 Hello friends, so we have taken another question to solve here and we are again solving a quadratic equation by Sridhra Charya's method which is also called quadratic formula method. So we have to solve this given equation where abx square plus b square minus ac times x minus bc equals to 0. So if you notice here there are no numerals in the equation all are in terms of constants represented by ab and c and the variable is x correct. So how do we solve this equation? So before that let's revise what is Sridhra Charya's rule. So there we understood that the solution to a given quadratic equation will be of the form of minus b plus under root d upon 2a and x is equal to minus b minus under root d upon 2a. Now here don't get confused with the a and b which is given in the equation here and the a and b we are writing here. So in the formula a is coefficient of coefficient of x square and b is coefficient of coefficient of x and c is coefficient of or it's a constant term correct so don't get confused. So a b and c here are in the formula this formula when we are talking about this formula then we are talking about abc which are coefficient of x square of bx and constant term respectively okay. But in the given question ab in this equation ab these are different from what we are talking about here okay so hence let us now try to solve it. So first of all d what is d? d is nothing but b square minus 4ac correct. So just to avoid confusion we can write small a as a so hence you can say in this case let us say this is capital A this is equal to capital B and this is equal to capital C where the equation is ax square plus bx plus c equals to 0. So hence what is a? a is given as in this question a is nothing but if you see a is ab what's b? b is b square minus ac and what's c? capital C is minus bc right these are the few things a is ab b is b square minus ac and c is minus bc okay. So let's first find it out find the discriminant discriminant is b square minus 4ac okay so what is b square in this case let me just write I should not write the small letters anymore why because we are using capital letter for the equation right. So hence d is capital B square minus 4 capital ac okay in this case just to avoid confusion and for no other reason we are now treating it as capital A b and c okay because there are small a b and c in the given equation okay. So it is capital B square b square will be nothing but small b square minus ac whole square and whole square minus 4 times a a is ab and c is minus bc okay. So let us calculate this this is nothing but open the brackets up so you will get b to the power 4 minus 2 times b square times ac plus a square c square is it and here if you see this is plus 4 times b square ac 4 times b square ac and let's simplify further so it is b to the power 4 minus 2 b square ac plus 4 b square ac plus a square c square let's simplify further you will get b to the power 4 plus 2 b square ac plus a square c square is it which is nothing but if you look closely it is nothing but b squared squared plus 2 times b squared times ac plus ac whole square whole square this is nothing but ac whole square let me rewrite it properly so this is ac whole squared. So if you notice this is again giving us a feel of you know square of a binomial so hence this is nothing but if you see this is b square plus ac whole square so d is this much this was our d now what is the solution then x is equal to minus b plus root of d upon 2 a and x is equal to minus b minus root of d upon 2 a both would be the solution again so instead of a b and c I should be writing a sorry for this confusion so I will be writing plus b minus b plus root d upon 2 a and similarly here it will be minus b and here it will be 2 capital a right 2 capital a okay now what let's you know deploy the values and get the solution so x is minus b minus b is my friends minus b what is b here so minus capital b will be simply minus of b square minus ac so hence it is minus b square minus ac plus root of d d is b square plus ac whole squared upon twice of a and a was my dear friends what was a let's see what was a a was a b so a b yep if you see a is a b yep so now next solve this and similarly here x will be equal to nothing but minus b square minus ac then minus root of d so b square plus 4 sorry not 4 ac whole squared under root divided by divided by 2 a b fair enough now what is it this will be nothing but minus b square minus ac and this will be simply there is a square and there's a square root this square will be cancelled by the square root right so hence it is b square plus ac divided by 2 a b and similarly here just the same change of sign nothing else it is minus sorry this will be in the brackets yep minus b square minus ac minus b square plus ac upon 2 a b right so I'm finding both the solutions simultaneously so now open the brackets you'll get minus b square plus ac plus b square plus ac upon 2 a b first solution is this so this is x equals to x equals to and here x equals to minus b square plus ac minus b square minus ac upon 2 a b isn't it so hence what will you get you will get here minus b square plus b square goes so 2 ac sorry this is b here so 2 ac twice ac ac plus ac 2 ac divided by 2 a b and this is one x equals to this much and the other solution is x is equal to this ac and this ac will go so minus 2 b square y 2 a b we can simplify further we get x is equal to this 2 this 2 will go if a is not equal to 0 then this a will also go so hence x is equal to c by b and here x is equal to this 2 and this 2 will go if b is not equal to 0 then this this b and this b will go so minus b upon a okay so these are the two solutions so x is equal to c and c by b and x is equal to mine x equals to minus b by a is the solution so x equals to c by b and x equals to minus b by a is the solution