 Hello and welcome to the session. In this session, we discuss the following question which says solve square root of 2 minus x plus square root of 2 plus x, this whole upon square root of 2 minus x minus square root of 2 plus x is equal to 3. Before moving on to the solution, let's discuss the result of component O and dividend O. According to this, we have if four quantities A, B and C are in proportion that is A is to B is equal to C is to D or you can say A upon B is equal to C upon D then on applying component O and dividend O, we have A plus B upon A minus B is equal to C plus D upon C minus D. This is the key idea that we use in this question. Let's proceed with the solution now. We have square root of 2 minus x plus square root of 2 plus x, this whole upon square root of 2 minus x minus square root of 2 plus x is equal to 3. We need to solve this. It means that we are supposed to find the value of x. Now applying component O and dividend O, square root of 2 minus x plus square root of 2 plus x plus the denominator that is square root of 2 minus x minus square root of 2 plus x, this whole upon square root of 2 minus x plus square root of 2 plus x minus the denominator. So we get square root of 2 minus x plus square root of 2 plus x is equal to, now since in the right hand side we have just 3, it means it is 3 upon 1. So here we have on applying component O and dividend O to the right hand side we have 3 plus 1 upon 3 minus 1. Now here in the left hand side in the numerator square root 2 plus x and minus square root 2 plus x cancels then in the denominator square root of 2 minus x and minus square root of 2 minus x cancels. So we are left with 2 times square root of 2 minus x upon 2 times square root of 2 plus x is equal to 4 upon 2. Then this 2 cancels with 2 and 2 2 times is 4. This means square root of 2 minus x upon square root of 2 plus x is equal to 2. Now squaring both sides we have 2 minus x upon 2 plus x is equal to 4. Now again applying component O, dividend O we get 2 minus x plus 2 plus x this whole upon 2 minus x minus the denominator. So we have 2 minus x is equal to, now this 4 means 4 upon 1. So applying component O and dividend O to the right hand side we have 4 plus 1 upon 4 minus 1. Now here this x and minus x cancels in the numerator and 2 and minus 2 cancels in the denominator. So we have 2 plus 2 that is 4 upon minus x minus x that is minus 2x is equal to 5 upon 3. Now here 2, 2 times is 4. So this means we have x is equal to minus 6 upon 5. So therefore the final value of x is minus 6 upon 5. This completes the session. Hope you have understood the solution of this question.