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      • (cont. )Equations – 2
      • (cont.) Equattions 3
      • (cont) Equations – 4
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(cont) Equations – 4

Purana Apurnabhyam Sanskrit Name: पूरणापूरणाभ्यां English Translation: By the completion and non completion (of square, cube, fourth power, etc). Prerequisites: Paravartya Sutra. It uses the technique of completion of polynomials with Paravartya sutra to find its factors. Examples: 1) 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 Hencex = -4,-1 2) x3 + 7x2 + 14x + 8 = 0 i.e. x3 + 7x2 = - 14x – 8 We know that (x+3)3 = x3+9x2+27x+27 = 2x2 + 13x + 19 (Substituting above step). i.e. (x+3)3= 2x2 + 13x + 19 Now we need factorize RHS in terms of (x+3). So apply Paravartya sutra. So Dividing 2x2 + 13x + 19 by (x+3) gives 2x2 + 13x + 19 = (x+3)(2x-7)-2 i.e. (x+3)3 = (x+3)(2x-7)-2 put y = x+3 So, y3 = y(2y+1) -2 Solving gives y = 1,-1,2 Hencex= -2, 4, 1, -1 3) x4 + 4x3 – 25x2 – 16x + 84 = 0i.e. x4 + 4x3 = 25x2 + 16x – 84 We know that (x+1)4 = X4 + 4x3 + 6x2 + 4x +1 = 31x2 + 20x – 83 (Substituting above step). i.e. (x+1)4 =31x2 + 20x – 83 Now we need factorize RHS in terms of (x+1). So apply Paravartya sutra. So Dividing 31x2 + 20x – 83 by (x+1) gives 31x2 + 20x – 83 = (x+1)(31x-11) – 72 i.e. (x+1)4 = (x+1)(31x-11) - 72 put y= x+1 So, y4 = y(31y-42) – 72 i.e. y4 – 31y2 + 42y + 72 = 0 On Factorization (Sum of coeff of even power = coeff of odd power. So (y+1) is 1 factor). Dividing above biquadratic eq by (y+1) will give cubic eq, Further factorization gives y = -1,3,4 or -6 Hence x = -2, 2 ,3 or -7 Vyasti Samasti Sutra Sanskrit Name: व्यष्टि समष्टि English Translation: Individuality or Totality(Vyasti: Individual; … [Read more...]

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