(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 = 31×2 + 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; Samasti: Totality)

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