**What is Square Edging?**

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

).The only limitation of Square Edging is that it is only applicable in taking the square roots of numbers called perfect squares. The square roots of perfect squares are always in a “whole numbers”, never as fractions or decimal numbers. In fact, from 1 up to 100, there are only 10 perfect squares. While from 101 up to 10,000, there are only 90 perfect squares. The rest (9,900 other numbers) are useless in dealing with this format of S.E. But TRUST ME, it will WORTH A LOT.

2) 5D/6D.SE

3) 7D/8D.SE

**Four Digit Square**** Edging** (3D/4D.SE)

__WARNING:__

To easily understand this method of Square Edging, I highly recommend that you first study the SSQ Method.

__Step 1__

Count the digits of the given number. Re-group them by twos, starting from the last digit.

__Step 2__

Find an index square, equal to or nearest to but less than the first group of digits of the given number. Write down the equivalent square root, below this first group of digits.

4

__Step 3:__

Find a pair of index squares ending with the same last digit, as to the last digit of the given number. Write down below the last group of digits, their corresponding square roots.

__4__ and 6__4__. Their equivalent square roots are, 2 and 8.

4 2

_ 8

__Table of Complementary Index Squares__

^{2} = 0__1__

9^{2} = 8__1__

1 + 9 = 10

^{2} = 0__4__

8^{2} = 6__4__

2 + 8 = 10^{}

^{}

^{}3^{2} = 0__9__

7^{2} = 4__9__

3 + 7 = 10

^{2} = 1__6__

6^{2} = 3__6__

4 + 6 = 10

^{2} = 00

5^{2} = 25

NO PAIRS

__Step 4__

Copy the first digit of the upper square root to complete the lower square root.

4 2 ← first possible square root

4 8 ← second possible square root

__Final Step__

One way to find out which of the two is the true square root, apply the 2D.SSQ

^{2} = 16’04 ← PSL

__+4x2x2 = 1'6 __← SP1 (provided that 6 of 16 aligned to 0 of PSL)

............. 17’64 ← T-Sum (provided that T-Sum aligned to PSL)

....48^{2} = 16’64 ← PSL

__+4x8x2 = 6'4 __← SP1 (provided that 4 of 64 aligned to the second 6 of PSL)

............ 23’04 ← T-Sum (provided that T-Sum aligned to PSL)

__Tip 1__: Check the first digit of any of the two possible square roots. Multiply it to a number next to it, higher by 1. Consider the product as our “square root indicator”.

The first digits of of both possible square roots of 42 and 48 are the same and that is 4

Sq. Rt. Indicator = 20

__Tip 2__: Compare the ‘square root indicator’ to the first group of digits of the given number

__First Condition__:

If the first group of digits is less than the square root indicator, pick the square root with lower value as your final answer

__Second Condition__:

If the first group of digits is greater than the square root indicator, pick the square root with higher value as your final answer.

__Step 1__:

Re-group by twos

Take note, that if we re-group 729, it will appear as 7’29. But in the general rules of SSQ - in the process of squaring a number, the count of digits of a number must be doubled. So, a two-digit number must become four-digit number. To obey this rule, we should write 729 as 07’29.

__Step 2__

2

__Step 3__

2 3

_ 7

__Step 4__

2 3

2 7

__Final Step__

Next to 2 is 3.

2 x 3 = 6

Sq,Rt, Indicator = 6

27 > 23

__Author’s Note:__

The truth is, this 3D/4D.SE in not really new. There are similar ideas that were posted in google and you.tube (this is one good example by Z-Math

http://www.ehow.com/how_2322332_square-root-number-mentally.html

).I wish to give the credit to a certain Prof. Barbosa, for the Helpful tips (watch:

**http://www.youtube.com/watch?v=WNJ2dCavUrA&feature=related**

square root of 665

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