 Hello and welcome to the session. In this session, we will discuss quadratic equations. Now the general form of quadratic equation in X is AX square plus BX plus C is equal to 0, where A is not equal to 0 and A, B, C are the constants. That is if A is equal to 0, then this equation does not remain a quadratic equation. So this equation is called the general form of the quadratic equation in X. Now here X is a variable and in this equation, that is in the quadratic equation, the highest power of the variable is 2. And now let us discuss the solution of the quadratic equation. Now the general form of the quadratic equation is AX square plus BX plus C is equal to 0, where A is not equal to 0 and we can write it as AX square plus BX is equal to minus C, which further implies now dividing throughout by A, it will be X square plus B by A into X is equal to minus C over A. Now we will complete the square on the left hand side. For this, we will add and subtract the square of half the coefficient of X on the left hand side, which implies X square plus B by A into X plus B square by 4A square minus B square by 4A square is equal to minus C over A. Now this implies X square plus, now this term can be written as 2 into X into B by 2A plus B by 2A whole square, that is this term. And now moving on this term on the right hand side, this is equal to minus C over A plus B square over 4A square. Now on this side, the square is completed, that is this is equal to X plus B by 2A whole square, which is equal to, now on this side taking the LC, that is 4A square, it will become B square minus 4AC. On the right hand side, we are getting B square minus 4AC whole upon 4A square. Now taking the square root, this implies B over 2A is equal to plus minus square root of B square minus 4AC whole upon, now this further implies X is equal to minus B over 2A plus minus square root of B square minus 4AC whole upon 2A, which further implies equal to minus B plus minus square root of B square minus 4AC whole upon 2A. We were supposed to find out the solution of this equation, that means we have to find out the value of at this we have to find out the roots of an equation, that is this equation, now the roots are the values of the variable satisfying a given equation. Now we have got the values, that is we have got the roots of the given quadratic equation. So the roots of the given quadratic equation B square minus 4AC whole upon 2A minus B minus square root of B square minus 4AC, the given quadratic equation AX square plus BX plus C is equal to 0. And now let us discuss reducible to quadratic equations. The type 1, the equation of the type A into X raised to power 2N plus B into X raised to power N plus C is equal to 0, quadratic equation. Now to reduce the given equation to a quadratic equation N is equal to Z in the given equation and then we get AZ square plus BZ plus C is equal to 0. So the given equation is reduced to a quadratic equation in Z. Now let us discuss this type with the help of an example. Here we have to solve X raised to power 2 by 3 minus 4 is equal to 0. Now for this illusion it is equal to Z, that is this equation can be written as X raised to power 1 by 3 whole raised to power 2 minus 4 is equal to 0. And now 1 by 3 is equal to Z here so it will be Z square minus 4 is equal to 0. Now it is a pure quadratic equation in Z equal to 4 which implies that is equal to plus minus 2. Now X raised to power 1 by 3 is equal to Z implies power 1 by 3 is equal to, now taking that is equal to plus 2. Now Q being both side is equal to minus 2 we get X is equal to minus 8. AZ plus V by ENC are the constants reduced that is this type of equation is reduced to a quadratic equation. Let us discuss this type with the help of an example. Now here 2 plus 1, 2 into X plus 3 the whole minus 3 is equal to 0. Solution is 3 whole upon 2 X plus 1 is equal to Z. By taking the reciprocal on both sides 1 whole upon X plus 1 by Z. So putting these values that is putting these values to 0 which implies 2 minus 3 is equal to 0 which implies plus 2 is equal to 0. Now the given equation is reduced in the form of quadratic equation in Z. And now we will solve this quadratic equation by splitting the middle term. So this implies Z square minus 2 Z minus Z is equal to 0 which further implies Z into Z minus 2 the whole is equal to 0. Which further implies Z minus 1 the whole into Z minus 2 the whole is equal to 0 which gives Z is equal to 1 Z is equal to the value of for Z is equal to 1 the whole upon 2 X plus 1 is equal to 1. Which implies X plus 3 is equal to 2 X plus 1 which further gives whole upon 2 X plus 1 is equal to 2 which implies X plus 1 by 3. Solution of the given equation that is the equation of the form X plus A the whole into X plus B the whole C the whole into X plus D the whole plus K is equal to 0. That is when the quantities is equal to the other two. Now by quantities we solve these type of equations that is the equation in this form in which the sum of any of the two quantities is equal to the sum of the other two quantities. Now let us discuss this type with the help of an example to solve X plus 5 the whole 1 the whole into X minus 1 the whole into equal to 0. The sum of the two quantities is equal to the sum of the other two quantities. That is 5 plus 1 is equal to 6 and minus 1 plus 3 is equal to 2 that means the sum of these two quantities is not equal to the sum of these two quantities. Now consider and minus 1 equal to 5 minus 1 which is equal to 4 and 1 plus 3 is also equal to minus 1 is equal to 1 plus 3 that means the sum quantities is equal to the sum of the other two quantities. Now for the solution we will multiply X plus 5 the whole 1 the whole here we will multiply X plus the whole and X plus 3 the whole. Why to test X plus 5 the whole in this 1 the whole into X plus 1 the whole their multiplication will give X minus 5 the whole into X plus 3 the whole equal to 0. Now here and this implies Y minus 5 the whole into Y plus 3 the whole plus 16 is equal to 0 which further implies on solving this Y squared minus 2 Y plus 1 is equal to 0. So we have reduced the given equation in the form of quadratic equation this is Y minus 1 whole square so this implies Y minus 1 whole square is equal to 0 which further implies Y is equal to 1 and 1. X is equal to 1 minus 1 is equal to 0 by using the formula for finding out the roots minus b that is minus 4 plus minus square root of b square that is 16 minus 4 is equal to plus 4 that is 2. So this is equal to minus 4 plus minus root 20 and we will get X is equal to minus 2 plus minus root 5 at the solution for the given equation. And now let us discuss the nature or the vector often have already discussed that for the quadratic equation AX square plus BX plus C is equal to 0 where A is not equal to 0 these are the roots of this quadratic equation. Now here denoted by the value of D decides the nature or the vector of the roots. Now these are the various possibilities that is D is greater than 0 and if D is greater than 0 then this that is if D is greater than 0 and D is a perfect square that is roots are rational and if D is not a perfect square D is equal to 0. And if D is equal to 0 then the roots of the given equation where in that case the roots are imaginary. Now let us discuss the sum and product b by root solution that is the quadratic equation AX square plus BX plus C is equal to 0 where A is not equal to 0 and ABC are constants is equal to that is by the formula. Minus B plus square root of B square minus 4 AC will be equal to A is equal to minus B minus square root of B square minus 4 AC will be equal to A. Then the sum of the roots is equal to B plus Q which is equal to B plus square root of B square minus 4 AC minus B minus square root of B square minus 4 AC. These terms are cancelled with each other and this is equal to minus 2B over 2A which is equal to minus B over A. And now product that is B into Q is equal to minus B plus square root of B square minus 4 the whole minus square root of B square minus 4 AC whole upon 2A the whole. Which is further equal to in the numerator it can be written as minus B whole square which is B square minus 4 AC whole square which is equal to minus 4 AC the whole. Whole upon 4 AC over 4A square which is equal to the same equation it is clear that C is the absolute term and B is the coefficient of X and A is the coefficient of X square. The sum of the roots is equal to minus B equal to minus the coefficient the given quadratic equation and we are letting the product of the roots is equal to C over A that means coefficient of X square in the given quadratic equation. So in this session we have got about quadratic equation the equation is equal to quadratic equation then nature or corrector of the roots session.