User:Celestianpower/Square root

To start with, write the following (where 1215 is your number):

| 1215 |

Then split it into sets of 2 digits, starting at the "units" column:

--- | 12 | 15 |

Then you think to yourself, "What is the biggest square number that is smaller than the leftmost set of 2 digits?" (in our case 9), write this number below the leftmost pair (12 in our example) and subtract:

--- | 12 | 15 |   9   ---    3

Then write the square root of the biggest square (3 in our case - square root of 9) both to the side (we'll call this the "carry number") and above the line above the original number:

3  ---   | 12 | 15 | 3    9     ---      3

So we now know that the square root of 1215 is 30-something. Now double the "carry number" (the 3 in our case doubles to 6), write it below the result of our subtraction (the other 3) and bring down the next 2 digits to sit next to the result of our subtraction (hence in our example making 315):

3  ---   | 12 | 15 | 3    9      315 6

Now you must think "60-something, times something is the largest number that is still smaller than the result of our subtraction and the remainder. What is 'something'?" (we'll call it the "remaining number" - in our case the 315). You may need to use paper to the side to work out multiplications. In our case, 65 x 5 would give 325: too big so 'something' must be "4". 64 x 4 = 256. Write this number next to the "carry number" and write the result below the "remaining number":

3   ---    | 12 | 15 | 3     9       315 64    256

Now subtract this number (the 256) from the "remaining number" (the 315) and write this underneath. Add the next 2 digits and this becomes the new "remaining number". Also add the number you added to the "carry number" to the end of the answer. We know now that the square root of 1215 is 34 point something:

3 4   ---    | 12 | 15 | 3     9       315 64    256       -        5500

So we continue in this manner. Double the last digit of the "carry number" (we'll call it the "leading digit"), add it to the first digit of the previous "carry number" and write the resulting number below the previous "carry number" (So, in our example, 4 doubles to 8 so you write 68). Then you think "6 hundred and forty something times something is the largest number that goes into your 'subtraction number' (in our case, 687 x 7 = 4809 and 688 x 8 = 5504 - too big). What is something?". Write this below the "subtraction number" and write the "leading digit" next to the answer (and the decimal point in the last pair of digits you brought down were zeros):

3 4 . 8   ---    | 12 | 15 | 3     9       315 64    256       -        5500 687    4809

Subtract to gain your next "subtraction number" and repeat.