http://easysqrtsforkids.blogspot.com...re-edging.html
I searched the google and looked for simple ways of taking the square roots of numbers.

There are those who suggesting the Guess-Divide-Average Method while there are also those who suggesting the algorithm.

Let us take the method of algorithm.

**Disadvantages of Manual Algorithm**
1) One difficult part is ‘subtracting’ the numbers. As the process goes on, the numbers involved to be subtracted are getting longer (count of digits increases)

1) It looks “easy in a paper”, doing the instructions of how to get the “next digits” but in reality, this part, “looking for the correct next digit” is the hardest thing to do. Admit it, the tricky steps of “multiplying by 2 and add the square of so and so to this and to that”, takes a lot of effort and mental torture. How much more, for a grade school kid?

I discovered a method of eliminating the subtraction of numbers and the tricky way of finding the next digit via “trial-by-error” by using the what I called "

**Universal Square Edging **(or

**MSM-3 format**)

Example, let’s take the square root of 2.

__Step 1__:

Find a square value, equal or nearest to 2. The answer is 1 and its equivalent square root value is 1.

Take note that 2 is in-between the squares “1 and 4”

**1^2 = 1**
**2^2 = 4**
So, the next digit could either be,

**1, 2, 3, 4, 5, 6, 7, 8 or 9**. Also, there are nine possible combination and only one is the appropriate one –

**1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8 and 1.9**
__Step2:__
“

**Parameter Checking”**
I came up with the idea of knowing first, the “middle square” value of 1^2 and 2^2

**1.5^2 = 2.25**

Take note that 2.25 is greater than 2.00. We assume then that the next digit is below 5

1.5^2 = 2.25

**1.4^2 = ?**

1,3^2 = ?

1.2^2 = ?

1.1^2 = ?
1.0^2 = 1.00

From nine combination, we slim it down into “four”.

__Step 3:__
“

**Digit Locating**”

Still, “four is too many”. Squaring them one at a time is a slow process. So I came up with an idea of averaging the value of 1.5^2 and 1.0^2

First: Add 1^2 and 1.5^2

**1.5^2 = 2.25**
**1.0^2 = 1.00**
............

**3.25**
Second: Divide the sum by 2

**3.25 / 2 = 1.625 **or simply

**1.62**
Third: Always subtract the quotient by “6”

**1.62 – 6 = 1.56**
You maybe asking, why subtract by 6?

The average value 1.56 corresponds to 1.25^2, which is the middle part of 1^2 and 1.5^2

**1.5^2 = 2.25**
1.4^2 = ?

1,3^2 = ?

**1.25^2 = 1.56’25 **(to be exact)

1.2^2 = ?

1.1^2 = ?

**1.0^2 = 1.00**

To simplify the equation, I just write them down as follows:

**(4, 3) / 2.25**
.........

**\ 1.56** This is the middle value taken from 1.62 - 6 (or average)

.

**(2, 1).. 1.00**
...........

**3.25 /** 2 ( “____/ 2” means divided by 2)

...........

**1.62 – 6 **(quotient and then subtracted by 6)

Take note, 2.00 (or 2, which is the given problem) is in-between 2.25 and 1.56 and the digits corresponding to this area are 3 and 4, to be specific…

**1.3 and 1.4**

__Step 4:__
**Digit Sampling**
Since there are only two numbers involved, we can square them, one at a time

Using the

**SSQ** method

...

**1.4^2 = 01.16**
.

**(+)1x4 =** .....**8 **.
.................

**1.96** Acceptable

Squaring 1.3 is not necessary, the nearest to 2.00 is 1.4^2

Therefore;

**1.4^2 = 1.96** (draw a square and enclose these data)

Now, the next problem is, knowing the next digit.

__Step 5:__
Do the second parameter checking

Take note;

**1.4^2 = 1.96**
Use this "Multiplication Technique" (I called it L.A.L. Multiplication Technique)

First: Put a

**5**, after

**1.4 **and

**25 **after

**1.96**

**1.4**__5__ ^2 = 1.96’__25__
Second: Multiply the already known digits (1.4) by the "doubled value" of the next "involved’ digit" (at this time, the next digit is 5)

To be strict, mathematically, the next digit 5 = .05

.

**05 x 2 = .10**

1.4 x .10 = 0.14
Third: Add the new product to the modified (partial), square value of 1.45

..

**1.4**__5__^2 = 1.96’__25__
__1.4 x 0.1 = 0.14 .__
...............

**2.10’25**
Take note that

**2.10’25 is greater than 2.00**, giving us the idea that the next digit is

**below 5 – either 1, 2, 3 or 4.**

__Step 6:__
Do the second ‘Digit Locating’

**1.45^2 = 2.10 **(middle limit square value)

..

**1.4^2 = 1.96** (lowest limit square value)

..

**(4, 3)2.10**
........

**/ 2.03** This is the middle value (or average value)

__(2,1) \ 1.96__
...........

__4.06__/ 2
..........

**2.03**
Take note, in this second digit locating, there’s

**no need to subtract by 6**
**2.03** (the middle or average value), is still above 2.00, so we choose 1 and 2 as the next digits.

__Step 7:__
Do the second digit sampling:

**1.4**__2__ ^2 = 1.96’__04__
**1.4x .04= **.....**5’6 **.
.............

**2.01’64 > 2 (over)**

**1.4**__1__ ^2 = 1.96’__01__
**1.4x .02= **.....**2’8 **.
.............

**1.98’81 < 2 Acceptable**

Therefore

**1.41^2 = 1.98’81**

For the next digits, repeat the process of:

1) Parameter Checking

2) Digit Locating

3) Digit sampling