Question

Prove Sridharacharya formula with example.

Write application of it

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Answer 1

AnswEr :

Consider a quadratic equation of the form :

$\sf \: {ax}^{2} + bx + c = 0$

Transposing the constant term to RHS,

$\longrightarrow \: \sf \: a {x}^{2} + bx = - c$

Dividing by "a" on both sides of the equation,

$\longrightarrow \: \sf \: x {}^{2} + \dfrac{b}{a} x = - \dfrac{c}{a}$

Adding (b/2a)² on both sides of the equation,

$\longrightarrow \: \sf \: {x}^{2} + 2 \times \dfrac{b}{2a}x + \dfrac{b {}^{2} }{4 {a}^{2} } = \dfrac{b {}^{2} }{ {4a}^{2} } - \dfrac{c}{a}$

The LHS of the equation is of the form : (a + b)² = a² + 2ab + b²

Thus,

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

Applying square root on both sides of the equation,

$\longrightarrow \sf x + \dfrac{b}{2a} = \pm \: \dfrac{ \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: x = \dfrac{ - b \pm \: \sqrt{ {}D} }{2a} }}$

Here,

D = B² - 4AC

D is known as the Discriminant

D determines the nature of the roots .

• D > O [Real and Unique]

• D = 0 [Real and Equal]

• D < O [Imaginary]

The above expression is known as Quadratic Formula or ShreeDharaCharya's Rule in honour of the mathematician who calculated it

The above expression is used to find the roots of a Quadratic Equation

Answer 2

He was known for two treatises: Trisatika(sometimes called the Patiganitasara) and the Patiganita. His major work Patiganitasara was named Trisatika because it was written in three hundred slokas. The book discusses counting of numbers, measures, natural number, multiplication, division, zero, squares, cubes, fraction, rule of three, interest-calculation, joint business or partnership and mensuration (the part of geometry concerned with ascertaining lengths, areas, and volumes).

He was one of the first to give a formula for solving quadratic equations. He discovered the formula :-

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(Multiply by 4a)