Ginti Vedic

Vedic Square Root Calculator

Type a number and watch its square root read off from its digits, with every step and a check by squaring.

Number

Try a classic example

√5,776

Found with Vilokanam

Sutra used

Vilokanam

विलोकनम्

“By mere observation”

5,776 is a perfect square, so its root can be read off: the last digit narrows the units digit to two choices, and the leading part gives the rest.

Watch it worked out: Vilokanam

Step 5 of 5

Check by squaring: 76² = 5,776, so √5,776 = 76.

What you entered

Number: 5,776

1
Split off the last two digits
5,776 → 57 | 76= leading part 57
2
The last digit shows how the root can end
last digit 6: 4² = 16, 6² = 36= ends in 4 or 6
3
The tens: the largest square not above 57
7² = 49 ≤ 57 < 64 = 8²= tens part 7
4
Choose 74 or 76: compare 57 with 7 × 8
57 ≥ 7 × 8 = 56= the larger one, 76
5
Check by squaring
76² = 5,776= √5,776 = 76

Result

Square root: 76

5,776 is a perfect square, so its root can be read off: the last digit narrows the units digit to two choices, and the leading part gives the rest.

Square roots read off by observation

For each perfect square, the last digit narrows the root's last digit to two choices, the leading part (everything before the last two digits) gives t, the part of the root in front of its last digit, and one comparison with t x (t + 1) picks the answer.

Number	Root ends in	Leading part	Compare with t x (t + 1)	Root
144	2 or 8	1, so t = 1	1 < 2: smaller	12
289	3 or 7	2, so t = 1	2 >= 2: larger	17
576	4 or 6	5, so t = 2	5 < 6: smaller	24
729	3 or 7	7, so t = 2	7 >= 6: larger	27
1764	2 or 8	17, so t = 4	17 < 20: smaller	42
2209	3 or 7	22, so t = 4	22 >= 20: larger	47
3025	5	30, so t = 5	only one choice	55
5776	4 or 6	57, so t = 7	57 >= 56: larger	76
7744	2 or 8	77, so t = 8	77 >= 72: larger	88
9801	1 or 9	98, so t = 9	98 >= 90: larger	99
15129	3 or 7	151, so t = 12	151 < 156: smaller	123

t is the largest number whose square is not above the leading part: for 5776 the leading part is 57 and 7^2 = 49, so t = 7. A number ending in 2, 3, 7 or 8 is never a perfect square, and one ending in 5 or 0 has only one choice. For a number that is not a perfect square the calculator finds the root digit by digit with the duplex method instead.

Reading √5776 from its digits

Split off the last two digits: 57 | 76. Only 4² = 16 and 6² = 36 end in 6, so the root ends in 4 or 6. The largest square not above 57 is 7² = 49, so the root is 74 or 76. To choose, compare 57 with 7 × 8 = 56: 75² = 5625 starts with 56, and 57 is past it, so the root is the larger one, 76. Check: 76² = 5776. No long division, just two small facts you can see in the digits.

What the last digit of a square tells you

A perfect square can only end in 0, 1, 4, 5, 6 or 9, so a number ending in 2, 3, 7 or 8 is never one. A square ending in 1 has a root ending in 1 or 9; in 4, a root ending in 2 or 8; in 6, 4 or 6; in 9, 3 or 7; in 5, always 5; and in 0, always 0 (a perfect square ending in 0 ends in 00). The two choices always add up to 10, one below 5 and one above, which is why a single comparison with t × (t + 1) picks between them.

Frequently asked questions

Does this work for numbers that are not perfect squares?

Observation only reads the root off a perfect square. For any other number the calculator switches to the duplex method, which finds the root digit by digit: for 2 the first digit is 1 with remainder 1, every later column is divided by 2, and the columns give 1.4142. It also shows the one-step estimate from the squares either side: for 50, 7 + 1 ÷ 14 ≈ 7.0714, against the true 7.071068.

How do I find the square root of a five-digit number like 15129?

Split off the last two digits: 151 | 29. It ends in 9, so the root ends in 3 or 7. The largest square not above 151 is 12² = 144, so the root is 123 or 127. 12 × 13 = 156, and 151 is less than that, so it is the smaller one: 123. Check: 123² = 15,129.

Why compare the leading part with t × (t + 1)?

Of the two possible roots, one is below 10t + 5 and the other above it, and (10t + 5)² = 100 × t × (t + 1) + 25. So if the leading part is at least t × (t + 1), the number is past (10t + 5)² and the root is the larger choice; otherwise it is the smaller. For 5776, t is 7, 7 × 8 = 56 and 57 is at least 56, so the root is 76 rather than 74.

Do the sutras really come from the Vedas?

Jagadguru Swami Bharati Krishna Tirthaji attributed them to an appendix of the Atharvaveda in his 1965 book Vedic Mathematics, but they have not been found in any known Vedic text, and historians of mathematics regard them as his own modern formulations. The methods are correct arithmetic either way, and every root here is checked by squaring it.