Question

What is the sridharacharya method?

Answer 1

Hint: In the above question that has been given we need to explain the sridharacharya method which is none other than the equation that we used to solve the quadratic equation i.e. the equation we use to find the roots of the quadratic equation.

Complete step by step solution:
 In the above question we are being asked about the sridharacharya method, Sridharacharya method is also widely known as a formula that we use to find the roots of a quadratic equation: let us take an example so that we can see the application of the quadratic formula or the sridharacharya method.
Let us take the our quadratic equation to be \[a{{x}^{2}}+bx+c=0\], now multiply both the sides by 4 and the coefficient of \[{{x}^{2}}\] that is 4a and we get our equation as
\[\Rightarrow 4{{a}^{2}}{{x}^{2}}+4abx+4ac=0\]
Now we will add the square of coefficient of x that is \[{{b}^{2}}\] and when we add this to both sides of the equation we get it as
\[\Rightarrow 4{{a}^{2}}{{x}^{2}}+4abx+{{b}^{2}}+4ac={{b}^{2}}\]
Now when we see on the left hand side of the equation we can see that the whole equation has become a formula for \[{{\left( a+b \right)}^{2}}\] so when we convert the left hand side into \[{{\left( a+b \right)}^{2}}\] form we will get it as
\[\Rightarrow {{\left( 2ax+b \right)}^{2}}={{b}^{2}}-4ac\]
Now we will take the square from the left hand side to the right hand side of the equation and we get it as
\[\Rightarrow \left( 2ax+b \right)=\pm \sqrt{{{b}^{2}}-4ac}\]
Now to find the final equation of for x we subtract b (coefficient of x) on both the sides and then divide both the sides by 2a which will leave only one variable on the left side i.e. x and on the right side we will get an equation which is also termed as quadratic roots equation so finally we get it as
\[\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
So this is how sridharacharya method came into act which later was also known as roots for the quadratic equation the roots for the quadratic equation being \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Note: In the above stated question we needed to explain the sridharacharya method in this method while square root we generally forget to use the \[\pm \] symbol which signifies of two values one positive and one negative and due to which we only get one root for the equation. After getting the roots always use it in the question to check whether the roots are correct or not.