Solution:
Split the radicand into pairs of digits.
√7I89
Determine the largest integer whose square is less than or equal to 7.
2√789
Square the value above the line and place it under the term like in a long division problem.
Subtract 4 from 7 to get 3
.
Bring down the next two digits of the radicand.
Double the number above the radicand (2⋅2=4) and write it down followed by a blank space (n), inside parentheses. Then, multiply it by n.
| | 2 | | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | |
| | 4 | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | |
| | | | | | ( | 4 | n | ) | · | n |
Find the largest integer value that could replace the nn and be placed in the next position above the radical so when the two numbers are multiplied it is less than the current remainder of 389. In this case 8 works because 48⋅8=384, which is less than 389.
| | 2 | | 8 | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | |
| | 4 | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | |
| | | | | | ( | 4 | 8 | ) | · | 8 |
Multiply the last digit above the radical (8) by the number inside the parentheses (384) and insert the product under the last term.
| | 2 | | 8 | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | |
| | 4 | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 |
Subtract 384 from 389 to get 5.
| | 2 | | 8 | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | |
| | 4 | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 |
| | | | 5 | | | | | | | |
Bring down the next two digits of the radicand. Since there are no more significant digits, insert 00.
| | 2 | | 8 | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | |
| | 4 | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 |
| | | | 5 | | 0 | 0 | | | | |
Double the number above the radicand (2⋅28=56) and write it down followed by a blank space (n), inside parentheses. Then, multiply it by n.
| | 2 | | 8 | | | | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | | | | | |
| | 4 | | | | | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 | | | | |
| | | | 5 | | 0 | 0 | | | | | | | | |
| | | | | | | | | ( | 5 | 6 | n | ) | · | n |
Find the largest integer value that could replace the n and be placed in the next position above the radical so when the two numbers are multiplied it is less than the current remainder of 500. In this case 0 works because 560⋅0=0, which is less than 500.
| | 2 | | 8 | | | 0 | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | | | | | |
| | 4 | | | | | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 | | | | |
| | | | 5 | | 0 | 0 | | | | | | | | |
| | | | | | | | | ( | 5 | 6 | 0 | ) | · | 0 |
Multiply the last digit above the radical (0) by the number inside the parentheses (0) and insert the product under the last term.
| | 2 | | 88 | | | 00 | | | | | | | | |
√ | | 77 | 88 | 99 | | | | | | | | | | | |
| | 44 | | | | | | | | | | | | | |
| | 33 | 88 | 99 | | | | | | | | | | | |
| | 33 | 88 | 44 | | ( | 44 | 88 | ) | · | 88 | | | | |
| | | | 55 | | 00 | 00 | | | | | | | | |
| | | | | | | 00 | | ( | 55 | 66 | 00 | ) | · | 00 |
Subtract 00 from 500500 to get 500500.
| | 2 | | 8 | | | 0 | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | | | | | |
| | 4 | | | | | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 | | | | |
| | | | 5 | | 0 | 0 | | | | | | | | |
| | | | | | | 0 | | ( | 5 | 6 | 0 | ) | · | 0 |
| | | | | 55 | 00 | 00 | | | | | | | | |
Bring down the next two digits of the radicand. Since there are no more significant digits, insert 0000.
| | 2 | | 8 | | | 0 | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | | | | | |
| | 4 | | | | | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 | | | | |
| | | | 5 | | 0 | 0 | | | | | | | | |
| | | | | | | 00 | | ( | 5 | 6 | 0 | ) | · | 0 |
| | | | | 5 | 0 | 0 | 0 | 0 | | | | | | |
Double the number above the radicand (2⋅280=560)(2⋅280=560) and write it down followed by a blank space (n)(n), inside parentheses. Then, multiplyit by nn.
| | 22 | | 88 | | | 00 | | | | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | | | | | | | | |
| | 4 | | | | | | | | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 | | | | | | | |
| | | | 5 | | 0 | 0 | | | | | | | | | | | |
| | | | | | | 00 | | ( | 5 | 6 | 0 | ) | · | 0 | | | |
| | | | | 5 | 0 | 0 | 0 | 0 | | | | | | | | | |
| | | | | | | | | | | ( | 5 | 6 | 0 | n | ) | · | n |
Find the largest integer value that could replace the nn and be placed in the next position above the radical so when the two numbers are multiplied it is less than the current remainder of 50000. In this case 8 works because 5608⋅8=44864, which is less than 50000.
| | 2 | | 8 | | | 0 | | 8 | | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | | | | | | | | |
| | 4 | | | | | | | | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 | | | | | | | |
| | | | 5 | | 0 | 0 | | | | | | | | | | | |
| | | | | | | 0 | | ( | 5 | 6 | 0 | ) | · | 0 | | | |
| | | | | 5 | 0 | 0 | 0 | 0 | | | | | | | | | |
| | | | | | | | | | | ( | 5 | 6 | 0 | 8 | ) | · | 8 |
Multiply the last digit above the radical (8) by the number inside the parentheses (44864) and insert the product under the last term.
| | 2 | | 8 | | | 0 | | 8 | | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | | | | | | | | |
| | 4 | | | | | | | | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 | | | | | | | |
| | | | 55 | | 0 | 0 | | | | | | | | | | | |
| | | | | | | 0 | | ( | 5 | 6 | 0 | ) | · | 0 | | | |
| | | | | 5 | 0 | 0 | 0 | 0 | | | | | | | | | |
| | | | | 4 | 4 | 8 | 6 | 4 | | ( | 5 | 6 | 0 | 8 | ) | · | 8 |
Subtract 44864 from 50000 to get 5136
| 2 | | 8 | | | 0 | | 8 | | | | | | | | | |
√ | | 7 | 8 | 9 | | | | | | | | | | | | | | |
| | 4 | | | | | | | | | | | | | | | | |
| | 3 | 8 | 9 | | | | | | | | | | | | | | |
| | 3 | 8 | 4 | | ( | 4 | 8 | ) | · | 8 | | | | | | | |
| | | | 5 | | 0 | 0 | | | | | | | | | | | |
| | | | | | | 0 | | ( | 5 | 6 | 0 | ) | · | 0 | | | |
| | | | | 5 | 0 | 0 | 0 | 0 | | | | | | | | | |
| | | | | 4 | 4 | 8 | 6 | 4 | | ( | 5 | 6 | 0 | 8 | ) | · | 8 |
| | | | | | 5 | 1 | 3 | 6 | | | | | |
Since there are 22 pair of numbers to the left of the decimal place in 789789, the decimal of the answer is put 22 place from the left side of the answer.
28.08