Quantitative Reasoning

I am searching in Google about how to calculate the square root of numbers in an easier way beside the long hand division method. I found one, which is an easy instruction by Z-math. But I found it, seems tricky when the perks (a short name for 'perfect square numbers'), involved, are in 6 or in 8 digits. I developed a much easier method, I called MSM-1 and would be glad to share to you. First, I recommend you to read Z-math instructions https://www.ehow.com/how_2322332_square- ... tally.html (if the Admin permit it), to shorten the discussion.

Let's start with a 6-digit perk, example \/66,564.

Step 1: Regroup the digits by twos

\/06'65'64

Step 2: Find the nearest square, equal or less than the first group of digits (that is, 06). Write down below it the equivalent square root.

\/06'65'64
2

Step 3: Find two squares that ends with 4. There are always a pair, 4 and 64.
Write down their equivalent square roots, 2 and 8.


\/06'65'64
2 _ 2
2 _ 8

Step 4: Notice that our problem is, what would be the middle digit? There are still ten choices, 0, 1, 2, 3, 4, 5, 6, 7,
8 and 9.

Step 5: Write down N as the missing digit.


\/06'65'64
2 N 2
2 N 8

Step 5: Apply the SSQ Method.


SSQ METHOD

SSQ stands for 'Systematic Squaring'. It is based on this popular algebraic expression, (x + y)^2 = x^2 + 2xy + y^2.

But to simplify the presentation, I recommend that all 'index squares' should be in two digits, so instead of 1^2 = 1, it should be 1^2 = 01. The same as to 2^2 and 3^2, they should be 04 and 09, respectfully. There are three main parts for SSQ, namely - PSL (Partial Square Line), SP1, SP2, SP3 and so on (Sub-product or sub-products that depend on how many digits involved, in squaring that number), the ST1, ST2, SE3 and so on (sub-totals, again, depend on how many digits involved) and TSUM (or the total sum).
...TO BE CONTINUED

(Note: I cannot find the square and sqaure root signs , so simply consider \/ as square root sign and ^2 as to the power of 2 or 'square of'.)