Square root in Sadratnamālā

Here we will be knowing a square root of a number as mentioned in Sadratnamālā.

Verse

शुद्धवर्गस्य मूलेन द्विघ्नेनावर्गतो हृतम् ।

तदादिमूलं तद्वर्गः शोध्यो वर्गात् पुनस्तथा ॥ १४ ॥

(Having deducted the maximum possible square from the last square place) divide the non-square place by twice the square root (of the maximum square earlier deducted). Deduct the square (of the quotient) from the next square place. Repeat this (to get the square root).

The number whose square root is to be determined, is denoted by placing the digits in a line. The odd places counted from right to left are the square places and the even places are the non-square places, as mentioned earlier. The maximum possible square (of numbers one to nine) is subtracted from the digit or digits in the last square place and keep the square root in a separate place. This is prathamaphala (the first result). Place the digit of the next non-square place on the right of the remainder and divide by twice the first result. This is dvitiyaphala (the second result). Place the digit of the next square place on the right of the remainder and deduct square of the quotient from it. Place the digit of the next non-square place on the right of the new remainder and divide by twice the second result. This is tṛtiyaphala (the third result). Place the digit of the next non-square place on the right of the remainder and divide by twice the third result. This is repeated until all the digits are exhausted.

Example: Square root of 877969

Denote the the odd places from right to left are square (S) places and the even places are the non-square (N) places.

Note: Prathamaphala = First result; Dvitīyaphala = Second result; Tṛtīyaphala = Third result

	N	S	N	S	N	S	Step details	Result
	8	7	7	9	6	9
- 92 (9 - Prathamaphala)	8	1					Subtract the maximum possible square (81 = 92) from the last square place (87) . Here 9 is Prathamaphala.	9
÷ 2 X 9 = 18	18)	6	7	(3 (3 - Dvitīyaphala)			Place the digit of the next non-square place (7) on the right of the remainder (6). Now the number is 67 and divide this by twice the first result (9) = 2 X 9 = 18
		5	4				Subtract the above number from the maximum possible number 18 X 3 = 54 . Here the quotient is 3. 3 is Dvitīyaphala	9 3
		1	3	9			Place the digit of the next square place (9) on the right of the remainder (13), Now the number is 139
- 32				9			Deduct square of the quotient (3) = 9 from it.
÷ 2 X 93 = 186		1	3	0	6	(7 (7- Tṛtīyaphala)	Place the digit of the next non-square place (6) on the right of the new remainder (130). Now the number is 1306 and divide by twice the second result (93) = 186.
		1	3	0	2		Subtract the above number from the maximum possible number 186 X 7 = 1302. Here the quotient is 7. 7 is tṛtiyaphala.	9 3 7
					4	9	Place the digit of the next square place (9) on the right of the remainder (4), Now the number is 49
- 72					4	9	Deduct square of the quotient (7) = 49 from it.
						0

Since the remainder is zero the given number is a perfect square.

Square root of 877969 = 937

Example: Square root of 11943936

Denote the the odd places from right to left are square (S) places and the even places are the non-square (N) places.

Note: Prathamaphala = First result; Dvitīyaphala = Second result; Tṛtīyaphala = Third result

	N	S	N	S	N	S	N	S	Step details	Result
	1	1	9	4	3	9	3	6
- 32 (3 - Prathamaphala)		9							Subtract the maximum possible square (9 = 32) from the last square place (11) . Here 3 is Prathamaphala.	3
÷ 2 X 3 = 6	6)	2	9	(4 (4 - Dvitīyaphala)					Place the digit of the next non-square place (9) on the right of the remainder (2). Now the number is 29 and divide this by twice the first result (3) = 2 X 3 = 6
		2	4						Subtract the above number from the maximum possible number 6 X 4 = 24 . Here the quotient is 4. 4 is Dvitīyaphala	3 4
			5	4					Place the digit of the next square place (4) on the right of the remainder (5), Now the number is 54
- 42			1	6					Deduct square of the quotient (4) = 16 from it.
÷ 2 X 34 = 68			3	8	3	(5 (5 - Tṛtīyaphala)			Place the digit of the next non-square place (3) on the right of the new remainder (38). Now the number is 383 and divide by twice the second result (34) = 68
			3	4	0				Subtract the above number from the maximum possible number 68 X 5 = 340. Here the quotient is 5. 5 is tṛtiyaphala.	3 4 5
				4	3	9			Place the digit of the next square place (9) on the right of the remainder (43), Now the number is 439
- 52					2	5			Deduct square of the quotient (5) = 25 from it.
÷ 2 X 345 = 690			690)	4	1	4	3	(6 (6 - caturthaphala)	Place the digit of the next non-square place (3) on the right of the remainder (414). Now the number is 4143 and divide by twice the third result (345) = 690
				4	1	4	0		Subtract the above number from the maximum possible number 690 X 6 = 4140. Here the quotient is 6. 6 is caturthaphala.	3 4 5 6
							3	6	Place the digit of the next square place (6) on the right of the remainder (3), Now the number is 36
- 62							3	6	Deduct square of the quotient (6) = 36 from it.
								0

Since the remainder is zero the given number is a perfect square.

Square root of 11943936 = 3456

See Also

सद्रत्नमाला में 'वर्गमूल