 Hello friends, so welcome to this session on quadratic equation and we will be trying to revise the concepts of quadratic equations in two hours time. We have some slides prepared for you and so to begin with I'll need who all are present so I can see the few students who are already there. So let's begin our class. So quadratic equation is an important chapter in your 10th board portion. So basically we start with polynomial. So what is a quadratic polynomial? So a quadratic polynomial is nothing but a polynomial of degree 2 and is called a quadratic polynomial. A quadratic polynomial with degree 2 is called a quadratic polynomial. The general form of a quadratic polynomial is a x square. So hence the general form of a quadratic polynomial is a x square plus b x plus c where a b c are real numbers. So a b and c all belong to the set of real numbers and a cannot be zero. Why? If a becomes zero then this polynomial is reduced to a linear polynomial. So hence a cannot be zero. So this is an example or let's say this is a definition of a quadratic polynomial. So what's next? So let us go to the next slide. We'll try to cover the topics quickly and then we will be going to problem solving. Now the second thing is if p x equals a x square plus b x plus c and a not equal to zero is a quadratic polynomial and alpha is a real number then p of alpha is equal to a alpha square plus b alpha plus c is known as the value of the quadratic polynomial. So hence the example will be let's say p x is equal to a 2 x square plus x plus 1 let's say then p of 1 is called the value of the polynomial at x equals to 1. So at p equals to 1 will be nothing but 2 times 1 square plus 1 plus 1 which is nothing but 4. So hence this is meant by what is meant by the value of the polynomial at any given x. In case during this you will be a little fast in revising so in case you have any doubt you can always post either as a WhatsApp query or in the chat sessions also. Next is a real number alpha is said to be a zero. A real number again we are dealing with real numbers a real number alpha is said to be a zero of the quadratic polynomial. So let's say if I have a quadratic polynomial p x equals a x square plus b x plus c please remember always this is the general form of the quadratic polynomial and if alpha is a zero then that is if alpha is a zero of this of p x then alpha is that value which will make p x zero. So hence that means if I plug in alpha in place of x you will get p of alpha zero. That means a times alpha square plus b alpha plus c equals zero. So let's take an example if p x is let's say a square minus 2 x or rather we will be taking it as x square. So x square minus 2 x plus 1. So if you see p of 1 that means if I plug in 1 in place of x I will get 1 square minus 2 times 1 plus 1 which is zero. So hence we can say zero or we can say 1 is the zero zero of p x. This is meant by what is meant by zero of a polynomial. Next move to the next slide. Next slide is yeah so next one is card number four which says if p x is equal to a x square plus b x plus c is a quadratic polynomial then p x equals to zero that is so this is definition of a quadratic equation. So equate any quadratic polynomial to zero you will get a quadratic equation. There is nothing much to delve into it so quadratic equation definition is there is a polynomial which is a x square plus b x plus c. So a x square plus b x plus c let's say if you equate it to zero or for that matter any other number if you also equate it to and then express in this form a x square plus b x plus c equals to zero then it is a quadratic equation mind you a cannot be zero a cannot be zero. Why if a becomes zero then it is reduced to b x plus c equals zero which is nothing but a linear equation in one variable linear equation. Okay so please be mindful of this. Let's go to the next slide next slide what is there the next slide next slide says a real number alpha. A real number alpha right a real number alpha is said to be a root mind the word so if real number alpha is said to be a root. Of the quadratic equation a x square plus b x plus c equals to zero if a alpha square plus b alpha plus c equals zero. This is you know time and again you have done it so no point investing more time into it so let's see alpha is the root. Then if you deploy that alpha in the equation you will get both sides LHS as well as RHS to be equal to zero right. A real number alpha is said to be a root of the quadratic equation if and only if alpha is zero of the polynomial. So these are you know you know we are trying to relate to things so let's say we studied zero of a polynomial zero of a polynomial. What is the zero of a polynomial that particular value of the variable which makes the polynomial zero. So if zero of a polynomial will become nothing but root of the equation root of the equation which equation. If you equate let's say the polynomial was px then if you equate that px to zero then that becomes root of root of this polynomial right root of px. So same value let's say if alpha is zero of polynomial px then alpha will be the root of equation px equals to zero. Okay so this is about root and zeros let's go to the next slide. Next slide is yeah so okay so this is if a x square plus b x plus c a not equals to zero is factorizable into a product of. Two linear factors then the roots of the quadratic equation can be found out by equating each factor to zero. So this is nothing but now we are we are heading towards how to solve or sorry wait a minute okay yeah so it's something but how to solve. How to solve quadratic. A quadratic equation how to solve a quadratic equation. So method number one is factorization method factorization factorization method. So in your exams there will be categorical questions where they will be saying using factorization method solve this quadratic equation and we'll see problems related to that a little later. But or we can also parallely solve you know questions like that but then the question will be factorize using factorization method solving. Another is something called completing the square method completing the square method square method this is the second method through which you can solve any quadratic equation. And third is third is my quadratic formula quadratic formula right which actually quadratic quadratic formula. Now quadratic formula is a direct fallout of completing the square method only so it is they are linked so quadratic formula comes from this but then now we have a. Direct formula to solve any quadratic equation correct now we'll take up factorization method one hour and completing the square method and quadratic formula one by one. So what is factorization method let me take you know this slide now quadratic let's say factorization method so what do you do in factorization method. Okay so let's say you have and we'll use examples let's say one question is given and the question is. Question is using quadratic factorization method solve this very simple question x square plus six x plus five equals zero will take you know difficult ones as well so x square plus six six plus five plus two zero. In factorization method which you have already learned we try and split the middle term right so split the middle term such that let's say if you have a general equation a x square plus b x plus c equals zero. What do you need to do you first multiply a and c find the value. Okay and then try to break b into b one plus b two a sum of b one plus b two such that b one b two the product of b one and b two should be equal to a and c a c right this is how this is how we do the factorization method solution. Okay now so x square plus six x now what is a here value of a let me write is one value of b is six and c is five so a c is equal to five. Correct now b is equal to six now I have to break six into two terms into two parts such that b one plus b two is six and b one b two must be equal to a c which is five. Now a and c are integers okay and five is a prime number so the only way you can factorize five is you can write five into one right. So hence b one becomes five and b two becomes one if you see it clearly satisfies this relationship so hence from here we can say it can be written as x square plus five x plus x plus five equals zero. So hence x times x plus five take x common plus one times x plus five equals zero so hence it is x plus five times x plus one equals zero. Now there are two two product there are two factors whose multiplication of product is zero so it is possible only when x plus five equals to zero or x plus one equals to zero. That means if this is true then x equals negative five and if this is true then x equals negative one okay so this is the solution so hence x equals to negative five and x equals to negative one. Okay you also remember guys for quadratic equations the maximum number of solution you can have or maximum number of roots you can have is two. Either the quadratic equation will have no solution or it will have only one solution which we say as equal solutions or it will have two real solutions. So hence what do I say the quadratic equation will have no real solution we'll see later on what are the condition no real solution no real solution or it can have two equal solutions two equal solutions and three is third is two unique or two unequal equal solutions. Now we are talking in terms of only real numbers we are not going into the realm of complex numbers so all are so if these are valid then these are valid for as the solutions are in real domain. Okay so we will see some more problems on this okay let's see what else let's take up some more questions on factorization method then we will proceed. Okay so the question is factorize solve the following quadratic equation by factorization method and the question is this question is x square plus two root two root two x minus six equals zero. Again what is A? A is one what is B? B is positive two root two and C is negative six okay now I have to what is AC? AC is six okay and B I have to split in such a way that B one plus B two is equal to two root two and B one B two is equal to six. Is it a B one sorry not six negative six okay now if you see there is a irrational irrational term here two root two is an irrational number and but B one B two is you know which is equal to AC sorry AC is negative six here so negative six right or AC is or this is equal to AC sorry this is equal to AC right. Now we have to split in such a way so we have to split two root two in such a way that the product is rational. Now how can that be possible that some of two numbers is irrational but product is rational that can be possible only when you eliminate the irrational part from here and that is possible only when I multiply two root two by at least root two. But now the sum must be yeah the sum must be two root two but the product must be negative six. Now another thing to be taken care of is if the sign of B one plus B two is not same as B one B two then what do we do is we express B one and B two as subtraction of a larger number minus a smaller number. So hence two root two if you see can be expressed as three root two minus root two and why did I do this so if you multiply these two number you will certainly get six negative six rather and the sum also is two root two. So hence the splitting of the middle term in this case will be three root two X three root two X minus root two X minus six equals zero. So X is common to it so X plus three root two and then here again if you see this is minus root two common here so this is X plus three root two X plus three root two this is equal to zero. Now if you if you see this is nothing but X plus three root two into X minus three X minus root two equals zero. So hence if you create this this is X plus three root two equals to zero and X minus root two equals to zero so hence from the first one you get X equals to negative three root two and here you let X equals to root two. Okay, this is how you have to solve. Let us take a little bit more. Yeah, so let us take a CBC 2013 CBC 2013 question. What is CBC 2013 question? What does it say? Okay, so let us say let us say we are we are trying to solve CBC CBC 2013 question. Okay, send CBC 2013 question the question says solve again by factorization method and the equation given is one upon X minus two plus plus plus plus two upon X minus one. Two upon X minus one is equal to six upon X. This is CBC 2013 some question asked in CBC 2013 examination. So one upon X minus two plus two upon X minus one equals to six by X. Now, clearly this is not a quadratic equation in this form, but it can be reduced to one. So let us take LCM and try and reduce it to a quadratic form. So hence it is nothing but X minus two and X minus one as a common denominator. So here is one. So one times X minus one and here is two. So two times X minus two and this must be equal to six upon X on the right hand side. Now, the problem is reduced only for simplification. So if you simplify this is X minus one plus twice of X minus four divided by if you see this is X square minus two X minus X plus two and this is equal to six upon X. Okay, now let us cross multiply cross multiplying I will get I will get X times X. So this is nothing but if you simplify further this is X plus two X which is three X. And minus one minus four is minus five. Okay, and this has this has to be equal to six times. Here if you simplify you'll get X squared minus three X plus two. Correct. So if you see simplify further this is three X squared minus five X. And this equals six X squared minus 18 X plus 12. Okay, now it is again, you know, we just need to reduce it further and then you will get a quadratic equation. So if you simplify everything you'll get six X square minus three X square is three X squared minus 18 X. And then minus five X go on to the other side it becomes minus 13 X and here is plus 12 and this is equal to zero. Okay, so hence we get an equation. We get an equation three X square minus 13 X plus 12 equals to zero. Now again, what is a here? A is equal to three B equals to negative 13 and C equals 12. So AC is nothing but 36. Now I have to split B in such a way that B1 plus B2 is negative 13 and B1 B2 is equal to AC which is equal to 36. Now how to do that? In such cases we find out all the integral pair of factors of AC. So hence I'll do this work here. So let's say 36 can be expressed as one into 36. Okay, clearly one and 36 will not yield you 30. So basically negative one into negative 36 will have to do like that. And then 36 can also be written as two into 18, right? But 18 and two also will not give you 13 here. So 36 can also be written as three into 12. Okay, but this again is not going to yield minus 13. So 36 now is nothing but four into nine and this is something which is interesting. Why? Because nine plus four is 13. So hence we now get the splits. So hence basically it can be written as three X square minus nine X minus four X plus 12 equals zero. If you now see negative nine into negative four is 36 and negative nine plus negative four is negative 13. So hence it is now you have to just the question remains to just find out the comments. So three X is gone. So it is X minus three. And then here four is common for X minus three equals zero. So this is reduced to X minus three times three X minus four equals zero. So hence either X minus three equals zero or three X minus four equals zero. So hence from this you will get X equals to three and from this you get X equals four upon three. Okay, so this is a trick. For example, if you see here B was negative and A and C were positive. In such case we arrest a shot that the roots will be positive. So hence you can take down as a trick. So let's say A was positive, A was greater than zero, B was less than zero and C was again greater than zero. In such cases either both the roots will be positive or both the roots will be negative. You can take that as a trick. Now just to check whether you have solved the equation correctly. Now having seen this, let's go back to our this thing. So we now learned what was our factorization method. So this is done. We will see completing the square method in quadratic formula in a little while. Now let's go back to next slide. So the roots of a quadratic equation can also be found by using the method of completing the square. So let us do the completing the square method quickly though the proves are not asked in the exam. But you must know because the process is important now. Let's go to completing the square method. Okay, now so what is the completing the square method? Let's see. So completing the square method, let us say, so we are now dealing with completing the square method. So as the name suggests, we need to complete some square. What kind of a square? Here it is a binomial square we are going to complete. How? Let us say the equation was a x square plus b x plus c equals zero, where a clearly is not equal to zero. Now what we can do is divide the equation, divide the equation by a. So what will you get? You will get x square plus b upon a x plus c upon a equals zero. Now if you see guys, there is a square term and there is a term which contains x and another factor and there is a constant. So what we can do is we can try and complete this square. It means what? Let's say this is half a rectangle I have shown. Why? Because we have two yet to complete this square. What does it mean? It means that if I somehow get a form x plus alpha square, what is the form? If you expand it, you will get x square plus two alpha x plus alpha square, isn't it? So if you see these two terms resemble the first two terms of this expansion of a square. So hence if we somehow add alpha square, then we can complete the whole square. That is what it means. So let us see how we do it. So hence I am saying it is x square plus b upon a x plus c upon a equals zero. Now if you see, I can say this is x square plus since I need a factor of two here. So then what we can do is we can multiply and divide by two. So basically we multiply and divide this term by two. So hence I can say this as two into x into this x and this I have added an extra two here, which I will take out by doing this b upon w. If you see now this whole term, this whole term is nothing but b by a x, but I have just manipulated it to write like that. Now this gives you a sense of x square plus two a b kind of thing. So where this is b, so hence I can very well write b by two a whole square, isn't it? But then if I have added something extra to the equation, I can get back to the original equation by subtracting the same quantity, isn't it? So this I can do now. This addition and subtraction nullifies the effect. And then finally I can add c by a which was already there. Now why did I do that? If you see, you can clearly see a pattern here. This is a square. So hence if you see, this is nothing but x plus b upon two a whole square. x square plus two x times b by two a plus b by two a whole square can be expressed as x plus b upon two a whole square. And this is equal to, I can take all these extra terms to the right-hand side and I can say this is b square by four a square plus c upon a, isn't it? Now I am writing it here. I am writing it here so that I don't lose the continuity. Now if you see from here I can say x plus b upon two a whole square is equal to b square plus, if you take the LCM and simplify it, it will be four a c by four a square. So that means x plus b upon two a will be nothing but plus minus under root b square plus four a c upon four a square. So hence simplifying you can say x is equal to minus b upon two a plus minus this is nothing but under root b square plus four a c upon twice of a. Because under root four a square will come out as twice of a. So hence it is nothing but minus b plus minus under root b square plus four a c upon twice of a. So this completing the square method itself leads to let's say solving the quadratic or finding the quadratic formula. If you see this is nothing but the quadratic formula. quadratic formula where we now know that x directly I can find out by deploying all the values here in terms of a b and c and you can directly find x. Oh I see I'm really sorry yeah there is some error this is when I take c by a it is it is it is sorry for this error this is not plus this is minus thanks for correcting. So this is clearly minus so hence here it is minus I'm sorry so all the pluses plus c will be minus c. Yeah so that's the correction thanks for correction. So this is plus plus plus everywhere it is minus minus minus minus minus and minus yeah so minus b plus minus thanks for correcting minus b plus minus under root b square minus four a c by twice of a. That is what is quadratic formula. So if you see there are two roots possible there are you know if you see there are plus and minus there are two roots possible one is called let's say alpha so alpha is minus b plus of under root b square minus four a c upon twice of a and beta is minus b minus under root b square minus four a c upon twice of a. These are the two these are the two. Okay these are the two roots let's say if we have some previous year question on solving the quadratic equation by completing the square. So yes so it is okay so there are lots of NCRD questions but we will pick up something which has been asked in the previous year. So that okay so okay never mind so let's solve this. Okay the question is solved by method of completing the square. What is the question? The question is solved by completing the square. So the question says this 4x square plus 4x square plus 3x 4x square plus 3x plus 5 plus 5 equals 0. Again 4x square plus 3x plus 5 equals to 0. Okay so first step is divided by a so hence it will be x square plus 3 upon 4x plus 5 upon 4 equals 0. Okay so now what do I do? x square plus twice of x into 3 by 8. Why? Because I have to multiply doing the denominator. Okay but since and then the b part here is 3 by 8 so hence you have to write 3 by 8 whole square but then as you added something you have to subtract also so 3 by 8 whole square and then here it is plus 5 upon 4 which is equal to 0. Okay so hence what do I get? I get x plus 3 by 8 whole squared is equal to 3 by 8 whole squared minus 5 upon 4 minus 5 upon 4 so let us simplify this. This is 9 upon 64 right minus 5 upon 4 which is nothing but 64 will be the LCM and hence it is 9 and hence it is 9 minus 16 4 just 64 so 16 5 is the 80 so here it is negative 71 by 64. Now here is the problem. The problem is that there is a square term on the LHS so LHS is greater than 0. Why? Because squares are always either 0 or let's say positive value but RHS, RHS is less than 0 it's a negative number so this is not possible so hence we say there is no real root. There is no real root of this equation. Okay so there is no real root for this equation. So this is what is called completing the square method guys so this is picked up now next is quadratic formula. So we saw quadratic formula hand in hand and now please remember the quadratic formula is what? Minus B plus minus under root B square minus 4 AC upon. Now B square minus 4 AC guys is called discriminant. What is it called? This is called D and D is discriminant. Discriminant. Okay we will be using this thing a little later just keep this in mind right what is meant by D and how we use D for various or let's say finding out the nature of root. Okay let us take up one another good problem on using solving the quadratic equation using completing the square method. Okay so that you will get a little bit of more hands on. Okay now so the question is what is the question? Question says find the root of the equation A square X square minus 3ABX 3ABX 3ABX plus 2B square plus 2B square is equal to 0. This is the yeah there are no numerals here it is a purely you know lots of variable constants so A square X square minus 3ABX plus 2B square equals to 0 and we will be solving it by completing the square method. Now again what do we do? We divide the entire here in this case the coefficient of X is A square. So let us divide the entire equation by A square. So dividing dividing the equation the equation by A square you will get X squared minus 3ABX upon A squared plus twice of B squared upon sorry A squared and this is equal to 0. I will write it a little clearly so that avoid overwriting so that you don't get confused yourself and you know it's always a better practice to write as neatly as possible because you yourself should not be confused. So X square minus 3ABX by A square plus 2B square by A square equals to 0. So hence I can say this is nothing but let's you know so hence X square I have to now do what? Multiply by 2 and keep X here so that 2X term is here and the rest of the term is 3AB so 1A will be cancelled if you see 1A will get cancelled by this so hence I will write 3B by A or in fact 2A because this 1 2A over here so I'll have to add to or multiply denominator by 2 as well so 3B by twice of A then now this becomes my B term so hence I'll have to complete this square hence I can write 3B by twice of A whole square right so if you do this the square gets completed but since you have added an extra term to the equation you have to remove it from it as well so 3B by 2A whole square plus 2B square by A square and this is equal to 0 so hence what can I say from this thing so this is X minus 3B upon twice of A whole square because if you see this is the expansion this is the expansion of this first three term first three terms these three terms here is the square term right and now this must be equated to so take everything on the other side you'll get 9B square so 3B square is 9B square and then minus sign becomes positive it's 4B square and then here it is negative 2B square by A square I'm sorry this is not B square in the denominator this is A square this is A square okay so yeah A square I'll write it properly so that you don't get confused you should also avoid overwriting yeah 9B square by 4A square minus 2B square by A square so simplify this what you get 4A square is the LCM so hence it is 9B square and A square and multiplied by 2 so it is 8B square so it is B square by 4 A square yeah so now if you remove the square sign you take what do square root on both sides you'll get X minus 3B upon twice of A is equal to plus minus B upon 2A okay so hence what do I get a solution so X is nothing but 3B upon twice of A plus minus B upon twice of A so which is nothing but so X alpha is nothing but 3B plus B upon twice of A which is 4B by twice of A which is 2B by A this is solution number one and solution number 2 beta is equal to 3B by twice of A minus B upon twice of A which is nothing but 3B minus B upon twice of A which is nothing but B upon A so if you see the solutions to this equation would be X equals 2B by A and Y equals B by A now please remember whenever you have such equations in the examination paper always put a star mark against that question because you need to spare some time towards the end of the question paper as we have discussed multiple times to check the solution so while you check the solution you can put any of these if let's say if you have lesser time to revise put any of these solutions which one which you think is a simpler one and see whether it went while plugging into the equation you get zero or not that will be a good enough indication whether you have solved it correctly yes if you have enough time you can go through the entire solution once again so use your judgment during the exam if you have lesser time plug in the solution and see whether you have solved it correctly or not so but during the rush hour that means towards the last ten minutes you will not be knowing which question was to be revised once again so hence I told you that put a mark in the in the answer script somewhere or you can write down towards the end that these questions need to be revised once again ok fair enough so let's move on to our next discussion so here is so we discussed this roots of a quadratic equation can also be found by using the method of completing this square and next is this quadratic formula now here is important things which is anyways is discussed in the next slide so let me go to the next slide now here is where we are talking about nature of roots of quadratic equation ok so now there will be questions on nature of roots of quadratic equation yeah so how do we find out so if you see in our quadratic formula let me just delete this so we are now dealing with quadratic formula as I told you previously we will come back to the discriminant thing and here is what I meant so let's say let's say your quadratic formula alpha was equal to minus b plus under root d upon twice of a is it and beta was also equal to minus b minus under root d upon twice of a right these are the two solutions or two roots of the quadratic equation a x square plus bx plus c equals to 0 now the problem is if d what was d d is nothing but b square minus 4 of a c right now there is a root of d so hence d must be greater than equal to 0 for alpha and beta to be real is it why because there is no solution to you know square root of negative value in real number set so hence d must be greater than equal to 0 if d is less than 0 then we say that there are non real solutions right non real non real solution those are those lines in the realm of complex numbers but in your cbsc 10th grade portions we are not going to deal with them so hence if d is less than 0 then there will be non real solution and if d is equal to 0 that means d equals to 0 then both the roots are equal isn't so if you see alpha is also equal to minus b upon 2a and beta is also equal to minus b upon 2a so hence in this case we say equal real and equal roots real roots are definitely real but they are equal they are equal and third is d is greater than greater than 0 in this case you have two distinct values of alpha and beta and we say that equations have real and distinct are real and different roots two real and distinct this is the term distinct roots so you will get two roots which are different from each other so please remember these three conditions questions will be asked on these the question will be you know find out some value of a b n c given that the roots are either distinct and real equal real and unequal or real and equal or not real imagine right this can be question we'll see such questions later on okay now let's go to the next slide okay so this summarized you know so we kind of you know these are the theory portions related to quadratic equation now the questions would be either of you know let's discuss what type of questions would be could be asked so in our experience in previous year papers also you have seen the questions asked are either of this form they will give you solving the equation solving the equation okay usually it is of two and three marks the solving the equation will be one right and now they will ask you categorically so please be very very careful while you are solving the equation as in what kind of method they are asking so if you see in the previous slide I'll show you so if you see here in the next slide they are saying solve by factorization method we just solved this question this is actually a CVSE question so it is categorically saying that solve by factorization method so when and be very very careful whenever you know during excitement what happens people think that we know how to solve quadratic equation and they miss on the instruction given on the question paper right so don't miss on the instruction given on the question paper when it says factorization method you cannot apply quadratic formula and when it is you know when it's nothing is mentioned then it is your choice whether you can use your factorization method you want to complete the square or you want to use quadratic formula most of the questions which will be there in the CVSE board paper they will be you know kind of you know you can solve it by factorization method mostly most of them ok so hence it's always good to solve through factorization method because it is you know simpler also and lesser problem calculation mistakes but yes it is always a good practice to come back and check it using a quadratic formula whether you have got the correct you know I'm saying in the in the rough side you can always do a quickly you know quick check and see whether the solution which you have arrived at is correct or not right see all the theory which we discussed you would be knowing it already now the problem would be when you'll be solving the questions either you will misread the question and hence lose marks or while solving it you will make careless mistake and then again miss marks and then as we were discussing these were the types of question which will be asked especially in the word problem if you misinterpret the question then you might lose marks so hence please be very very careful about it so hence what type of questions I talk I told you either it will be solving the equation so hence be careful what is the method is what is the method which is being asked whether it is a factorization method so you have to the moment it is a factorization method you have to use the spreading the middle-down method so you can't use the quadratic formula yes you can always check it through a quadratic formula then they will ask you categorically completing the completing the square method so then you know what to do you divide the coefficient by the coefficient of x square complete the square and then take under root and do not forget to take both plus and minus of you know the square root of the term which you will be obtaining so there is one mistake people generally do then third is a quadratic formula where it is always a good practice to write the value of ABC by the side of the paper so please do write what is A what is B what is C in this in this what happens is let's say there is a negative number and they will try to trick you with giving some negative number and usually people you know forget that negative science is always good practice to write a what is A then B and then C then write the quadratic formula and then then write the full you know whenever you are deploying the number you should be careful for the science so hence these are the three question three type of questions it could be asked direct questions solving differential equations sorry quadratic equations now second point is questions on nature of rules so second type of question will be nature of rules so you know on nature of rules so you know what is it we have to deal with discriminant yeah so discriminant is B square minus 4 AC so be careful so they will be you know do not write B square plus 4 AC which I did some time ago you must be very very careful of the science which you are using and you know remember it properly so hence it equals to be B square minus 4 AC and then either it is greater than equal to 0 or less than 0 basis that they will ask you under what conditions the nature of rules would be equal unequal or unreal or not real imaginary all those things would be there they will for example if I go back to the slide let me see I have put one question which was a one marker usually a one marker or two marker question is then always a word problem never mind so we'll come to the word problems as well so you know so the one marker question could be that in fact I have shot a few videos if you go if you go to our you know YouTube channel you can see all those questions you know solved there so you can always go through the previous question questions there yeah so nature of rules third will be third will be of word problems right so there could be word problems on on quadratic equations right so they will give you so let us take one word problem which was asked in previous year and let us try and solve that question so hence here is the question the question says the numerator of a fraction is three less than the denominator I have saw in this question also and have posted the solution online on our YouTube channel so you can go there and see that so the numerator of a fraction is three wait a minute I'm sorry three less than yeah so the numerator of a fraction is three less than the denominator okay the numerator of a fraction is three less than the denominator so I will highlight this three less than denominator so if I now know if x is the numerator the numerator of a fraction is three less than the denominator that means denominator is x plus 3 okay this is the fraction If 2 is added to both the numerator and the denominator, so let us add 2, so hence it is x plus 2 and the denominator becomes x plus 5. And this is the case, 2 is added to both the numerator and the denominator and the sum of the new fraction and the original fraction is 29 by 20, so hence what can I say? Okay, so hence it is, let me, you know, so just 29 by 20 is the sum, so let me go to this slide and then solve it here. So let us say, so, yeah, so yeah, so we will solve it here, okay, so let us say, they are saying that numerator is 3 less than the denominator and 2 is added to both numerator and denominator and the sum of the 2 is 29 by 20. Let us please read the question, at least 2 times to make sure that you do not misread the question. The numerator of a fraction is 3 less than the denominator, so x upon x plus 3 and 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is 29 upon 20 and you have to find the original fraction, very good. So basically now we have to find out x upon x plus 3, clearly this is not a quadratic equation in this form, but it can be reduced to 1. How? Let us do take the LCM, so hence if you say take the LCM x plus 3 and x plus 5 on the denominator and here it is x times x plus 5 and here it is x plus 2 times x plus 3 and this is equal to 29 upon 20, guys 29 is a very hard, it is a prime number and the moment you see number like 29, there is a thought in the mind that it will be difficult to multiply with 29, so be very very cautious, so hence there. So one thing which happens in this question is, what is that? That is you are prone to do careless mistakes or calculation mistakes, please always remind yourself that there is a possibility of careless mistakes, so be very very cautious. So let us try and solve this, so x times x is x squared plus 5x, then here if you know the identity you can use, what is the identity, let us say x plus a into x plus b, I know the identity is nothing but x square plus a plus b x plus a b, if you know that then you can easily do this multiplication or you can do a proper step by step multiplication. So I will say if you do not remember then you multiply step by step, hence it is x square plus 3x plus 2x plus 6 and it is always a good practice to count the number of terms also so that you do not have, you should miss any particular term, so product of two factors which two terms each will give you four terms, so see I have got four terms, and in the denominator again I should get four terms, what is it x square plus 5x plus 3x plus 15, so four terms, and then here it is 29 upon 20, now simplify, x square plus x square is 2x square and 5x plus 3x plus 2x is plus 10x, isn't it, and then it is 6 and then in the denominator it is x squared plus 8x plus 15 equals 29 upon 20, now is the time to do cross multiplication, so hence it is 20 times 2x square plus 10x plus 6, I hope I am not making any mistake, 29 anyways we will be checking, so 29x square plus 8x plus 15, exactly this will be the scenario where you will be writing the exam, so let us now calculate it is nothing but 40x square, then it is 200x, isn't it, 20 to 2 is 40, so 20 to 200 and then it is 120, and this is 29x squared plus 8 into 29 is nothing but 8 into 30 minus 8, so this is 232 I believe, so 8 minus 72, 7 that is 7, 16 plus 7 is 232 and there is an x and 29 into 15 is 30 into 15 which is 450 minus 15 that is 435, so 435, right, hence if you simplify 40 minus 29 is 11x squared, 200x plus or minus 232x is minus 32x I believe, I hope I am not, yeah, and then 120 minus 435 is, how much, it is 235, sorry this is 215, right, 125 and 123 and sorry 315, right now 120 minus 435 is, so check multiple times so that no worries, you know, because if you make one mistake and then the entire thing will go for a toss, yeah, 315 equals 20, okay, so this is my equation, let me write the equation afresh or let me get some space to write off, okay, so the equation is, what is the equation, the equation is 11x squared minus 32x minus 315 is equal to 0, now either we can go for a quadratic formula but then I am, I say, you know, let us try and see if I can get up, so 11 into 315, this is a negative 11 into 315, this is ac, right, now we have to split in such a way that I get a sum of 32, right, how do we do it, 11 is a prime number, so let's factorize 315 which is nothing but 3 into 105 or 9 into, yeah, so it is nothing but 9 into 35, right, so hence it is nothing but minus 11 into 3 squared into 7 into 5, correct, this is my, yeah, so now I need to find out, I need to break it in such a way, now again, if you see ac is negative, so now if ac is negative, I have to break this b in such a way that it is a sum of two numbers but one number has to be greater than the other in this case, okay, this is the case, right, now let us see, so 9 and 7, 9, 5, 45 and 63, okay, so if you see 77 and 9, 77 and 45 I think, so 77 and 45, yeah, so if you see 77 minus 45, if you see 11 into 7 is 77 and 9 into 5 is 45, 7 minus 5 is 2, so 32, right, so it works, so hence I can say it is 11 x square minus 77 x plus 45 x minus 315 equals 0, okay, so this means, this is 11 x x minus 7 plus 45 x minus 7 equals 0, check with 7 into 45 is 315, yes, 7 into 5, yeah, 315, right, it is correct, so hence it is x minus 7 and 11 x plus 45 equals 0, right, this implies either x equals to 7 or x equals to negative 45 by 11, now clearly the fraction was, you know, the term has to be positive, right, it was said that it is our integer, let us go back to the question once and see what was the question, so the numerator was 3 less than the denominator, so hence it has to be, you know, an integer, so hence clearly I get one integer, so x equals to 7 is one solution, so hence the fraction will be 7 upon 7 plus 3 that is 7 upon 10, now quickly check whether this is correct or not, so 7 by 10 plus add 2 to numerator and denominator must be equal to 29 by 20, let us check whether that is correct or not, so 10 and 12 if you see 60 is the LCM, so this is 6 into 7, 42 and then this is 12 into 5 and 95 is the 45, so hence it is 87 by 60, so divided by 3 you will get 29 by 20 that means it is correct, it is correct, so hence the fraction is 7 upon 10, this was our previous year question paper, so these are usually the questions which are asked in, you know, quadratic equation topic, let us take a few more and see whether we are, okay, so let me give you, let me solve another problem which is, yes, okay, so let me take another CVSC problem, now the question is let me go to one slide, okay, so now the question is find the values of k for which the given equation has real and equal roots and this is CVSC 2015 question, even if the pattern has changed the nature of question will be similar, so the question is find the value of k, find the value of k, k such that, such that, such that, such that k plus 1, k plus 1 x squared minus 2 times k minus 1, k plus 1 x squared minus 2 times k minus 1 x plus 1 equals 0, again, read the question carefully again and again, don't misread the question, so find the value of k such that the equation k plus 1 x squared minus twice k minus 1 x plus 1 equals to 0, 0 has real and equal roots, equal roots, now this is another way of testing, you know, what they will do is they will give you a quadratic equation within a quadratic equation in terms of, you know, so you, in this question also though you have to find out the coefficients here, but actually it will lead to let's say, you know, use of what solving quadratic equation itself, so what is the condition for real and equal roots, you know that condition number 2 that means d must be equal to 0, what was d? d was b square minus 4ac, so b square minus 4ac must be equal to 0, so what did I tell you? write a, b and c, so what is a? a is k plus 1, b is negative twice k minus 1 and c is clearly 1, c is clearly 1, now let us find out the value of d, so b square, so hence write 2k minus 1 and do not hesitate to put as many braces as possible, so twice square minus 4 into a, a is k plus 1 and c is 1, so 1 and this must be equal to 0, so for real and equal roots this must be satisfied, now if you see what is it, if you square it, the negative sign has got no effect, so it is 4k minus 1 whole squared minus 4k minus 4 must be equal to 0, is that it? I have expanded it and now let us expand this, what is it? 4 times k square minus 2k plus 1, a plus b plus, you know, a plus b whole square is a square plus twice a b plus b squared, so hence you do this minus 4k minus 4 equals 0, so if you simplify this, this is 4k square minus 8k plus 4, keep checking again, immediately check also helps a lot to eliminate any careless mistakes, so you quickly go back and see 4k square minus 8k plus 4 quickly and then minus 4k minus 4 equals 0, now what is it? This is 4k squared and then minus 8k minus 4k is minus 12k, do not make mistakes in adding with signs and then 4 and minus 4 is 0, so it is 0, right? This makes our job a little easier because now you don't need to go for any, you know, complex method of solving the equation, why? Because the coefficient is 0, so hence what you can do is, you can plug it, plug out what? 4k is common, so hence it is k minus 3 equals 0, so hence you get either k must be equal to 0 or k must be equal to 3, okay? So either k must be equal to 0 or k must be equal to 3, let us check whether that is actually the case, so if you put k equals to 0 in this equation, what will you get? You will get x square minus x square plus 2x plus 1 equals to 0, which is an identity, if you see, if you put k equals to 0, this is a check, right? If you put k is equal to 0, you will get x square plus twice of x plus 1 equals 0, which is nothing but x plus 1 whole square is equal to 0, that means it has two roots both are equal, what are the roots? x equals to 1 and 1, both are equal, so hence for k equals to 0 it is valid, let us check whether it is valid for k equals to 3 also, so in case x k equals to 3, then what will happen? It will be 4x square minus 4x plus 1, indeed it is 2x minus 1 whole square is equal to 0, so x equals to half and x equals to half are the two equal embryo roots, so hence both the solution that is k equals to 0 and k equals to 3 will satisfy the condition and the equation will then have real and equal roots, okay? So let us take another problem from previous year paper, so usually these will be, let's say, you know, either in one marker or a two marker question mostly, okay? Now the question is, there's an interesting question and though it is not a previous year question but then it could be asked in the exam, so let us say the question says, show that the equation, the question is show that the equation show that the equation x squared plus ax x squared plus ax minus 4 equals to 0 has real and distinct roots, distinct roots, has real and distinct roots for all real values of a for all real values of a, again read the question once again, make it a habit, show that the equation x squared plus ax minus 4, sorry not ax, x squared plus ax minus 4 equals to 0 has real and distinct roots for all real values of a, okay? So here's, so what is the condition for real and distinct roots, you know, d must be greater than 0, right? Not even equal, it has to be greater than 0, okay? So let us find out d, d is nothing but for a quadratic equation it is b square minus, sorry, 4ac, isn't it? Now what is b here? The b value is a, if you see, don't get confused, here the b value is a, so hence I will write a square minus 4ac, so minus 4 times 1, a is 1 in this question, so this, don't get confused by this a, yeah, this a is nothing but b in our general equation and here the a value is 1 and c value is negative 4, so hence negative 4 is equal to a square plus 16, correct? Now a is a real number, a is a real number, isn't it? So hence a square is always greater than equal to 0 because real number squares cannot be negative, right? So a square is always greater than equal to 0, so clearly a square plus 16 is always greater than 0 or it cannot be said as equal to 0 because 16 is added to a non-negative number, so a square plus 16 is always greater than 0, whatever is the value of a, isn't it? So hence b is always greater than 0, hence it is a condition for real, sorry, it is a condition for real and distinct, distinct roots. Fair enough, so this is on nature of roots, let us now take few more different type of question, okay, another question, let's say let's take another previous year question, so the latest or other, you know, so let's take, let's take up CBC 2014, okay, now the question again is on the nature of roots, nature of roots and the question is asking this, find the value of K, so again find, find the value of K, find the value of K for which, for which, for which, for which the roots, for which the roots, roots are, roots are real and equal, real and equal for the following, for the following equation, okay, and what is the equation given? In CBC 2014 this question was asked and this is Px times x minus 3 plus 9 equals 0, this is which year? This is CBC 2014, not that, you know, not long back. Now again read the question once more, the question says find the values of K for which the roots are real and equal, again real and equal, for E for following equation, what is the equation? Px times x minus 3 plus 9 equals 0, now if you see there is no direct x square term which you can see but it is, it is, right, so you know, don't get confused, you can see that if you expand this you will get an x square term, so if you see it is nothing but Px squared minus thrice Px, 3 times Px plus 9 equals 0, now for real and equal what do I know? D must be equal to 0 for real and equal roots, is it right? So that is so, then what is D? D is B square minus 4 AC must be equal to 0, now what is B? B in this case is negative 3P, if you see negative 3P is B, so B square, that means square of this minus 4 into P into 9 must be equal to 0, so if you see this is 9P square minus 36P equals 0, you can, you need not multiply it here also because there is a, you know, you can extract a common factor, so you can now say 9P into P minus 4, isn't it? Equals 0, so this is the case then either P equals to 0 or P must be equal to 4, again there are two solutions to this, so let us check if P equals to 0, actually P cannot be 0, if you see P cannot be 0, why? In a quadratic equation the coefficient cannot be 0, so hence you have to write P cannot be 0 or you can write P cannot be 0, cannot be 0, cannot be 0, because you name it as 1 and say because, because 1 is a quadratic equation, but hence let us check whatever if P equals to 4, so if P equals to 4 the equation becomes 4x square, let us check, so let us check 4x square minus, this check you don't do while solving the question in the main sheet, you can always do it in the rough, so 4x square minus 3 into P, P is 4 again into x plus 9, isn't it? So which is nothing but this is, yeah, so this is 2x whole square minus, yeah, so this is 2 into 2x into 3 if you see, yeah, 4x into 3 is this plus 3 square, this is equal to 0, this is equal to 0 and hence it is 2x minus 3 whole square is equal to 0, so hence there is only one solution, equal rules, right, what is that? 3x equals to 3 by 2, so hence P is equal to 4 is the right answer, okay guys, so we solved this question also on nature of roots, let us go and solve some word problems now, okay, there are different types of word problems, we will take up one by one, you know, we just saw numerator and denominator kind of a, yeah, yes, let us take up some previous year questions, first is on digits of 1, yeah, so let us take up some previous year questions, next is now, meanwhile guys, if you have any question in your mind or let's say if you are not able to solve anything, you can always post it here or through WhatsApp and we will be able to solve it and post it here, okay, let us take another question, so question was asked, this question was asked in 2006, question was asked in 2006 CBC, what is the question? Question is a two digit number, a two digit number, two digit number is such that is such that the product of its digit that the product of its digits, digits is 18, okay, when 63, when 63 is subtracted, when 63 is subtracted from the number, from the number, from the number, the digits interchange their places, the digits interchange their places, their places, so you have to find the number, find the number, once again, read the question, once again, so that you don't make a mistake, okay, the question says a two digit number is such that the product of its digits is 18, when 63 is subtracted from the number, the digits interchange their places, find the number, now this is interesting, interchanging of the digits, right, so let's say two digit number is such that the product of its digit is 18, so if one digit let, you write like this, let one of the digits be x, then the other digit will be 18 upon x, now in such questions you know that digits will be an integer, right, what integer, from 0 to 9, but clearly x is not 0, why, because if x were 0, I'm sorry guys, so if x is 0, then you will not get the product of the two digits as 18, also x cannot be 18, why, so you can do, this thing should go in your mind parallely, you don't need to stop and think like this, but I'm just giving you a hint to how to eliminate your, or how to get to answers, and this also can help you in let's say checking whether you have done it correctly or not, so x clearly cannot be 18, because if you see 18 is the product of two digits, so 18 can be 1 into 18, 18 can be equal to 2 into 9, so here is one possibility, 18 can be also written as 3 into 6, other possibility, right, and then 6 into 3, and then 9 into 2, and then 18 into 1 again, so hence if you see either it will be, you know, so either of these solutions will be there, okay, but let's go ahead and see whether we are doing it correctly or not, so yeah, now the second is when 63 subtracted from the number, so what is the number guys, let's say x is unit space, okay, so let's say x is in unit space or one of the digits bx, let's say you also write unit space, unit space, okay, so what is the number, so the number is, the number will be nothing but 10 into 18 upon x plus x, this is the number, right, why because if you are having a two digit number, let's say ab, then the value of the number is a into 10 plus b, all of you, no this I am not writing, I am explaining it again, so this is the number 10 into 18 upon x plus x, now what are they saying, if 63 subtracted from the number, the digit interchanges, so let us subtract some, 63 from it, so 180 by x plus x minus 63, so we can do that, that the number, the digits interchanges space, so units becomes 10s and 10s becomes units, so it will be now 10x plus 18 by x, so this is the question, now once you are done with formulating an equation, your job is half done, you are now, you have to just solve the equation, so let us see, let's try and solve, okay, so once the, you know, it is a better practice to check once again whether the formulation of the equation is correct or not, so if you see this is 10x, so 10 times the digit at the 10s place plus the units place digit minus 63 is equal to the reversal of digits, okay, looks good, so let us now solve it by collating all the x terms together, so if you see, this is nothing but 180 by x and so let's take everything on the right inside, this means that 10x minus x is 9x, okay, so this x will go on to the right inside, if you are not comfortable with that then you please write full step, full step will be this, that 10x plus 18 by x and then this becomes minus 180 by x, this becomes minus x and plus 63 was 0, okay, but directly also you can write as what, so 9x and then 180 minus 18, so this is nothing but 162 by x, right and then plus 63 equals 0, is it it, so 9x minus 162 by x plus 63 equals to 0, you can make it, it is not a quadratic form right now but you can reduce it to 1, so it is 9x squared minus 162 plus 63x equals to 0, right, so hence it is x 9x square, yeah, so if you see you can actually take out 1, 9 from everywhere, isn't it, all are multiples of 9, all the coefficients are multiple of 9, so you can say x square strike of 9 minus, this is 18x for 18 and this is 7x equals 0, yeah, so hence it is x square plus 7x minus 18 equals 0, this implies you can write x square minus 9x plus 2x, sorry the other way around, so hence this is what I was saying, please be very very every step you just do a mental recalculation, so plus 9 and minus 2x minus 18 equals 0, this implies you can take x as common for the first two terms, so x plus 9 and this is negative 2 common from the second last two terms, x plus 9 equals 0 and hence I am writing it here, now it is nothing but x plus 9 times x minus 2 equals 0, x equals to negative 9 or x equals 2, right, clearly x equals to negative 9 is not a solution because we are talking about digits, digits are between 0 to 9 both inclusive, so x equals to 2 will be the right answer, so when x equals to 2 the number is 18 by 2, 92, so hence the number is 92, right, 92, now if you subtract 63 from here, what will you get, you will get 29, you should get 29, so if you see 92 minus 63 is actually 29, so hence the number the digits are interchanging, it is present, right, so hence our solution is correct, so this was another type of problem, let us go to another one, another different type, so let us solve another, yeah, now quadratic equations for solving problems on time and distance, we will take up one time and distance problem also, this is also very common commonly asked, okay, so let me use the space, yeah, so the question again, question is this and I will take up a previous year paper question, so this is again 2006 paper which is there, so there is absolutely no problem in solving a little older questions also, why, because the pattern of the question remains the same or rather we can take up a recent one, so 2000, the question is a motor boat, a motor boat, motor boat who speed in still water, who speed in still water, still water is 18 kilometers per hour also, be mindful of the units, okay, takes one hour more, one hour more to go to go 24 kilometer, 24 kilometer upstream, upstream, upstream, then to return down, return downstream, downstream to the same spot, to the same spot, okay, find the speed of the stream, fine, you have to find the speed of the stream, okay, so once again, a motor boat who speed in still water is 18 kmph, 18 kilometers per hour, it takes one hour more to go 24 kilometer upstream, that to return downstream to the same spot, find the speed of the stream, okay, so will, so whatever is required you assume that to be the variable, that is the common trend or common practice, find the speed of the stream, so I'll say let the, let the speed, speed of the stream be x kilometers per hour, write the units here itself, so that even if you miss writing units literally, you know, in the last step, then you know, it should not make or you should not attract any analytics, now, a motor boat who speed in still water is 18 kilometers per hour, obviously when it is going downstream, its speed will be more, why, because the velocity of the water will aid the velocity of the motor boat and while it is going upstream, it will have a less speed, is it not, so now it is saying one hour more, so what is, what is, you know, what is constant here, constant is the speed, sorry, the distance, right, so this is traveled upstream is equal to distance traveled downstream, so what is distance traveled, distance, oh, okay, 24 kilometers is given actually anyways, one hour more to go 24 kilometer upstream, okay, so when your distance is 24 kilometers, so distance is equal to speed into time, you know that, in case the, if speed is constant, then distance is speed into time, so hence 24 kilometers is equal to speed of, let's say while going upstream, it takes time p, okay, so the speed was x and it is a drive, so hence, so what will happen, so let us write this thing separately, so hence velocity upstream, that means when the, or the speed upstream and the steamer or the boat is going upstream, the velocity will be 18 minus x kilometers per hour, is it not, kilometers per hour, why, because the velocity will be reduced, because the motor boat has to go against the speed of the stream, okay, now and velocity downstream, downstream is equal to 18 plus x kilometers per hour, is it not, 18 plus x kilometers per hour, now, so while it is going upstream, so distance is 24 into, yeah, so time, so hence we will say time upstream, okay, time upstream is equal to nothing but distance by speed, is it not, so distance is 24 and speed for upstream is 18 minus x, okay, time downstream is equal to 24 upon 18 plus x, now if you see in both the t up and t down, 24, the numerator remains same, denominator is more in the second case, so hence t down is less, which is quite obvious, but denominator is more here, denominator is less here, so hence t down, that means going downstream the speed, sorry time taken will be lesser, now there is a difference between the two times, obviously this is more, is it not, t up will be more than t down, but there is a relationship given, what is that, it takes 1 hour more to go up, that means I can write t up is equal to t down plus 1, is it not, so hence what is t up, so let's now form the equation, so 24 upon 18 minus x is equal to 24 upon 18 plus x plus 1, so time, I calculated time up, I calculate time down and then I know time up is 1 hour more to go upstream, right, so this is my, so you will get 50 percent of the marks here itself when you are writing the equation correctly, now the question remains to solve this, the other half is to solve the question correctly now, so hence let us solve this, now how to solve this, you have to write 24 upon 18 minus x minus 24 upon 18 plus x equals 1 after rearranging, take LCM, so it is and 24 is common, so you can write 24 into 18 plus x and then here it is and the denominator first write the denominator, so 18 minus x times 18 plus x denominator and hence 24 is common, so 18 plus x minus minus 18 minus x like that, equals 1, now what is it, so if you see 18 minus 18 will get cancelled, it will become 2x in the denominator, so 2x, so hence it is 24 into 2x upon 18 minus x times 18 plus x equals 1, okay, so hence if you see, if you see the next is solve this, this is 48x is equal to 18 minus x and 18 plus x, okay, so this is 48x is equal to 18 into 18, so I am not multiplying, I am just keeping it like that because you know, we will see if we can factorize it later on and then oh so it is 18 square minus x square, then and simple, 18x minus 18 plus x is a square minus b square, now the final equation is x square minus sorry plus 48x minus 18 into 18, so I am purposefully keeping it, I am not multiplying because anyways you have to factor it, okay, so let us see how do I split 48 to get 18 into 18, okay, so hence what is 18 into 18, 18 into 18 is 81 into 4, so if you see this is nothing but 18 into 18, what is it, it is nothing but 3 square into 2 into 3 square into 2, right, so now you know 81 into 4 that which is nothing but 4 324, yeah, but you have to break this 18 into 18 in such a way that you get 48, so obviously since it is a negative number, so you have to break in such a manner that it is 48 is the difference of two numbers whose product is 18 into 18, okay, so let us try and see if we can get 48 here, so if you see it is how do we, so let us say 9 into 4, so yes, 6, so 9 into 4 is, yeah, so how do we break it, so if you see 48, so hence I have to, let us break them, so hence clearly 3, I am doing it here, so 18 into 18 can be written as 3 into or let us start with 2, so 2 into what is left, 81 into, 9 into 9, 81 into 2 is 162, this will not work, then 3 into, this is also not going to work, why, because it is 2, 2 bigger number for having a difference of 48, so hence we have to find, let us say 3 into 4 is 12 and here it is, no, we will have 6 into 6 and rest is 54, 54 and 6 is the number, if you see, how 3 into 2 is 6, so 6 if you see 6 and 54 is the other factor, right, so 6 into 4 is, 6 into 54 is 324, correct, so hence this will work and 54 minus 6 is 48, so hence I get x square plus 54x minus 6x minus 324 equals 0, so this implies, this implies x into x plus 54 minus 6 times x plus 54 equals 0 and hence, so hence if you see, this is nothing but x plus 54 and x minus 6 equals 0, so either x equals to negative 54 or x equals to 6, but speed cannot be, speed cannot be negative, right, this cannot be negative, so hence we say speed cannot be negative, can't be negative, so hence solution is x equals to 6 kilometers per hour, right, this is the solution, this was asked in 2014, CBC, let's take up another take, now it is nothing but on ages, let's say there is a question on age, okay, so this is another type you'll get, let us solve this, this is again a CBC 2010 question, CBC 2010 question, yeah, question says, a girl is twice old as her sister, a girl is twice as old as her sister, yeah, you have to convert these into expressions, right, mathematical expressions, four years, four years hence, now in four years hence the product of their ages, the product of their ages, the product of their ages will be, it's mentioned in years, will be 160, 16, 0, find their present ages, find their present, present ages, present ages, present ages, so a girl is twice as old as her sister, so hence you have to be careful about, you know, years hence, please be careful, many of you, what you do is you augment the case of the product to add to the age of the product, right, so now, so what is the question, find their present ages, okay, so we're saying, find their present ages, present ages, present ages of the sister's be X, X and X plus, sorry twice X, right, so X into X, so girl is twice as old as her sister, right, X into X, four years hence, the product of their ages will be 160, so four years hence, four years hence, the ages of sisters would be, how much, would be X plus four and two X plus four, another product they're saying, so X plus four and two X plus four is equal to 160, so you have to find out what is the value of X, so let us simplify this, this is two X squared plus four X, two X squared plus four X, plus eight X, plus eight X, plus eight X and then 16 equals 160, then quickly check whether it is correct or not, so what is it, two X squared, that's correct, then four X, that's correct, four into two eight X, that's also correct and four into four is 16, which is equal to 160, so this implies our two X squared plus 12 X plus, sorry, minus 144 equals zero, you can eliminate two from here, you can eliminate two from here, this is the quadratic equation, so half the marks will be given here itself if you have done it correctly, so hence eliminating two you'll write X squared plus six X minus 72 equals zero, you can write dividing, dividing the equation by two, isn't it, now it boils down to solve this question, so hence it is, if you see 72, so 72 is to be broken in such a way that, such that the product is 72, when the sum is six, so there's no brainer here, you can do it as plus or plus 12 X minus six X minus 72 is equal to six, zero, why, because 12 X is 72, if you now do this take common, this is X plus 12 minus six X plus 12 equals zero, so this is X plus 12 into X minus six equals zero, so hence, so hence my dear friends, X is equal to negative 12 or X equals to six, but clearly this is not a feasible solution, not a feasible, feasible solution, why, because age cannot be, age cannot be negative, so X equals to six is the right solution, okay, so the ages of the sisters will be ages of the, or you should write present ages, present ages of the sisters are six and twice of six, two into six equals to 12 years, this is about, you know, age is problem, okay, now there will be few application of the same in geometry, you know, you can use Pythagoras theorem, there will be illustrations in that, then in mention they can also combine a problem in geometry or menstruation, so we'll take up one of that sort, so let us see what is other type of other application, so one is, you know, again CBC 2000, CBSE 2015 question, now this question says the area, the area of an isosceles triangle, isosceles triangle is 60 centimeters square and, and the length, and the length of, of each one of its, of its equal, equal size, equal size is 13 centimeter, is 13 centimeter, okay, find its base, find its base, so let's read the question once again, the area of an isosceles triangle is 60 centimeters square and the length of each one of its equal size is 13, find its, find its space, very good, let's try and solve this problem, so isosceles triangle, we'll have to make, okay, so this is my isosceles triangle, so in geometry problems, please do not forget to make a drawing map, okay, now the area of an isosceles, so what is an area of an isosceles triangle, so you're going to draw a particular function, so always draw, always draw a figure whenever you are solving a geometry problem, okay, so this is a rough diagram where you do, you know, you can use it for this purpose, now what is it given, right, so each of the equal size, so this is 13 centimeter, 13, so let's say A, B, C, so it's given, what is given, given is A, B is equal to A, C is equal to 13 centimeter, okay, and area of triangle A, B, C is equal to 60 cm square, okay, now I have to find, to find what, to find what, B, C, okay, now construction, what did it do, construction, drop, are you right here, A, D, A, D perpendicular to B, C, okay, now clearly in an isosceles triangle, you know that BD is equal to, BD is equal to DC, right, you can write the reason in an isosceles triangle, the perpendicular from the vertex, vertex bisects the non-equal side, okay, so let us say BD is equal to, right, so let us say BD, let BD is equal to X, okay, so to find BD, now if BD is equal to X, then AD is equal to nothing but under root 13 square minus X square, isn't it, AD is 35 and you write the reason by Pythagoras here, okay, now what do we now do, so I now know the X, so area, area of triangle, so half into 2X, half into base into height, that is AD will be equal to 60, that is what is given, okay, so hence half into 2X, so I can just cancel this thing off, so X into AD, what is AD under root is equal to 60, isn't it, so hence if you, so and then you'll write, what will you write, you will write squaring both sides, these steps please do not forget, squaring both sides, right, you'll get X square into 13 square minus X square is equal to 60, isn't it, so hence it is nothing but 13 X square minus X to the power 4 is equal to 60 or you can write X to the power 4 minus 13 X square is equal to, oh sorry, plus 16 is equal to zero, so I hope we are doing the 13 square minus X square, yeah, half into 2X, half into, oh wait a minute, yeah, half into 2X into AD which is 13 square minus X square is 60, so hence, hence you will get X to the power 4 minus 13 X square, oh no, wait a minute, this will be, this is a mistake, this is square, so this will not be 60, this will be, sorry, this will be 3600, plus 3600 equals to zero, so please be very, very careful, oh this is also 13 square, right, so please be very, very careful when you are doing this, so hence it is nothing but X to the power 4 minus 169 X square plus 3600 equals zero, okay, so again either you can go for quadratic formula thing and also it is a bi quadratic equation, so you can say let X square be Y, you can solve it here itself or you can simplify by saying let X square equals to Y, so the equation is reduced to Y square minus 169 Y plus 3600, okay, so now if you see, if you try to split 3600, so 36 into 100 is 1, but 100 plus 36 is not 136, so we will have to do a little bit more, so if you see it is also, this is actually 44 into 125, yes, yes, so 125 into 44 actually will give you 3600, yes that's fine, I think it is correct, I'm sorry not 44, no, no, no, sorry the other way around, 144 into 25, 125 doesn't divide 3600, so it is 144 into 25, yeah, that makes sense, so 144 into 25, yeah, is this 34, yeah, so this is equal to zero, so hence what will you say, you will say Y square minus 44 Y minus 1, sorry 144 Y minus 25 minus 25, write it properly, don't overwrite, so yeah minus 25 Y plus 3600 equals zero, okay, yes, so hence it is Y times, so I'm writing it here, Y times Y plus 144, yeah, so Y times Y plus 144 and then Y plus 144 anyways will be taken out and here it will be simply minus 25 equals to zero, am I right, not, oh sorry wait a minute, there is this thing, yeah, so that's what the problem is, please write it very clearly, so it will not be plus, it will be minus, right and here also this is minus, so hence it is Y minus 144 into Y minus 25 equals zero, so either Y is 144 or Y is 25, these are the two possible solutions, so if Y is 244 guys then what will be X, X square was equal to Y isn't it, so hence we can write, that means X square is 144, so X equals 12, I am neglecting minus 12, why because negative 12 cannot be dimension of a triangle, so hence X equals 12 and from here X square equals 25, so X will be simply 5, again I am ignoring the negative values, ignoring X equals to negative 12, these will definitely be the solution to these equations but they will not be feasible solution, X equals to negative 5, ignoring these, right, so X is 5 or X equals to 12 centimeter, so hence you have to find out the base, so for the lack of space I am writing it here, so the base BC will be twice of X, that is either 10 centimeter or 24 centimeters, okay, this is how you will be solving these sums, so I thought, I think we have covered a lot of problems today, we also covered the theory part, so I think you guys should be good now and as I told you there will be typically three types of problems which will be asked, one is solving the equation by any of the methods, second will be on the nature of rules and third will be world problems where again solving of equation will be important as you know very important ingredient of the problem, so if you are you know solving all the previous year questions around 20 odd questions you know before you take the exam I think you should be good and in case you have any difficulty, any doubt or any concept has to be revisited, please reach out to me and if needed we will be doing another dedicated session on the same topic, I hope these revision process are helpful to you, so please attend these, if you are not getting time or because of let's say your other mock exams going on school, if you are not able to find time I would recommend that whenever you get time you please go through the youtube channel and see it, watch it at your own pace, so we will call it a day and I hope this session was useful to you and we will be back again with another class tomorrow, so thanks for attending the session guys, all the best