 Hello and welcome to the session. In this session we will discuss the following question and the question says solve 1 upon 5 minus x minus 1 upon 5 plus x is equal to 0.21. Let's start the solution now. We have given this quadratic equation and we have to solve it. So the given quadratic equation is 1 upon 5 minus x minus 1 upon 5 plus x is equal to 0.21. First we will take the LCM on the left hand side. So by taking LCM on left hand side we get the numerator becomes 5 plus x the whole minus 5 minus x the whole. This whole divided by denominator is 5 minus x the whole into 5 plus x the whole. This is equal to 4 upon 21. This implies we will now open the brackets so the numerator becomes 5 plus x minus 5 plus x whole divided by the denominator becomes 25 minus x square is equal to where upon 21 5 and minus 5 get cancelled in the numerator. So this implies upon 25 minus x square is equal to 4 upon 21. We will now do cross multiplication. So on cross multiplication we get 21 into 2x into 25 minus x square the whole. This implies left hand side becomes 32x is equal to 4 into 25 which is 100 minus 4x square. We will now transpose 100 and minus 4x square on the left hand side. So this implies 4x square plus 32x minus 100 is equal to 0. Now we will divide both the sides by 2 so dividing by 2 we get plus 21x minus 50 is equal to 0. So in this way we have obtained this quadratic equation. Now we will split the middle term so this implies 2x square minus 4x plus 25x minus 50 is equal to 0. We have split the middle term that is 21x into minus 4x and 25x. This implies taking 2x common between the first 2 terms we get 2x into x minus 2 the whole. Now taking 25 common between the last 2 terms we get plus 25 into x minus 2 the whole is equal to 0. This implies the whole into x minus 2 the whole is equal to 0. So we have factorized the given quadratic equation into 2 factors. This implies 2x plus 25 is equal to 0 or x minus 2 is equal to 0. This implies from the first equation we get x is equal to minus 25 by 2 or from the second equation we get x is equal to 2. So these are the two possible solutions to the given quadratic equation. We will now check if these two values of x satisfy the quadratic equation. For this first we will substitute x is equal to 2 in the given quadratic equation. So the left hand side becomes 1 upon 5 minus 2 minus 1 upon 5 plus 2. This is equal to 1 upon 3 minus 1 upon 7. Taking the LCM this is equal to the numerator becomes 7 minus 3 divided by the denominator is 21. This is equal to 4 upon 21 which is the right hand side of the quadratic equation. So x is equal to 2 satisfies the given quadratic equation. We will now check for x is equal to minus 25 by 2. So we will substitute x is equal to minus 25 by 2 in the quadratic equation. So the left hand side becomes 1 upon 5 plus 25 by 2 minus 1 upon 5 minus 25 by 2. Taking the LCM this is equal to 2 upon 10 plus 25 minus 2 upon 10 minus 25. This is equal to 2 upon 35 plus 2 upon 15. Taking the LCM this is equal to 6 plus 13 whole divided by 105 which is equal to 20 upon 105 which is equal to 4 upon 21 which is the right hand side of the quadratic equation. So x is equal to minus 25 by 2 also satisfies the given quadratic equation. So we can say that x is equal to minus 25 by 2 and x is equal to 2 is the final answer of the given quadratic equation. With this we end our session. Hope you enjoyed the session.