1.5 Division method for finding square root:
We have learnt how to find the square root of a
perfect number by factorization method, where we list all the factors of
perfect numbers and then take square root of the factors
Ex : 484 = 2*2*11*11 = 2^{2}*11^{2}
Therefore _{} = 2*11
Factorisation method is time consuming when the
number whose square root to be found, is too large. Because of this reason we
follow another method called division method to find the square roots.
In this method we pair the digits whose square root
has to be found, from the right side (units place).
If the number of digits in the number is even then
all the groups will have 2 digits.
For example the number 219024 is grouped into three
pairs of (21),(90) and (24).
The number 34567890 is grouped as (34,56,78,90).
If the number of digits in the number is odd then
the first group will have one digit and rest will have two digits.
For example the number 19024 is grouped in to three groups of (1),(90) and(24).
Similarly 3456789 is grouped as (3),(45),(67) and
(89).
1.5.1 Finding square root of whole
numbers:
1.5.1 Problem 1: Find square root of 219024 by division method
Solution :
Step1
: Group the pair of digits from right
side(unit place). The three groups are 21,90,24. Step2
: Find the Largest square number less than or equal to the first group (21). Since
5^{2}>21 and 4^{2}<21. 16 is the number. Step
3 : Take the square root of 16 = 4 Step
4: Place 4 as quotient above the first
group, Step
5 : Place 4
also as divisor Step
6 : Subtract from the first group (21), the product of divisor and quotient=16(=4*4)
:The reminder is 5(2116). Step
7 : Consider this remainder and the
second group (=590) as the new dividend. Step
8 : Add divisor and the digit in the unit place of divisor.(4+4= 8). Find a digit x such that 8x multiplied by itself(x) gives a
number < or = the new dividend (=590). We find that 86*6 = 516 which is
less than the dividend 590. Therefore x=6 . 86 will be the
new divisor. Write 6 above the second group at the top. Step
9 : Subtract 516, the product of this new divisor(86)
and 6
from the dividend 590  86 *6 =74 Step
10: Consider this remainder (74) and the next group (24) as the new dividend
(7424) Step
11: Add divisor and the digit in the unit place of divisor.( 86+ 6= 92).Find a digit x such that 92x
multiplied by itself(x) gives a number < or = the new dividend (=7424). We
find that 928*8 = 7424 which is equal to dividend. Therefore x=8.
Write 8
as quotient above the third group at the top. Step
12 : Subtract 7424, the product of this new divisor(928)
and the quotient 8 from the dividend 7424 928*8 =0 Step
13 : Continue this process till
there are no more groups for
division. We
stop the process here as there are no more groups remaining for division. 

_{}_{} = 468
Verification:
468 is of the form460+8 and we know the identity (a+b)^{2}=a^{2}+2ab+b^{2}
_{}468^{2}=460^{2}+2*460*8+8^{2}
= 211600+7360+64 = 219024
1.5.1 Problem 2: Find square root of
657721 by division method
Solution :
Step 
Divisor 
8 1 1 
Explanation 
2,3,5 
8 
_{} _{} _{} 
64<65<81 , _{} =8 
6 
+8 
64 
64=8*8 
7,8 
161 
1 77 
6564=1 16 =8+8 
9 
+1 
161 
161*1 =161 
10,11 
1621 
1621 
161+1 =162:
177161=16 
12 

1621 
1621*1 =1621 


0 

We stop the process here as there are no more
groups remaining for division.
_{}_{} = 811
Verification:
811 is of the form800+10+1 and we know the identity
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)
818^{2}= 800^{2}+10^{2}+1^{2}+2(800*10+10*1+800*1)
= 640000+100+1+2*(8000+10+800) = 640000+101+17620
=657721
1.5.1 Problem 3: Find square root of
49244 by division method
Solution :
Since the number has odd digits, there will be only
one number (4) in the first group
Step 
Divisor 
2 2 2 
Explanation 
2,3,5 
2 
_{} _{} _{} 
4=4<9 , _{} =2 
6 
+2 
4 
4=2*2 
7,8 
42 
0 92 
44=0 4 =2+2 
9 
+2 
84 
42*2 =84 
10,11 
442 
884 
42+2 =44:
9284 =8 
12 

884 
442*2 =884 


0 

We stop the process here as there are no more
groups remaining for division.
_{}_{} = 222
Verification:
222 is of the form200+20+2 and we know the identity
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)
222^{2}= 200^{2}+20^{2}+2^{2}+2(200*20+20*2+200*2)
= 40000+400+4+2*(4000+40+400) = 40000+404+8480
=49284
1.5.1 Problem 4 : A
Solution:
Since,
the gardener is left with 5 plants after arranging in rows. Number of plants
used for planting =64009
(=640145). Because
the plants are planted in rows of perfect square, we need to find the square
root of 64009. Since
64009 has odd number of digits, there will be only one number (6) in the
first group and other groups are (40) and (09).


_{}_{} = 253
Hence the gardener had arranged the roses in 253
rows with 253 plants in each row
Verfication:
253 is of the form250+3 and we know the identity (a+b)^{2}=a^{2}+2ab+b^{2}
253^{2}= (250+3)^{2}= (250)^{2}+2*250*3+(3)^{2}
= 62500+1500+9 = 64009
1.5.1 Problem 6 : Find the least number which must be added to 9215 to make it a perfect square.
Solution:
Step 
Divisor 
9 6 
Explanation 
2,3,5 
9 
_{} _{} 
81<92<100 , _{} =9 
6 
+9 
81 
81=9*9 
7,8 
186 
11 15 
9281=11 18 =9+9 

+6 
11 16 
185*5 =925,186*6
=1116 


1 
11151116 = 1 
We stop here as there are no more groups to be
considered.
We have seen that in case of perfect numbers, the
reminder in the last step has to be zero, which was not the case in the above
problem.
Since it was given that 9215 is less than the
nearest perfect square, in the last step we had to have a reminder.
9215+1 is a perfect square and the _{} = 96
Verfication:
Verify that 96^{2}= 9216
1.5.1 Problem 7: Find the least number which must be subtracted from 5084
to make it a perfect square.
Solution:
Step 
Divisor 
7 1 
Explanation 
2,3,5 
7 
_{} _{} 
49<50<64 , _{} =7 
6 
+7 
49 
81=9*9 
7,8 
141 
1 84 
9281=11 18 =9+9 

+1 
1 41 
141*1 =141,141*2 =282 


43 
184141=43 
We stop here as there are no more groups to be
considered.
We have seen that in case of perfect numbers, the
reminder in the last step has to be zero, which was not the case in the above
problem.
Since it was given that 5084 is larger than a perfect square, in the last step, we
had to have a reminder.
5041= 508443 is a perfect square and thus _{} = 71
Verfication:
Verify that 71^{2}= 5041
Observations:
Number 
Its
Square 
3(1
digit) 
9(1
digit) 
4(1
digit) 
16(2
digits) 
31(2
digits) 
961(3
digits) 
32(2
digits) 
1024(4
digits) 
316(3
digits) 
99856(5
digits) 
317(3
digits) 
100489(6
digits) 
3162(
4 digits) 
9998244(7
digits) 
3163(4
digits) 
10004569(8
digits) 
.3(1
place after decimal) 
.09(
2 places after decimal) 
.01(2
places after decimal) 
.0001(4
places after decimal) 
.001(3
places after decimal) 
.000001(
6 places after decimal) 
Conclusion: If a number has n digits, then
its square root will have n/2 digits if n is even and (n+1)/2 digits if n is
odd. If a number has n decimal places after it, then its square root will have
n/2 digits after decimal places.
1.5.2
Finding the square root of decimals:
We follow the same procedure that we followed in
the case of whole numbers. The main difference is the way we form groups. In
case of decimal numbers:
The grouping of whole numbers is done from the left
side in to groups of two numbers. In the case of numbers after the decimal
point, we form groups of two numbers to the right of the decimal number.
Ex: the grouping for 205.9225 is done as (2),
(05), (92), (25)
The division process is separately carried out for
whole numbers and decimal numbers.
1.5.2 Problem 1: Find square root of 235.3156 by
division method
Solution :
We group 2 and 35 as
groups for whole numbers
We group 31 and 56
as groups for decimal numbers
Step 
Divisor 
1 5. 3 4 
Explanation 
2,3,5 
1 
_{}_{}._{}_{} 
1<2<9 , _{} =2 
6 
+1 
1 
1=1*1 
7,8 
25 
1 35 
21=1 2 =1+1 
9 
+5 
1 25 
25*5 =125 
10,11 
303 
1031 
255+5 =30:
135125 =10.Put a decimal point at the top after
15 as we have started taking groups from decimal part 
12 
+3 
909 
303*3 =909 

3064 
122 56 
303+3=306 


122
56 
3064*4 =12256 


0 

_{}_{} = 15.34
Verfication:
Verify that 15.34^{2}= 235.3156
ALTERNATE METHOD:
First, follow the division method to arrive at _{}=1534. We know that 235.3156 = 2353156/10000
_{} _{} =_{} =_{}/_{} = _{}/100 = 1534/100 =
15.34
Finding square root of numbers which are not
perfect squares.
Any number x
is same as x.0000
( 5 is same as 5.0000 and 11 = 11.0000)
In order to find the square root of a non perfect
number we rewrite that number as the same number with four zeros after the
decimal point and we then follow division method to find the square root of
that number with four decimals.
1.5.2 Problem 2: Find the approximate length to 3 decimal places of the side of a square whose area is 12.0068
sq. meters
Since we are required to find the length to 3
decimal places, we need to rewrite the number with 6 decimals.
Solution :
12.0068 = 12.006800
The group for whole number is 12 and groups for decimal
parts are (00), (68),and
(00)
Step 
Divisor 
3 . 4 6 5 
Explanation 
2,3,5 
3 
_{}._{}_{}_{} 
9<12<16 , _{} =3 
6 
+3 
9 
3*3=9 
7,8 
64 
3 00 
129=3 6 =3+3 Put a decimal point
at the top after 3 as we have started taking groups from decimal part 
9 
+4 
2 56 
64*4 =256 
10,11 
686 
44 68 
64+4 =68:300256=44, take next group 
12 
+6 
41 16 
686*6 =4116 

6925 
3
52 00 
686+6=692 


3
46 25 
6925*5 =34625 


5 75 

We may continue this division method to the
required number of decimal points
_{}_{} is _{} 3.465 rounded to 3.47_{}
Verfication:
Verify that 4.465^{2}= 12.006225
You may notice that the value of _{} to 14 decimal places
is = 3.46410161513775
Since 12 is not a perfect square the decimals in _{} is never ending.
Note: if it is required to find the square root of rational
number, then convert the number in to a decimal number and follow the procedure
as given above(for example to find the square root of 11 _{}, convert this number to decimal as 11.6666 and then
follow the procedure as in 1.5.2 Problem 2)
Alternatively convert the rational number in to a
number free from radical sign in the denominator by suitable multiplication and
then calculate the square root of numerator as worked out earlier.
For example 11
_{} =35/3 = (35*3) ÷ (3*3)
= 105÷9
_{} _{} = _{} = (_{})/3
1.5 Summary of learning
No 
Points studied 
1 2 
Finding
square root by division method Finding
square root of decimals 