TAKING THE SQUARE ROOT OF A NUMBER
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METHOD
 Separate the number's digits into pairs starting from the decimal point, moving to the left; and, then, moving from the decimal point, moving to the right.
For example, let's take the square root of the number 88532.423. Separating it into pairs yields the following:
________________  08'85'32.42'30 


 Determine the first divisor. What number can you square that will net you a result close to the first pair of digits (in this case the digits 08)? Two, yes? Because 2^{2}; = 4. 3 squared would yield 9, which is too large.
 Write that number above the 8 and to the left of the diagram, record the square of the number underneath the first pair of digits, do the subtraction, and then carry down the next pair of digits like this:
2 <=== "Root"  08'85'32.42'30 <=== "Radicand" 2  04  4 85 


To determine the next divisor, double the first digit of the root (the two up top) by 2 and write it to the left of the diagram.
2  08'85'32.42'30 2  04 4  4 85 


We need one more figure in the new divisor. We get that by envisioning a digit call it "y"that can be "tacked on" to the 4 such that the number 4y multiplied by y is close or exactly equal to the number 485. That number is 9. 49 x 9 = 441. Write a nine above the second pair in the radicand and write it also to the left of the 4 of our new divisor. Write the 441 under the 485, subtract, and bring down the next digit pair from the radicand.
2 9  08'85'32.42'30 2  04 49 4 85  4 41  44 32 


To determine the next divisor, we again multiply the already determined digits of the root by 2 (we'll always multiply the already determined digits by 2 as we work through the problem) and write that number (58) to the left of the diagram on the next level down. Then we determine what digit we can tack on to 58 such that 58y multiplied by y gets us a number closest to 44 32
2 9 7  08'85'32.42'30 2  04 49  4 85  4 41 587  44 32  41 09


Continuing the process, we get the following:
2 9 7. 5 4  08'85'32.42'30 2  04 49  4 85  4 41 587  44 32  41 09 5945  3 23 42  2 97 25 59504  26 17 30  23 80 16 2 37 14


At this point, we carry the decimal point up into the root to obtain our answer. As it turns out our answer's going to be a little off (the square of 197.54 is 88530.051 ) because we haven't carried our operations far enough. So let's add a zero pair of digits to the end of the number and carry on to get an additional measure of accuracy.
2 9 7. 5 4 3  08'85'32.42'30'00 2  04 49  4 85  4 41 587  44 32  41 09 5945  3 23 42  2 97 25 59504  26 17 30  23 80 16 595083  2 37 14 00  1 78 52 49 58 61 51


We're much closer. The square of 297.543 is 88531.84. As we did above, we could continue and add a few more pairs of zeros onto the number to obtain a more exact answer (a more exact answer is 297.54398).
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WHY DOES THIS METHOD WORK?
Our method is derived from algebra. Let's find the square root of an algebraic expression.
Suppose we have a number, call i(a + b) ^{2 }. We know that: (a + b) ^{2 }= a ^{2} + 2ab + b ^{2 } and that the square root of this number will be (a + b ). Let's take a look at this in the context of the method we used above. First we determine that a is the largest number squared which will most nearly equal a^{2 }. That number is a , and, in this case, when squared, equals a ^{2 }exactly. And we see that subtracting a ^{2 }leaves us with 2ab + b ^{2 }:
a a  a^{2 }+ 2ab + b^{2 }  a^{2}  2ab + b^{2 }


Now we know that b is the other part of the root. To get b , we could divide 2ab by 2a, which is twice the part of the root we've already determined.
a + b a  a^{2 }+ 2ab + b^{2 }  a^{2 }  2a  2ab + b^{2 }


So we write down the 2a and then determine the number y , such that 2ay times y is equal or nearly equal to 2ab + b ^{2}. In this case, we know we need to add a b.
"Wait a minute! Add b?" I can imagine you thinking. We tack on y , right? What does that mean? Well, take a look at the first problem we did above when we determined that 4 was the first digit of the new divisor and then we tacked on a 9 (our y ). In essence, what we did there was add 9 to 40, since the 4 was a tens "placeholder." Make sense?
Okay, so we see, in our algebraic case that the result is:
a + b a  a^{2 }+ 2ab + b^{2 }  a^{2}  2a + b  2ab + b^{2 }  2ab + b^{2 }


Magic, heh? This is the perfect method for determining a square root when your calculator's batteries die and no replacement batteries are at hand.
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