Sridharacharya

Sridharacharya (born: 750 AD) was a great mathematician of ancient India. He interpreted zero and rendered the corresponding formula for solving quadratic equation.

Our information about them is very small. Certainly nothing can be said about their time and place. But it is estimated that his lifetime was between 870 to 930E; They were born in present-day Hooghly district; His father's name was Baldevacharya and mother's name was Achchoka.

Works and contributions

He wrote about two famous books of 750 AD, Trishtika (also called 'Patiaginitasar'), Patigital and Ganitasarar, He made many important inventions of algebra. The rules invented by them to solve classical equations by making them a complete square are still popular in the name of 'Shridhar Rule' or 'Hindu Rule'.

'Mathematics, abstract abstract and trinity are their available compositions, which are basically related to arithmetic and field behavior. At the end of the algebra, Bhaskaracharya has said that the algebra of Brahmagupta, Shridhar and Padmanabhas have been elaborated and elaborate -: 'Brahma-शशश्री धरशप बीििििििि यतितिविविविविविविविृति' '' '' '' '' ''. '

It seems that Sridhar had composed a large text on algebra which is no longer available. Bhaskar has only quoted Sridhar's law for solving class equations in his algebra - At the time of quaterhood, the rupees are attributed. Unrealistic If a number is added to zero then the sum is equal to that number; If a number is reduced to zero then the result is equal to that number; If zero is multiplied by any number then the product will be zero. They have not said anything about what will happen if you divide into a number by zero. Mathematical compilation (Round diameter cube semi-self-eating part numerator mathematica) अर्थात V = d / 2 + (d / 2) / 18 = 19 d / 36 By comparing it with the volume of π d / 6 of the shell, it shows that they have taken 19/6 instead of the pie. Sadhricharya method to solve class equation

ax + bx + c = 0

4ax + 4abx + 4ac = 0; (Multiplied by 4a)

4ax + 4abx + 4ac + b = 0 + b; (B on both sides when added)

(4ax + 4abx + b) + 4ac = b

(2ax + b)(2ax + b) + 4ac = b

(2ax + b) = b - 4ac

(2ax + b) = (√D) ; (D = b-4ac)

So the following are the two root (root) of x -

पहला मूल α = (-b - √(b-4ac)) / 2a

दूसरा मूल β = (-b + √(b-4ac)) / 2a

Also see them

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