At the beginning, I thought only of creating or designing a method that would help grade school children, in their studies involving square roots of large numbers. The fact that I was been challenged by these following statements:

http://www.homeschoolmath.net/teaching/square-root-algorithm.php**
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**
**

**Somehow, I agree with them. **Why do you think a child should take (or extract), the square root of a large number, in millions, only to find out that there is “no end” in their attempt to find out the square root of such number?

**Square Edging,** in a sense, is more ‘logical’ and ‘practical’. Children are more interested in knowing “results” which are more reasonable than ‘trivial’ answers (such as, knowing the square root of 6,4967,028). So, **MSM-1** and **MSM-2** are practically designed, **obviously**, __for “perfect square” numbers in large values only__, **or else, use a calculator**.**
**

**
**

**MSM-3**, has its own “special” place.

Lets take this example:

**√20**

( http://www.itl.nist.gov/div897/sqg/dads/HTML/squareRoot.html )

__.... 4.__

**√20.00**

__- 16__

.... **4 00**

**(4 x 20 x ?) + ? ^{2} ≤ 400**

**Let ? = 5**

**(4x20x5) + 5 ^{2} ≤ 400**

**400 + 25 ≤ 400**

**425 (over) ≤ 400
**

**
**

**Let ? = 4 **

**(4x20x4) + 4 ^{2} ≤ 400**

**320 + 16 ≤ 400**

**336 ≤ 400 Accepted**

__.... 4. 4__

**√20.00**

__- 16__

.... **4 00**

.... ** 3 36** = (4x20x4) + 4

^{2}

....... **64 00**

This time,**
**

**
**

**(44x20x ?) + ?^{2} ≤ 64 00**

**Let us use “7**”

**(44x20x7) + 49 ≤ 64 00**

**6160 + 49 ≤ 64 00**

**6209 ≤ 64 00 Accepted**

**(44x20x8) + 64 ≤ 64 00**

**7040 + 64 ≤ 64 00**

**7104 (over) ≤ 64 00 **

__.... 4. 4 .7__

**√20.00 00 00**

__- 16__

.... **4 00**

.... ** 3 36** = (4x20x4) + 4

^{2}

....... **64 00**

....... ** 62 09** = (44x20x7) + 49

…......**1 91 00**

**number of digits involved to further increase** (this time, from two digits, 44, into three, 447), while **guessing the next possible digit seems to become very difficult to attain**

** (447x20x ?) + ?^{2} ≤ 19100**

This is the “main problem” of the traditional method of extracting the square roots of numbers ;

2) As you try to add another digit, the longer the multiplication and subtraction of numbers involved**
**

**
**

**
**

**APPLYING THE **

**√20**

__Step 1: __

Like in any rule, look for a square, equal or nearest the given number. Write down its square root value.

__.... 4.__

**√20.00**

__Step 2__:

Create a “**P-Chk**”

**P: 20**

**4.5 : 20.25 ↓**

**4↓**

__Step 3: __

Create A Digit Locator. (Note: it is similar to the Square Root Locator, except this time we’re simply looking for the next digit after 4) Please review :

http://easysqrtsforkids.blogspot.com/2010/08/five-digitsix-digit-square-edging.html

**/ 20.25 \**

...... **\ 18.06 /** 20

......... __16 ....__

......... __36.25__ / 2

......... **18.12**

The given number 20 is in-between 20.25 and 18.06. On the left side of the equation are the digits 3 and 4

__Step 4__:

Do the **Digit Sampling**

Let start with “4”

**4.4 ^{2} = 16.16**

__4x8 = ...3 2 .__

.......... **19.36** ≤ 20

** 4.4^{2} = 19.36** Accepted

Write down the 4 as our second digit

**4. 4**

**√20.00**

__L.A.L. Multiplication Technique____ __

Do you still remember the R.A.R. multiplication pattern? (See: http://easysqrtsforkids.blogspot.com/2010/08/new-method-of-squaring-numbers.html )

**R.A.R**. stands for, __R__eference and __A__ll the digits to the __R__ight but this time, you will learn another technique:

**L.A.L.** stands for __L__ast digit and **A**ll the digits to the __L__eft. I will not explain here how I got it (except if requested). It is still based on the same equation (A+B)^{2} = A^{2} + 2AB + B^{2}.

**4.4 ^{2} = 19.36**

If we put a digit next to 4.4, example, a digit “2”

**4.4 2^{2} = ?**

__Instruction 1__:

Write down the square of the original value (4.4^{2}) = **19.36** then write next to it the square of 2

**4.4 2^{2} = 19.36’04**

__Instruction 2: __

Following the LAL multiplication technique, multiply the last digit (2) to all to the left digits ( that is, 44, ignore the decimal point).

**44 x 2 = 88**

__Instruction 3:__

Don’t forget the DTP, "double the product”**
**

**
**

**88 x 2 = 176**

**As a practice, always double the value of the last digit before multiplying to all digits to the left.**

__Instruction 4__:

To make it clear, the square of 44.2 must end in 4. So, the last digit of 17__6__ must__ be aligned to the next digit to the left __of the last digit of__ __19.36’__0__4

....

**4.4**

__2__^{2}= 19.36’04__(+) 44x4 = .... 17’6 .__

.... **4.42 ^{2} = 19.53’64** (which is the true square root of 4.42, check in your calculator)

**our main discussion.**

We already know that the second digit of our square root is 4, so our “partial answer” will be equal to “4.4”. The next question is…what will be the next digit?

**4.4 ?**

__Step 6:__

Create a 2^{nd} P-Chk for the next digit after 4.4

In data above, the square value for 4.4 is

4.4

^{2}= 19.36

__2 ^{nd} P-Chk__

**P: 20.00 00**

**4.4 5^{2} = 19.36’25**

__‘’x10 = ____.... 44 ....__ (Ignore the decimal point)

............ **19.80’25** ↑ (the up arrow shows that P: 20 is in upper area)

**5↑**

**5↑** indicates that the next digit is **above 5** (There are four possible digits, **6, 7, 8 **or** 9 **and only one is acceptable).

__Step 7__:

To easily determine which group the next digit belongs (9, 8 group or 7, 6 group), use a 2^{nd} Digit Locator

__2 ^{nd} Digit Locator__

**(4.50 ^{2}) = 20.25’00** (already given, it is the same as to 4.5

^{2}= 20.25)

**(M.V.) = 20.02’62**

__(4.45 ^{2}) = 19.80’25__

..............** 40.05’25 / 2**

..............** 20.02’62**

Therefore, the possible digit belongs to “**7, 6**”**group**.

^{nd} Locator

__2 ^{nd }Digit Locator^{}__

**9, 8 / 20.25**

.......**\ 20.02 \ ↓**

........** 19.80 / 7, 6 **(20 is between 20.2 and 19.80)

........ __40.05’__/ 2

........** 20.02**

^{nd} digit locator gives us a clue that the next digit could either be **7 **or **6**__
__

__
__

__Step 8:__

Again, use **7** or **6** as sample, (**do the** **Digit Sampling**)

....** 4.4 7^{2} = 19.36’49**

__(+)’’ x14 = .... 61’6 .__

.... **4.47 ^{2} = 19.98’09
**

**
**

**“There is no need to solve for 6, because its answer would be much lesser than 4.47”**

__
__

__Making Sure of Your Choice of Digit:__

To avoid the possibility that the 7 is not yet the correct digit, choose 8 as a ‘sample’

....** 4.4 8^{2} = 19.36’64**

__(+)’’ x16 = .... 70’4 .__

.... **4.48 ^{2} = **... w/out completing the addition, you will notice that if we add 3 to 7, the carry 1 makes the 19 to become 20, giving us a clue that the 4.48

^{2}> 20.

**7** is the correct, next digit after **4.4**

^{2} = 19.98’09

__Step 9__:

For the next digits, simply repeat these __three basic rules of square edging__

1) Create a parameter checker **(P-Chk**)

2) Determine the correct digit by using the **Digit Locator**

3) To make sure which digit is the appropriate digit, do the **Digit Sampling**

__
__

__3 ^{rd} P Chk__

**P: 20.00 00**

**4.47 5^{2} = 19.98’09’25**

__‘’x10 = ____........ 4’47 ....__ (Ignore the decimal point)

.............. **20.02’56’25** **↓ **(the down arrow shows that P: 20 is in lower area)

**4↓**

__3 ^{rd} D Loc__

**20.02**

**20.>. \ ↓ **(The symbol “**20.>”** means, 20 point “something”)

** 19.98 / 2, 1 **(the digit locator is giving us a hint that the next digit could either be 2 or 1)

**40.00 / 2**

**20.00 **

__Digit Sampling__

**4.47 2^{2} = 19.98’09’04**

__....’’ x4 = ...... 1’78’8 .__

**4.472 ^{2} = 19.99’87’84 (acceptable)**

To make sure, check digit “3”

**4.47 3^{2} = 19.98’09’09**

__....’’ x6 = ...... 2’68’2 .__

**4.473 ^{2} = 20...0.......... > 20 (over)**

We are now sure that the next digit is 2, therefore;

**4.472 ^{2} = 19.99’87’84**

__P-Chk__

**4.472 5^{2} = 19.99’87’84’25**

**................. ......... 44’72 **

__...__

**.................20.00’32’56’25 ↓**

**4↓**

**20.00’33 (round-off value)**

**20.00’10 \ ↓**

__19.99’88 /__ 2, 1

__40.00’21 /__ 2

**20.00’10**

__D. Sampling__

**4.472 2^{2} = 19.99’87’84’04**

__‘’x 4____.... . = ..........17’88’8 .__

................ **20. ... 04’ **...... **> 20 (over)**

**4.472 1^{2} = 19.99’87’84’01**

__‘’x 2____.... . = ........... 8’94’4 .__

................ **19.99’96’78’41 (acceptable)**

If you want, you can continue the process by repeatedly doing the parameter checking, digit locating and digit sampling.

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