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(easy hangman)
(difficult)
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Use the algorithm to find the square root of a six-digit number
Example: Find √297 504
to two decimal places.
First group the numbers under the root in pairs from right to left, leaving
either one or two digits on the left. For each pair of
numbers you will get one digit in the square root.
To start, find a number
whose square is less than or equal to the first pair (29), and write
it above the square root line (5).
|
|
| 5 | | |
| √29 | .75 | .04 |
|
- 25 | | |
| (10_) |
4 | 75 | |
|
| 5 |
4 | |
| √29 | .75 | .04 |
|
- 25 | | |
| (104) |
4 | 75 | |
|
| Square the 5, giving 25, write that
underneath the 29, and subtract. Bring down the next pair
of digits.
|
Then double the number above the square root
symbol line (highlighted), and
write it down in parenthesis with an empty line next to it as shown.
|
Next think what single digit number something could
go on the empty line so that hundred-something times something would
be less than or equal to 475.
102 x 2 = 204
104 x 4 = 416, so 4 works. Write therefore 4 in the answer.
|
| 5 |
4 | |
| √29 | .75 | .04 |
|
- 25 | | |
| (104) |
4 | 75 | |
| - 4 |
16
| |
| | 59 | 04 |
|
| 5 |
4 | |
|
√29
|
.75 | .04 |
|
- 25 | | |
| (104) |
4 | 75 | |
| - 4 |
16
| |
| (108_) | 59 | 04 |
|
| 5 |
4 |
5 |
|
√29
|
.75 | .04 |
|
- 25 | | |
| (104) |
4 | 75 | |
| - 4 |
16
| |
| (1085) | 59 | 04 |
|
|
Calculate 4 x 104, write the product below 475, subtract,
bring down the next pair of digits.
|
Then double the number
in the answer (54), and write the doubled number (108) in parenthesis with an empty line next to it
as shown.
|
Think what
single digit number
something could go
on the empty
line so that thousand-eighty-something
times something would be
less than or equal to 5904.
1085 x 5 = 5425, so 5 works and goes to the answer.
|
| 5 | 4 | 5 | |
|
√29
| .75 | .04 | .00 |
|
- 25 | | | |
| (104) |
4 | 75 | | |
| - 4 |
16
| |
|
| (1085) | 59 | 04 |
|
| - | 54 | 25 |
|
| 4 | 79 |
00 |
|
| 5 | 4 | 5 | |
|
√29
| .75 | .04 | .00 |
|
- 25 | | | |
| (104) |
4 | 75 | | |
| - 4 |
16
| |
|
| (1085) | 59 | 04 |
|
| - | 54 | 25 |
|
| (1090_) | 4 | 79 |
00 |
|
| 5 | 4 | 5. | 4 |
|
√29
| .75 | .04 | .00 |
|
- 25 | | | |
| (104) |
4 | 75 | | |
| - 4 |
16
| |
|
| (1085) | 59 | 04 |
|
| - | 54 | 25 |
|
| (10904) | 4 | 79 |
00 |
|
|
To continue, we need to add extra decimal zeros to our number. The
steps continue in exact same manner: calculate 1085 x 5 = 5425, write
that below 5904, subtract, bring down the next pair of (decimal) digits
|
Then double the 'number' 545 which is in the
answer,
and write the doubled number 1090 in parenthesis
with an empty line next to it.
|
10904 x 4 = 43616, so 4 is the next digit in the answer. To find the
answer to two decimal places, we need to find the third decimal with the
algorithm, so we will know whether the answer to two decimals would be
rounded up or down. So two more rounds to go.
|
| 5 | 4 | 5 | 4 | |
|
√29
| .75 | .04 | .00 | .00 |
|
- 25 | | | | |
| (104) |
4 | 75 | | | |
| - 4 |
16
| |
|
|
| (1085) | 59 | 04 |
|
|
| - | 54 | 25 |
|
|
| (10904) | 4 | 79 |
00 |
|
| - 4 | 36 |
16 |
|
| | 42 |
84 |
00 |
|
| 5 | 4 | 5 | 4 | |
|
√29
| .75 | .04 | .00 | .00 |
|
- 25 | | | | |
| (104) |
4 | 75 | | | |
| - 4 |
16
| |
|
|
| (1085) | 59 | 04 |
|
|
| - | 54 | 25 |
|
|
| (10904) | 4 | 79 |
00 |
|
| - 4 | 36 |
16 |
|
| (10908_) | 42 |
84 |
00 |
|
| 5 | 4 | 5 | 4 | 3 |
|
√29
| .75 | .04 | .00 | .00 |
|
- 25 | | | | |
| (104) |
4 | 75 | | | |
| - 4 |
16
| |
|
|
| (1085) | 59 | 04 |
|
|
| - | 54 | 25 |
|
|
| (10904) | 4 | 79 |
00 |
|
| - 4 | 36 |
16 |
|
| (109083) | 42 |
84 |
00 |
|
|
|
|
|
| 5 | 4 | 5 | 4 | 3 |
|
√29
| .75 | .04 | .00 | .00 |
|
- 25 | | | | |
| (104) |
4 | 75 | | | |
| - 4 |
16
| |
|
|
| (1085) | 59 | 04 |
|
|
| - | 54 | 25 |
|
|
| (10904) | 4 | 79 |
00 |
|
| - 4 | 36 |
16 |
|
| (109083) | 42 |
84 |
00 |
| - | 32 |
72 |
49 |
|
10 |
11 |
51 |
|
| 5 | 4 | 5 | 4 | 3 | |
|
√29
| .75 | .04 | .00 | .00 | .00 |
|
- 25 | | | | | |
| (104) |
4 | 75 | | | | |
| - 4 |
16
| |
|
|
|
| (1085) | 59 | 04 |
|
|
|
| - | 54 | 25 |
|
|
|
| (10904) | 4 | 79 |
00 |
|
|
| - 4 | 36 |
16 |
|
|
| (109083) | 42 |
84 |
00 |
|
| - | 32 |
72 |
49 |
|
| (109086_) |
10 |
11 |
51 |
00 |
|
| 5 | 4 | 5 | 4 | 3 | 9 |
|
√29
| .75 | .04 | .00 | .00 | .00 |
|
- 25 | | | | | |
| (104) |
4 | 75 | | | | |
| - 4 |
16
| |
|
|
|
| (1085) | 59 | 04 |
|
|
|
| - | 54 | 25 |
|
|
|
| (10904) | 4 | 79 |
00 |
|
|
| - 4 | 36 |
16 |
|
|
| (109083) | 42 |
84 |
00 |
|
| - | 32 |
72 |
49 |
|
| (1090869) |
10 |
11 |
51 |
00 |
|
Since the last decimal we find with the algorithm is 9, it means the previous
decimal will be rounded up, and thus to one decimal place, √297504
= 545.44
See also:
Find square roots without a calculator
An explanation of why this square
root algorithm works.
A geometric view of the square root algorithm
Square roots by Divide-and-Average
Explanation and example of the Babylonian algorithm for approximating square roots.
Square Root Algorithms
Formulas for a recurrence relation and Newton's iteration that can be used to approximate square roots. For the mathematically minded.
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