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सूत्र ८. पूरणापूरणाभ्यां

(Sūtra 8. pūraṇāpūraṇābhyāṃ) - By completion, or non-completion.

The Sūtra: pūraṇāpūraṇābhyāṃ (By completion, or non-completion) is a technique used to factorize and solve quadratic, cubic and bi-quadratic equations.
 
As an illustration, let us use this Sūtra, in simple steps, to solve:
x2 + 3x + 2 = 0
  Steps Solution
1. Assume an expression of the form (ax+b)n, considering the power of x, that comes nearly as the expression. In this case, let us assume (x+2)2
2. Compare them, to check the deficient terms. In this case, we get:
(x+2)2 = x2 + 4x + 4
And, the expression to solve:
x2 + 3x + 2
Clearly, the deficient terms are:
x + 2
3. Add the deficient terms to both side of the equation, to complete the assumed expression. In this case, adding (x + 2):
(x2 + 3x + 2) + (x + 2) = 0 + (x + 2)
or, x2 + 4x + 4 = (x + 2)
or, (x + 2)2 = (x + 2)
4. Replace common terms with another variable, and solve. In this case, let y = (x + 2).
Then we get:
y2 = y
This solves, y = 0 and y = 1
5. Use solution to solve original variable. In this case, y = (x + 2).
y = 0, gives:
x + 2 = 0; or, x = -2
And, y = 1, gives:
x + 2 = 1; or, x = -1
 
Thus, we get:
x = -2, x = -1, which is the answer!
So, for a practitioner of Vedic Mathematics, for something like:
x3 + 6x2 + 11x + 6 = 0
  Let us assume (x+2)3:
(x + 2)3 = x3 + 6x2 + 12x + 8
 
Clearly, the deficient terms are: (x + 2)
 
Adding to both sides:
(x3 + 6x2 + 11x + 6) + (x + 2) = 0 + (x + 2)
or, (x + 2)3 = (x + 2)
 
Let, y = (x + 2). Then,
y3 = y solves for 0, 1 and -1
 
x + 2 = 0 gives: x = -2
Also, x + 2 = 1 gives: x = -1
And, x + 2 = -1 gives: x = -3
 
Thus, we get:
x = -2, x = -1, x = -3
 
Again, for something like:
x3 + 8x2 + 17x + 10 = 0
  Let us assume (x+3)3:
(x + 3)3 = x3 + 9x2 + 27x + 27
 
Clearly, the deficient terms are: (x2 + 10x + 17)
 
Adding to both sides:
(x3 + 8x2 + 17x + 10) + (x2 + 10x + 17) = 0 + (x2 + 10x + 17)
or, (x + 3)3 = (x2 + 10x + 17)
 
Let, y = (x + 3). Then,
y3 = y2 + 4y – 4
This solves for 1, 2 and -2
 
x + 3 = 1 gives: x = -2
Also, x + 3 = 2 gives: x = -1
And, x + 3 = -2 gives: x = -5
 
Thus, we get:
x = -2, x = -1, x = -5
Due to its ease of solution, this Sūtra is also discussed for factorization.
It follows easy steps: To equate with Zero, and solve using this Sūtra to get factors.
 
Following the example above, to factorize: x2 + 3x + 2
Solve: x2 + 3x + 2 = 0, which gives us x = -2 and x = -1
So, the factors are: (x + 2) and (x + 1)
 
Lastly, please remember that, as in any other form of mathematics, the mastery of Vedic Mathematics require practice and the judgement of applying the optimal method for a given scenario - a guideline of which, is presented in Applications »
 
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