 Hello friends, so far we have seen how to solve quadratic equations, there were multiple methods we have seen, we have seen factorization method, we have seen Sridharacharya's rule, you know completing the square method and others. Now thing is let us say we have an equation which is not quadratic, let us say here is an example, x to the power 4 minus 10x square plus 9 equals 0. So it is clearly a bicodratic equation, so the degree of the polynomial on the left hand side is 4, so will the rules which we have learned in quadratic equation help here? So it is clearly not a quadratic equation but if you see I can reduce it by some assumption to a quadratic form and what is it? So let us say, let us assume y to be equal to x squared, okay? So hence now this equation can be reduced to quadratic form and what will that be? So x squared square I can write this as this minus 10x square plus 9 equals 0 isn't it? Now what? You have to just simply put y in place of x square, so it will be y square minus 10y plus 9 equals 0. Now clearly it is quadratic, this is a quadratic form and quadratic in what? Which variable? y, variable here is y. We have been familiar with the solving equations with variable x but how does it matter whether you call it x, y, t, s, a whatever. If it satisfies this kind of a pattern then it is a quadratic equation, right? So now easily I can adopt any of the methods which we have learned either those literature is rule, quadratic formula rule, factorization rule. So in my experience I have seen that most of the quadratic equations related to let us say an exam or something until it is real, you know you are designing some factor equipment or something, a real life situation. Until it is a real life situation most of the time in any comparative exam you will see that the most of the quadratic equations will be solved by factorization method itself. So be very thorough with that method, okay? So here I can split the middle term and write that as y square minus 9y minus y plus 9 equals 0, right? Now you can take comments. So hence y times y minus 9 and minus 1 times y minus 9 equals 0. So hence this becomes y minus 9 times y minus 1 equals 0. That means y is either 9 or y is 1. Now what was y? We never wanted y, we wanted x, right? So hence we can now say x square is 9, right? Or x is 1, x square is 1, sorry x square is 1. That means x square minus 9 equals 0 or x square minus 1 equals 0 or I can say x minus, x square minus 3 square is 0 or x square minus 1 square is 0, right? Again I can factorize this as x minus 3, x plus 3 equals 0 or x minus 1, x plus 1 equals 0. Now you know the solution to these quadratic equations. So either x is equal to 3 or x equals to minus 3 or here x equals to 1 or x equals to minus 1. So there are four roots, obviously. Now you can see the initial equation was a power 4 equations, bichord radius power 4. So it was expected that it could be having four solutions and hence we are getting four solutions, what all? x equals to 3, s equals to minus 3, x equals to 1 and x equals to minus 1, right? Similarly, if I take another equation, let us say I am taking this equation 25 a to the power 4 upon x square plus x square is equal to 26 a square. Now clearly if you see it is not a quadratic equation, it is not a quadratic, quadratic equation. Why? Because x square is there in the denominator, right? So hence this is not a quadratic equation, right? But can it be reduced to? Yes, of course. But what to do? So let us say, let me say I am taking x square as y again. So what will this equation be reduced to? So hence you will see it is nothing but 25 a to the power 4, okay? 25 a to the power 4 divided by y, okay? Plus y is equal to 26 a squared. Now you take common denominator, so what will this become? 25 a to the power 4 plus y square upon y, isn't it? On the left hand side. And this is 26 a square. Now cross multiply and reduce it to 25 a to the power 4 plus y square is equal to 26 a squared y. And if you now see, by now you will realize that it is reduced to a quadratic equation like this, correct? Plus and I can write this as 5 a square square, isn't it? This is equal to 0, right? Or you can leave it to, leave it as 25 a squared a to the power 4 itself because this is going to help you, right? So hence write 25 a to the power 4. Now again, it is quadratic equation in y. So simply use any of the methods you know. I will use the factorization method. I will write y square minus a square y minus 25 a square y plus 25 a to the power 4 equal to 0. This implies y you take common, so you will get y minus a square. Then minus 25 a square common and you will get y minus a square is equal to 0, right? So hence you will get y minus a square times y minus 25 a square is equal to 0. And that means y is equal to a square and y is equal to r or y is equal to 25 a square can be written as 5 a whole square. That means x square is equal to a square or x square is equal to 5 a whole square, isn't it? That means x is equal to plus a or minus a like the previous case, you will get two values here and here x will be plus 5 a or minus 5 a. These are the four solutions to the given equation, right? So what is the learning? If you see, I can reduce any, you know, such type of equations into a quadratic form and solve according to the rules of quadratic equation or the methods adopted for solving quadratic equation and here also there are four roots, four roots, okay? So that is how these kind of equations could be solved.