 Welcome again friends so continuing in the series on quadratic equation in this session and subsequent few more sessions we are going to discuss now methods of solving a quadratic equation methods right so we had so far seen what is quadratic equation and what is meant by solution to quadratic equation now we were we are going to learn methods of solving a quadratic equation now there are you know lots of studies on the different methods of solving the methods which we are going to take up today are a solving by factorization solving by factorization this is one of the methods will study in detail then we have solving solving by completing the square method completing the square method don't worry we are going to discuss everything at length and third one is solving by solving by Sridharacharya's formula Sridharacharya's method so Sridharacharya happens to be a Indian mathematician who had somewhere in around 600 700 or rather 8th century AD he has given one method then I'm not very sure about the dates though so don't take the dates but he was an Indian mathematician who has given a method so solving by Sridharacharya's method we learn and then there is one more and that is nothing but quadratic formula quadratic formula however all these methods are quite similar to each other one leads to the other so we remind you know not much of a difference is there but then we it is you know important to know these methods separately so today let's start with solving by factorization method okay so what is that so let us take a quadratic equation ax square plus dx plus c equals 0 so we are studying here factorization method factorization method right so we are going to try to factorize the given polynomial on the left-hand side this one and then see how to solve it right so factorization method so hence let us say if we could factorize px which is equal to ax square plus bx plus c now we have learned a lot of factorization techniques in previous grades so we will be adopting one of them there also there is something called completing the square then splitting the middle term using algebraic identities whichever way you can use you factorize it and then convert this quadratic polynomial into two linear factors let's say px plus q and rx plus s okay this let's say we convert a quadratic a quadratic polynomial can have at max two linear factors and then let us equate it to 0 now if you see here if you pay attention on this part two linear factors product is 0 how is that possible only when either either of them is 0 so either px plus q equals to 0 or rx plus s equals 0 right if either of these conditions are fulfilled then the given quadratic equation or this this will hold isn't it so that means if you now this is a simple linear equation which which means x equals minus q upon p will be one of the solution or x equals to minus s upon r so hence these two become the solution right these two become the solution so let us take an example and understand let us say we have x square plus 6x plus 5 equals 0 this is an example we are going to solve this quadratic equation now if you see on the left hand side we have a quadratic polynomial and there are three terms so we know a very common method of factorizing it which is called splitting the middle term so we split the middle term those who do not know spreading the middle term method are advised to go through our previous lectures on factorization but nonetheless we'll also try and you know repeat the process so how is splitting the middle term work how does it work so basically what you need to need to do is let us say you have a x square plus bx plus c this is the polynomial okay so what you need to do is you multiply a with c okay a with c so hence you when you multiply a with c so and then next step is you have to split this b into two let us say b1 plus b2 so you have to do this split b into two parts such that b1 plus b2 is equal to b okay now b1 and b2 can be negative both negative both positive one negative one positive right whichever way you have to split this into two parts b1 and b2 which is and some is b such that b1 and b2 product is equal to ac this is what is the crux of splitting the middle term right so hence we say so b is split into two terms b1 and b2 and hence it is called splitting the middle term let us see how we work here so basically here b is equal to six so you have to you know break six into parts such that the product of those parts will be nothing but here there is one so one times five is ac so one times five so we have to break b into b1 plus b2 should be equal to six such that b1 b2 is equal to one into five which is five clearly if you see don't need to break your head on to this and clearly if you see b1 if you write b1 as one and b2 as five just by trial and error itself you can get the solution so hence I can split my polynomial like this x plus 5x now if you see x plus 5x is 6x and then plus five equals zero correct so if you now see it is you can take x common now from the first two terms it will become x plus one and then take five common from the other two terms it will become x plus one again this is equals zero and now from these two terms here you can take x plus one common because x plus one is in both and this is reduced to x1 plus x plus one plus x plus one times x plus five equals to zero now we have two factors here one and two so these two product is zero only when either x plus one is zero or x plus five is zero or both of them right so hence x equals to minus one from here or x equals to minus five so x equals to minus one and x equals to minus five is the solution let's check whether it is true so we know how to check whether this extra x equals to minus one is a solution or not put it back into the equation and see whether the LHS is equal to RHS so we had the equation x square plus 6x plus five equals zero isn't it let's put x equals to minus one so what are we doing we are doing check checking whether the solution is correct or not so minus one whole square plus six times minus one plus five which is clearly equal to zero equal to RHS so hence minus one is correct solution similarly for minus five x equals to minus five let's check so minus five square plus six times minus five plus five is nothing but 25 minus 30 plus five which is again equal to zero is equal to RHS so hence both our solutions are correct x equals to minus one as well as x equals to minus five let's take one more example we can take this equation nine x square minus 3x minus two equals zero so clearly here a is nine c is minus two so ac becomes minus 18 and b is minus 3 so you have to split b in such a way so that b1 plus b2 is minus 3 and b1 b2 is minus 18 okay so it is always advisable to factorize b1 b2 or ac so minus 18 is nothing but if you see it is minus 2 into 3 into 3 right now one combination of this only two combinations rather will give you b1 and b2 so if you see it can be written as minus 6 into 3 right and minus 6 plus 3 is clearly minus 3 so hence we got our b1 and b2 so minus b1 minus point b2 are there right so hence nine x square can be split and sorry minus 3x can be split as minus 6x plus 3x minus 2 equals to zero where did I get minus 6 from here and where did I get 3 from here so this is the these are the two split terms so now you can take what 3x common from the you know first two terms and it is left with 3x minus 2 and again if you see this is one times 3x minus 2 equals 0 so 3x minus 2 times 3x plus 1 equals 0 so again by the logic that either of the factors must be 0 so 3x minus 2 is 0 or 3x plus 1 equals 0 that means x is equal to 2 upon 3 or x equals minus 1 upon 3 so this is the solution for the given given equation so understood what did we do we took equation and then we you know tried to factorize it and then we equated to 0 each term must be equated to the equated to 0 and hence finally you've got the solution okay we'll take up more problems in next session