The method is done by reversing the process of squaring 2-digit numbers. Let's take an example from the image, square root of 169. Segregate the digits of 169:
1 6 9
First, think of a number that when squared will equal to 1 .
1*1 = 1
So, first digit of the answer is 1.
Next, double this number then divide from the next digit 6.
6/(2*1) = 3
So, last digit of the answer is 3.
Square root of 169 = + - 13 (positive and negative 13)
To check if perfect square, take the last digit 3 then square it.
3*3 = 9 This equals to the last digit of the problem 169. Therefore, perfect square.
More examples:
Square root of 1681 = ?
16 8 1
4*4 8/(2*4)
4 1
Square root of 1681 = + - 41
Square root of 441 = ?
4 4 1
2*2 4/(2*2)
2 1
Square root of 441 = + - 21
Square root of 3600 = ?
36 0 0
6*6 0
6 0
Square root of 3600 = + - 60
Square root of 1225 = ?
12 2 5
Here, 12 is not a perfect square. We will use a number that if squared will be near to 12, but not over it. If 3x3 = 9, less than 12. If 4x4 = 16, more than 12. Therefore, use 3, then there will be remainder of 12 - 3*3 = 3. Put this remainder beside the next digit 2 and become 32.
32/(2*3) = 5 with remainder 2
Square root of 1225 = + - 35
Rewriting the solution:
Square root of 1225 = ?
12 2 5
n*n R2/(2n) r5 R = 12 - 3*3 = 3
3*3 32/(2*3) r5
3 5 with r = 2 25
Square root of 1225 = + - 35
5*5 = 25 = r5 Therefore, perfect square.
Square root of 888 = ?
888 is not a perfect square. There is no perfect square ending with 2, 3, 7 and 8. This can be check on the squares of 1 to 10.
These are long to write but quick on the mind.
#Matamata
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