VEDIC MATHS-104

 

VEDIC MATHS

                           By OMKAR TENDOLKAR

Hello friends,

                      This is  post number 104 from the series of "Vedic maths" blogs. Here in this blog we will learn about "Applications of the Sutras & Sub-sutras - 4"

VEDIC MATHEMATIC :

"VEDIC MATHEMATICS" is the name given to the ancient system of mathematics, or, to be precise, a unique technique of calculations based on simple rules and principles, with which any mathematical problem be it arithmetic, algebra, geometry or trigonometry can be solved, hold your breath, orally!

What we call "Vedic Mathematics" is a mathematical elaboration of 'Sixteen Simple Mathematical formulae from the Vedas' as brought out by Sri Bharati Krishna Tirthaji.

Sutras: 

The system is based on 16 Vedic sutras or aphorisms, which are actually word formulae describing natural ways of solving a whole range of problems.

Some examples of sutras are "By one more than the one before", "All from 9 & the last from 10", and "Vertically & Crosswise". These 16 one-line formulae originally written in Sanskrit, which can be easily memorized, enables one to solve long mathematical problems quickly.

7. Sankalana – Vyavakalanabhyam ( By Addition and By Subtraction )

In two general equation such as, ax + by = p and cx + dy = q, where x and y are unknown values,
x = (bq – pd) / (bc – ad)
y = (cp – aq) / (bc – ad)

Example I:
3x + 2y = 4 and 4x + 3y = 5
x = (10-12) / (8-9) =  2
y = (16-15) / (8-9) = -1

Example II:

x + y = 4 and x - y = 2
Here we use By By Addition and By Subtraction,
Adding the equation gives 2x = 8, so x = 4,
and subtracting the equation gives 2y = 4, so y = 2.

8. Puranapuranabhyam ( By the completion or non-completion ) 

This is a method of completion of polynomials to find its factors.

Example:

x³ + 9x² + 24x + 16 = 0  i.e.  x³ + 9x² = -24x -16

We know that (x+3)³ = x³+9x²+27x+27 = 3x + 11 (Substituting above step).

i.e. (x+3)³ = 3(x+3) + 2 … (write 3x+11 in terms of LHS so that we substitute a term by a single variable).

Put y = x+3

So, y³ = 3y + 2

i.e. y³ – 3y – 2 = 0

Solving using the methods discussed (coeff of odd power = coeff of even power) before.

We get (y+1)² (y-2) = 0

So, y = -1 , 2

Hence, x = -4,-1

In next blog we will discuss about "Applications of the Sutras & Sub-sutras - 5".     

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