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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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