 welcome friends so again we have taken a quadratic equation now this doesn't look so beautiful here because of this 2 root 2 but this is what we are going to learn today if such equations are there how to solve such quadratic equations and that too by factorization method we are going to adopt factorization we can directly use the quadratic formula and get the solution but we are going to adopt factorization method which is clearly splitting the middle term here so you see we have to basically break 2 root 2 into two parts such that the product is 1 times 6 okay now how to do that so clearly if you see b here is an irrational number 2 root 2 and let me write it so let's say a equals 1 here c is minus 6 so ac is minus 6 a rational number isn't it but my b is 2 root 2 irrational number now how now you have to split b into two terms b1 plus b2 such that it is equal to b and if I split an irrational number into you know rational and an irrational then the product will never be rational so both b1 and b2 must be irrational and that too carrying root 2 only right so both of them must have a irrational part with root 2 right so hence you can assume there is something some fill in the blanks root 2 plus fill in the blanks root 2 and I have to just get this x and y let us say x and y such that this is equal to 2 root 2 okay but the product of this must be 6 that means we must have 3 in x and y either either of x and y must be having 3 so hence we can we can say that it is nothing but 3 root 2 minus root 2 if we see this is clearly equal to 2 root 2 and the product also is if you product these two you'll get minus 6 isn't it so we got the split of the middle terms again once again see we had an irrational number in b and rational number as ac so both the split terms should have root 2 in it only then when multiplied to irrational number when multiplied will become rational right only only one in one case what is that case that the irrational part must be same in the two parts right so hence both b1 and b2 must have root 2 in it so that when it get multiplied the root 2 into root 2 becomes 2 and hence the irrationality disappears you learn this by more practice but this is how it has to be done and then if you if you know split those like this 3 root 2 minus root 2 it works why because 3 root 2 minus root 2 is 2 root 2 and 3 root 2 times minus root 2 is minus 6 which we wanted as a you know as a last term right so hence how can we do this so now coming back to the equation it is x square plus 3 root 2 x and minus root 2 x and then minus 6 equals 0 now if you see what are the common factors here x that's it so hence it will be x plus 3 root 2 ok now without any thought you can just write this x plus 3 root 2 here because it has to be that common factor again now we have to just fill in this blank isn't it so how to do it so just check what should I multiply root x with to get this term and that is nothing but minus root 2 if I multiply here minus root 2 then we'll get minus root 2 x and the other will automatically come this 6 will automatically come how minus root 2 times 3 root 2 you see is minus 6 so our job is done now let us take the common factor so x plus 3 root 2 is common to both and the rest items are x and minus root 2 equals 0 so hence what will be the solution x plus 3 root 2 must be equal to 0 or x minus root 2 must be equal to 0 that means x equals minus 3 root 2 or x is root 2 so these are the two equation the best ways to you know deploy these values back and you will get to know whether you have found the solution correctly or not in this case I know that the solutions are 3 root 2 minus 3 root 2 and root 2 correct so this is how you solve the equation other way to check is you know just sum the this thing does the two solution yeah the two solutions which you got if you submit you will get the middle term negative of the middle term by the coefficient of x square anyhow we will we will you know make you learn this trick a little later when you deal with some of the roots and product of the roots till then you can park this mechanism of checking whether the solution is correct or not the alternative way is just deploy the found out solution back into the equation and it must be satisfied that is LHS must be equal to RHS