhere over here. Okay, by the way, if I'm making in the negative direction, that doesn't mean k is a negative number. It's just that it happened to happen in the diagram. Yeah, tell me the conditions that must be satisfied. Let's start with the first condition. Who will tell me the first condition? Write it down on the chat box. Okay, so Satyam says, let's start with d greater than equal to 0. Exactly, Satyam, roots must be real first of all. Okay, then only we can talk about it being greater or lesser than something, right? Second condition. Second condition. Who will tell me? You can use your f of x as a x square plus b x plus c if you need it. Can I say here a f of k will be positive again? But again, note that these two are necessary conditions but not sufficient. So what will you need? You will need the final nail in the coffin, which will make it sufficient as your minus b by 2a should be actually greater than k. Yes or no? Yes or no? So can I say these together, these three simultaneously must be satisfied if you want any quadratic equation to have both the roots greater than a given number k. Okay, here I would like to write again, your f of x is your a x square plus b x plus c. The quadratic polynomial part I am calling as f of x. Okay, is it clear or not guys? Is it making sense to you? This is a very important type of question being asked in many competitive exams. So let's take the same question. I think it has got eight sub parts. I think the second sub part is where we can sit and solve. Yeah, so the second part basically talks about our present scenario. Yes, anybody? Okay, Satya, very good. See first condition as I told you, d should be greater than equal to 0. By the way, we had already solved it and this gave us m lying between minus infinity to 1 union 9 to infinity. So we have already solved it, no need to solve it again because we are using the same quadratic here also. Second scenario is f of 2 should be greater than 0 because a is 1 anyways. So this also we had solved it in the previous slide. What was it? Just to, yeah, what was it? Are you not this one? The one before this? Yeah, this gave us m less than 10. Correct. So I'll be using that only. So save your time because you have already done half the conditions here. So m should be less than 10. m should be less than 10. Okay. And the third condition is minus b by 2a, that should be greater than 2. Correct. Okay. I think we had solved less than 2 in the previous slide. So we can use that information as well. So m less than 7 will now become m greater than 7 in this case. Correct. m greater than 7. So these three conditions now we have to take an overlap. Okay. Let's take an overlap. So again, critical values are 1, 9, 7, 1, 9, 7, 10. So the first fellow says I want to be lesser than 1 and greater than 9. The second fellow says I want to be less than 10. Third fellow says I want to be greater than 7. Where is the overlap happening? Overlap is happening here. So the answer is m should be greater than equal to 9. That means m should belong to 9 to infinity, open at infinity and close at 9. Oh, I'm so sorry. So sorry. So sorry. Thank you. It is only in this zone. Yeah. So 9 to 10. 10 is, yeah. Is there any questions here? Any questions? Any concerns? Clear? Understood? Okay. Comfortable with this concept now? All right. Chalo, we'll take other conditions as well slowly. Next condition that I would like you to tell me what should be the condition which should be satisfied if a given real number k lies between the roots? Lies between the roots? Okay. So again, let me make the two scenarios here. Yeah. So what should be the condition which must be satisfied if you want a number k to lie between the roots? Between doesn't mean center. Between means somewhere in between the interval alpha beta. Okay. So what conditions must be satisfied? Start writing them down on the chat box. Let me see who's the first one to write the first condition. No, sir. D should be purely greater than 0, not greater than equal to. Now, why all of us are only greater than, not greater than equal to because if a number k lies between the roots, roots must be distinct. No. Okay. They can't be real and equal. They have to be real for sure, but real and distinct. So don't include equal to 0. It will be greater than 0. Okay. What next? Right. The second condition is absolutely right. So a into f of k should be negative. Everybody agrees with this? Everybody agrees with this? Is there anything else needed or are these two sufficient? Sufficient or you need something more? Means if these two conditions are met, can I say the k value will be definitely between the roots? It can't be like yes, sir. Okay. I have become sir now. Thank you. All right. Let's take a question on this. Yes. Let's take a question on this. The same question just to make your life easy so that you don't have to recalculate the same things over and over again. Let's do the third part. One root is smaller than two and the other root is greater than two. It's all this. Everybody should now give me the answer. Yes. Anybody see one root is smaller than two, other root is greater than two means two lies between the roots. A fancy way of saying that two lies between the roots. Correct. So if two lies between the roots, okay, somewhere in between, I know the first condition will be satisfied, discriminant greater than zero. By the way, discriminant greater than zero will give you m belonging to minus infinity to one open brackets. Okay. You'll all use open bracket here because it is not greater than equal to it. It's just greater than. Okay. So pure inequality is there. Second condition is f of two should be negative. f of two should be negative. Now this will give you m greater than 10 if I'm not mistaken. Correct. Let's take an overlap of the two scenario. One, nine, 10. The first fellow says I want to be less than one, greater than nine. Second fellow says I want to be greater than 10. Yes or no. So where is the overlap happening? Where is the overlap happening? 10 to infinity? No. Yeah. So overlap is happening between 10 to infinity. So 10 open. Okay, infinity will definitely be open. Is it fine? Any questions, any concerns? So this becomes your answer to that given question. Is it fine? Any issues? So this was our situation where both the roots, one of the roots is higher than a number and other root is lesser than the same number. Okay. That means that number lies between the roots. Next, four situation. Four situation is exactly one root lies between two given numbers K1 and K2. Okay. Two given real numbers K1 and K2. Okay. Let's try to write down this situation. Let's try to analyze this situation. So what conditions universally will work for both the types of parabola irrespective of whether they open upwards or open downwards? Yeah. Oh, sorry. Okay. So exactly one root lies between K1 and K2. So let's say it could be a situation like this or it could be a situation like this. K1 is here, K2 is here. Okay. Right. Similarly, here also K1, K2 or K1, K2 something like this. Okay. So please ensure we have to take care of all the four situations here. Okay. So tell me, think carefully and answer what conditions must be true if you want exactly one root? Whether alpha, whether beta or beta doesn't matter. Whatever conditions you should write, it should accommodate both the situations. Okay. So think carefully and write. Yes, sir. Tell me. What absolute silence? Okay. So Satyam is saying D should be greater than zero agreed, right? Because your root should be real as well as distinct. Correct. Okay. Now, actually just to, you know, reduce the conditions. Can I just say f of K1 and f of K2 must be of opposite signs. So A becomes irrelevant there. A becomes irrelevant here. So all I can say is that whether you take this situation, both will be opposite in sign or you take this situation. Both will be opposite in sign or whether you take this situation opposite in sign or this situation opposite in sign. Correct. Are these two sufficient enough or do you think more is required? Are they necessary or are they sufficient? Are they just necessary or are they sufficient? What do you think? I believe they are sufficient. So there is no other thing required. If you honor these two conditions, you will definitely have one of the, exactly one of the roots lying between K1 and K2. Is it fine? Any questions? Any questions? Guys, I'm not asking you to mug up these conditions. Okay. It will come automatically to you when you analyze the situation. Okay. So there are six such situations we are right now on the fourth one. So I don't think so you'll be able to memorize these things and don't try to do that if you cannot. Rather analyze the situation as per the situation given, as per the question given to you. Okay. Let's take the fourth question here. Exactly one root lies between the interval 1 comma 2. Done. Anybody? Okay. Okay. So as I basically told you, first of all, discriminant should be greater than 0. That means again, m should be minus infinity to 1 union 9 to infinity. Second situation is your f of 1 and f of 2 products should be negative. By the way, f of 1 is just going to give you a 4. Okay. And f of 2 is going to give you 10 minus m. So this should be negative. That means m should be, m should be greater than 10. Okay. Now take an overlap. I think it should give you the same answer like what we had for the previous one, isn't it? Yeah. So 1, 9, 10. So the first guy says I want to be lesser than 1 or greater than 9. The second guy says I want to be greater than 10. Okay. Yeah. Yes. So overlap basically is between open interval 10 to infinity. So this becomes your answer to the question. Is it fine? Any questions? Any concerns? Okay. Next. Fifth condition is where both roots, both roots lie between K1 and K2. K1 and K2 are basically two given, given real numbers. Okay. Let me make a quick diagram. Yeah. So K1 and K2 lie between the, okay. Sorry. The roots lie between K1, K2. Sorry. So that means your K1, K2 should be lying here. Yes. Start telling me the conditions. I am waiting. Sir, we don't know beta and alpha. Oh, sorry. Madam, we don't know beta and alpha. You have just been given K1, K2. Alpha, beta, you have to find it out and that will come in, come in terms of your parameters. So can we think of a smarter way out? Okay. Discriminate should be greater than zero is what many people are saying, but I disagree with it because discriminant could be equal to zero also because roots are lying between K1, K2 and roots can be same also. Who is stopping roots from being same? Right? They can be same also. That is condition number one. What else? f of K1, f of K2 would be positive. Now, Satyam, this will lead to false positive because K1, K2 product would be positive even if they are within the roots. Let's say K1 is here, K2 is here, then still both will be negative and hence their product will be positive. And if let's say it is this situation also, both will be positive, hence their product will be positive. So that condition is going to lead to unwanted results. Okay. Can I say here that a f of K1 and a f of K2 both would be positive? That is for sure. Now, are these three sufficient or do you require more conditions? Sufficient. Okay. Harshitha, just to answer your, I mean, just to respond to your answer, even if K1, K2 was like this, okay, then the same three conditions will be satisfied. Now, so how are you ensuring that K1, K2 are on the opposite side? I mean, the roots are between those two. So these three are not sufficient. These three are not sufficient. They are necessary. Right? Okay. Because if this is satisfied, then intersection is going to give you those values or those situations where your roots will be between K1, K2 for sure. Guarantee. Is it fine? Any questions? Okay. Let's take a question. Hey, Monday you have a paper. Monday exam with you. Pumbudu project. Yes. So which question should we talk about? Talk about, talk about both roots lie in the interval 1, 2, 2 or sub part 5 we should solve. Both the roots lie between 1, 2, 2. Very good, Harshitha. Okay, Satyam. Okay. So some of you are saying there cannot be any real value of M for which this condition is going to be fulfilled. Okay. Great. Let's check it out. So the first condition says discriminant should be greater than equal to 0, which means again, M should lie between, when I change the color, it doesn't change. Yeah. M lies between minus infinity to 1 union 9 to infinity. Okay. Second situation, which we saw was f of 1 into A should be greater than 0. Correct? So f of 1 comes out to be 4. 4 greater than 0. This is true for all real values. Okay. So any M you take, this is going to be true. Okay. Third condition is A into f of k2. That means f of 2 should be also greater than 0, which means M should be less than 10. Correct? And finally, minus B by 2A should lie between 1 and 2. That means M minus 3 by 2 should lie between 1 and 2, which clearly means M should lie between 5 and 7. M should lie between 5 and 7. Now these four conditions written in yellow, let's try to take their overlap. So I need a 1. I need a 5. I need a 9. I need, sorry, 7 I forgot. I need a 10 also. Okay. Yeah. So first condition says, hey, I want to be less than 1 or greater than 9. Second condition says, hey, I can be any real numbers. Third condition says, I want to be lesser than 10. Fourth condition says, I want to be between 5 and 7. Is there any place where all the four are agreeing with each other? Illa, no place. Okay. So what does it mean? It means there is no real values of M. So answer is M is a null set. Okay. Is this fine? So it is a possibility that you cannot have any real number or this condition is going to be satisfied. Is it fine? Any questions, any concerns? Okay. Have a look at it. Tell me if you have any questions. Okay. So with this, we'll now move on to our six conditions. When that given K1, K2. Okay. So K1, K2, these are two given real numbers. They lie between the roots. They lie between the, between the roots. Okay. Again, let's quickly analyze by making the two situations here. Hey, what am I drawing? Sorry. K1, K2 should lie between the roots. Why am I drawing? Yeah. K1, K2 should lie between the roots. Yes. Tell me which conditions must be satisfied. Condition number one. Who will tell me? So the root should be distinct if this has to happen. So discriminant should be greater than 0. Correct. What else? What else? Can I say A into F of K1 should be negative. A into F of K2 should also be negative. Correct. Any other thing that I'm missing off or are these two good enough? Sufficient. Sorry. Are these three sufficient? I think so. Yes. These three are sufficient because any other situation if you take these, one of these three conditions will get violated. Okay. All right. So with this, we come to the next question on that particular list. Question number or subpart number six. One root is greater than two and the other root is smaller than two. One root is greater than two. Other root is smaller than two. That means one comma two is between the roots. Sorry. One root is greater than two and other root is smaller than one. Sorry. I mispronounced it. Do it and let me know the result. Okay. Satyam again. Okay. Fine. Let's check. Anybody else? Okay. Harshita. Just two people active today. Harshita and Satyam. Okay. Chalo. We'll discuss it out quickly. So the first condition says discriminant should be greater than zero. Right. Which clearly implies your M should be between minus infinity to one union nine to infinity. Correct. Second situation says A into f of k one should be less than zero. Correct. So here f of one should be less than zero, which basically gives you the shock when you realize that here itself it is violated. That means this condition will never be true in your life. That means here itself the story is over. Problem solved. Answer is going to be no real values is going to satisfy it. Is it fine? Any questions? Any concerns? I'll be telling you in the last five minutes, not to worry. Okay. So we'll take the seventh question also. Why to leave that off? Seventh and eighth we'll do it. This you will be solving it because with respect to theory we have covered everything. Now in this condition you want at least one root to lie between one and two. At least one root to lie between one and two means what? At least one root to lie in the interval one and two means both roots lie between one and two. Sorry. Both roots lie between one and two and exactly one root lying between one to two. So basically have to take the union of the two conditions. So please do it and tell me the result because I think both the conditions we have already taken up. Both roots lying between one, two was option five. I mean sub part five. Exactly one root was sub part four. Okay. So you basically have to take the overlapping condition of not overlapping union condition of four with five. Okay. So what was the overlapping condition of four with, what was the condition number four? So I think four was m lying between 10 to infinity. Right. Okay. And the fifth condition was if I'm not mistaken null set. Okay. So union of this will be, union of this will be 10 to infinity. Is it fine? So even if your question is different from the theory that we have learned, you know that it can be basically built up by using the theory that we have already discussed. Okay. Let's take the last question for this academic year. So request everybody to get this right. At least one root is greater than two. Again, at least one root greater than two is made up of union of two cases. Both roots greater than two or exactly one root greater than two. Exactly one root greater than two. That means two has to be between, two has to be between the numbers. Okay. So here the situation is like this. I'm just making it. Both roots are greater than two means two is here. Okay. And this is a situation where two is between. That's where exactly one will be greater than two. So basically it's a union of third and second. Isn't it? Third is one root smaller than two. Other is greater than two. That's where two will come in between. And second is both roots are greater than two. So it's a union of second and third. So second and third. Isn't it? So what was the second condition? Yeah. Second condition was m should be between nine to 10. Right. If I'm not mistaken, check it out. And the third condition was m should be from 10 to infinity. So union will basically be nothing but nine to infinity excluding 10. So this will be your answer to the question. Is it fine? All of you got it? Okay. Great. So with this, we come to the happy conclusion of our class 11th. But just to remind you, there are certain topics of class 11th, which I'll be covering in 12th, especially when it comes to conic section. Conic section you have learned at a very shallow level. To go into the depth, I would engage you on certain Sundays. Okay. So be prepared to be called for one and a half hours class at least. Okay. Not more. I will not call you for three and a half hours, one and a half hours for certain Sundays. Okay. Maybe I would need around 10, 15 Sundays of your upcoming academic year. Okay. Okay. So this is all with respect to our discussion of the topic.