Question

What is the formula of Shri Dharacharya​

Answer 1

Answer:

The quadratic formula, is of the form. x = \frac { - b \pm \sqrt{ b^2 - 4ac } } { 2a} . x=2a−b±b2−4ac . It is also known as Shreedhara Acharya's formula, named after the ancient Indian mathematician who derived it.

Answer 2

Proof

Consider the equation ax² + bx + c = 0 , Where a , b , c are real number and a≠0

Dividing the given equation by a ( a≠0) , we get

$\to \sf x^2 + \dfrac{b}{a}x+\dfrac{c}{a}=0$

By adding and subtracting (b/2a)² , we get

$\sf\to x^2+\dfrac{b}{a} x+\bigg(\dfrac{b}{2a}\bigg)^2-\bigg(\dfrac{b}{2a}\bigg)^2+\dfrac{c}{a} = 0$

$\sf\to\bigg[x^2+2\dfrac{b}{2a} x+\bigg(\dfrac{b}{2a} \bigg)^2\bigg]=\bigg(\dfrac{b}{2a} \bigg)^2-\dfrac{c}{a}$

$\sf\to\bigg(x+\dfrac{b}{2a} \bigg)^2 = \dfrac{b^2}{4a^2} -\dfrac{c}{a}$

$\sf\to\bigg(x+\dfrac{b}{2a} \bigg)^2 = \dfrac{b^2-4ac}{4a^2}$

$\sf\to x+\dfrac{b}{2a} = \pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$

$\sf\to x =\dfrac{-b}{2a} \pm{\dfrac{\sqrt{b^2-4ac}}{2a}$

$\sf\to x ={\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

This Formula for finding the roots of the quadratic equation is often referred to as the quadratic formula . as this formula was given by an ancient Indian mathematician Sridharacharya's formula