Seuanr Roor
Introduction
Thesquarerootofanumberaisthatnumberwhichwhenmultiplied
by itself, gives a as the Product'
In other words, if x is the square root of y then'
x2= y
i.e. * = .,F or
(v)4
in our secondary
You must have learnt two methods commonly taught
number'
any
of
school syllabus, to find the square root
They are (a) Factor Method
(b) Division Method
Boththesemethodsaretime-consumingandboring.VedicMathematics
providesseveralVedicSutrastofindthesquarerootofanygivennumber.
^some
not
of these vedic Sutra which we shall discuss in this chapter ale
energy' Besides
only easy to understand, but save our precious time and
shall also learn
we
Veiic Sutras to find the square root of any number'
similar to
quite
Arvabhatta's method of finding square root' which is
Dwanda-yogu (
6q frt ) v"di" tuttu'
t23
o
Vedic Sutra for Square Root
The best method taught in our school syllabus for finding the square
root is Long Division Method. Here we shall learn the best vedic Method
called Dwanda-Yoga (Fq +tr ) or Duplex Method to find the square roor
f'he termination of an exact square in 9 is possible, if and only if
square root ends in 3 or 7.
Aryabhatta's Method of Finding Square Root
qm
of a number just by small mathematical calculation.
(fr
The literal meaning of Dwanda-yoga
*.r) nas been discussed
Chapter-6 (Square), please refer to it for more information.
g,
i'
it can't be a perfect
square root.
If
a number ends
in an odd number of zeros, it can't be a perfect
square root.
Unit digit of a square root
Unit digit of a square
0
0
I
I or9
4
2or8
5
5
6
4or6
3or7
9
The above table shows:
r A complete square ending in I must have either 1 or 9, mutual
cotrplements from 10, as the last digit of its square root.
A square can end in 4, only if the square root enels in 2 or g, a
complement from 10.
o If a square ends with 5, its square root
too ends with 5.
o A square ending in 6 must have 4 or
6, a complement of 10, as
the last digit in its square root.
t24
fdgdn
qriq*t
q{t E+ yG ddi rErrrRt
Meaning of Vedic Sutra
Some Important Points to Remember:
If the unit digit of any number is 2, 3,'/, or
arMfu
qsrTt
I
I
This is the lburth sloka of Ganita Pada, written by the great Indian
astronomer and mathematician Aryabhatta born in 4j6 AD. The above
mentioned fourth sloka describes the method of finding the square root and
consists of the following steps:
o
o
Subtract the greatest possible square from the last odd place.
Always divide the even place by twice the square root of the
preceding odd place.
r
Subtract from the odd place (standing on the right) the square of
the quotient.
o
Repeat the process as long as there are
still digits on the right.
This method is illustrated with the help of some examples which are
self-explanatory in nilture.
Example: Find the squale root of 529.
Solution:
o First form the group of numbers. Mark the even places by
horizontal lines and odd places by vertical lines.
t_
I
52
o
o
Since
is 2.
9
4<5<9
or2.u[.3,
Subtracting the square
so that the integral part
of the root.
L,[b
-4
125
ofn[
Divide
by
2x 2 =
o
+f-]l| I
-- | 2'
9
Subtracting the square
ofquotient.
-
Since 9
<
"./lT it
3.
11
<
16 or 3
Subtracting the square
32
x
of quotient.
As no more places are left on the right, the
The required square root of 529 is 23.
Example: Find the square root of 1225.
of the root
t-l-
tt971
Subtracting the square
-2
process ends.
a2
e1
of quotient
.-42
Divide by 2x
Subtracting the square
Solution:
o
First form the group of numbers. Mark the even places by
horizontal lines and odd places by vertical lines.
Divide bv
'
3
2x34=68
t_l
t225
o
Since
is
o
9 < 12 < 16 or 3
.
^!12
of the root
,l'
-_9_
Divide bY 2x':'J-]-f-l
'
--30
the square ofquotient Z
S
5:
Hence the square robt
of
=d- 2 4 V_
57
411
40 8
X
1225
is
35
Example: Find the square root of lt97l6.
Solution:
o First form the group of numbers. Mark the even places by
horizontal lines and odd places by vertical lines.
t-t-r
tr9716
-62
Subtracting the square ofquotient
X
left on the right, the process
The required square root of 119716 is 346.
Example: Find the square root of 738915489
As no more places are
rz2sh
Subtracting
Ja
36
.rfiT < 4, so the integral part of
3.
Subtracting the square
< J:.jL.4, so the integral pa
ends.
Solution: Placing the horizontal lines on even places and vertical
on the odd places, we have
t-t-tl-l
738915489
The integral part of Jj i" Zas22 <7 <32.
Subtract the square
of the integral part
738915489
-4
Divide
by
2x2 = 4
JJ
28
5
58
Subtract the square ofthe quotient
t26
727
-2
Divide by
2x27
o
=rr]-=r -tr
-._ll
Working Frocedure of Duplex Method
451
Subtract the square ofthe quotient - f
Divideby
And conversely, if the given number has n digits, the square root
will contain nl 2 or (n + 1)12 digits.
The working of Duplex method is as simple as the straight division.
The Vedic Method oi l-inding the square root is a treatment over Aryabhatta's
method of extracting the square root.
2*ztt=ffi
Let us take a few exarnples to understand the modus operandi of this
quotient 82
Divide by 2*2jtB = 5436
1630
-16308
method.
Subtract the square ofthe
l
rl
i
Subtract the square ofthe quotient
i
-J
i
Example: Find the square root of 529 by Duplex rnethod.
8
Solution:
9
;2
o
e
Group the number S Z
9 bv
placing a bar over it.
Put horizontal and vertical lines as shown below.
X
l
Hence the required square root
=
Important Point to Note : When divided by 4 we avoid to take 8
as quotient, as at the next stage we will get 18 - 82 = - 46, a negative
l
ti
7183
< 5 < 32 , so the first digit of square root in quotient
column is 2. Double the quotient and set down as divisor
Since
2?
number.
I
Vedic Method for Square ftoot
I
'
Renrainde:r=-5-2?=l
The best and simplest Vedic Method for finding the square root is
Dwanda-Yoga or Duplex Method.
Fut this rernainrler bc'klw thc ncxt cliviclend digit. Hence the next
gross dividend = 12
Fundamental rules for extraction of square root of any number
The given number is first arranged in two-digit groups from right
to left and a single digit if any left over at the left hand end is
counted as a simple group by itself.
I
l
I
;
I
The number of digits in the square root will be the same as the
number of digit-groups in the given number itself including a
single digit if any such there be.
lf a square root contains n digits, the square must consist of 2n
or2n-1digits.
o
Divide 12 by 4 and put the quotient 3 in quotient column.
12=4=3
o
Next dividencl = 9
Subtract the square of quotient fiom the next dividend.
Netdividend=9-32-0
Since no more digits are left, so the square root
129
128
of
529
is
23.
Example: Find the square root
of
Since 12 < 2 < 22, so the first digit of square root in quotient
column is 1. Double the quotient and set down as divisor.
1225'
Solution:
i
zs ,
by placing a bar over it.
a
Group the number
a
Put horizontal and vertical lines as shown below.
Remainder-2-Iz=I.
Put this remainder below the next dividend digit. Hence the next
gross dividend = 10
< 12 < 42, so the first digit of square root in the quotient
column is 3. Double the quotient and set down as divisor'
Since
32
r
Remainder
2
l0+2=5,Q=5andR=0
=12-32=3
Since remainder cannot be taken as 0 until the whole operation
completes, hence we need to take the quotient less than 5. This is
because we callnot consider the remainder zero in the middle of
the next
Put this remainder below the next dividend digit' Hence
gross dividend = 32.
6lr2l
Divide 10 by
325
division.
aF]--
= 4 and Revised Remainder =
Hence the Revised Quotient
and
Divide 32 by 6 and put the quotient 5 in the quotient column
the remainder 2 below 5.
2
Next dividend = 2J
.
Next dividend = 25
Subtract the square of quotient from the next dividend'
Net dividend = 25 - 52 = 0
t225 \s
Since no more digits are left, so the square root of
2 U 36, by placing
a bar over it'
Put horizontal and vertical lines as shown below'
130
'
p 27
6
45 '3
Solution:
Group the number
35
of quotient lrorn the next dividend.
Net dividend = 27 - 42 = ll
Divide ll by 2 and write the quotient (Q
(R = 1) at their respective place.
212l
Example: Find the square root of 20736
o
o
Subtract the square
Gross dividend
=
13
- (Duplex of 45)
=13- 2x4x5
=13*40=-27<0
Net dividend
=
13
131
= 5) and remainder
(Since Net dividend is less than zero, hence we can't take the quotient
(Q = 5) as taken above. Now for
lI+
Put this remainder below the next dividend digit. Hence the n
gross dividend = 29
2, Revised Q = 4 and Revised
R=3.)
zlzlp
Qlll4
Gross dividend
=
336
4
Divide 29 by 6 and put the quotient 4 in the quotient column
=33-2x4x4
i
remainder below 7.
33
Net dividend = 33
=33-JL
27
(Duplex
of
44)
a^
Next dividend = 57
-1
of quotient from the next dividend.
Net dividend = 51 - 42 = 57 - 16 -- 4l
Divide 4l by 6 and write the quotient (6) and remainder
Subtract the square
Divide Lby 2 and put the quotient Q = 0 and remainder R =
in its proper place.
1
(5.
their respective place.
&ft#'144
.
Next gross dividend = 16
Net dividend = 16 - Duplex of 4
Gross dividend
=16-42 -0
Net dividend = 51
Since no more digits are left, so the square root of 20736
Example: Find the square root of 119716 by Vedic method'
is
51
- (Duplex of 46)
=51 - 2x4x6=3
144
Divide it by 6 and put quotient (0) and remainder (3) in its prc
place.
Solution:
o
o
=
Group the number
n W G, by placing a bar over it'
Put horizontal and vertical line as shown below'
o
Since 32 < 11< 42, so the first digit of square root in quotient
column is 3. Double the quotient and set it down as divisor'
Example: Find the square root of 6990632I
Solution:
o
Remainder=11
-32=2
132
Next gross dividend = 36
Net dividend = 36 - duplex of 6 = 36 - 62 = O
Since no more digits are left, so the square root is 346
Put horizontal and vertical lines as shown below.
II
Since 82
<
69 <
So first digit
of
Divide 0 by
92
square root
16
Q=0R=0
=8
Divisor = 2 x 8 = 16 Remainder = 69
Next gross dividend = 59
-
82=
16l69 l s9 110 s6
5
al8l
43 12 ol
361.000
Since no more digits are left so the square root is
Example: Find the square root of 389376
8361.
Solution:
Divide 59 by 16 and set quotient and remainder in their respective
columns. The Q = 3 and R
16 169 | 59
o
=
11
n06321
The one line Vedic solution has the following steps:
Next gross dividend = 110
Net dividend = 110 - 32 = 110 - 9 = 101
Divide 101 by 16 and place Q = 6 and R = 5 at proper place.
16l69l s9 rO s632r
Next gross dividend = 56
Net dividend = 56 - Duplex of 36
=56-2x3x6=56-36=20
Divide 20 by t6 and write Q = 1 and R = 4 at proper place.
.
r
n
'
r
.
'
.
16l69l 59 n0 s6 432r
36t
First digit = 6
Remainder = 38 - 62 = 2. Write 2 below 9.
New dividend = 29 and Divisor = 2 x 6 = 12
For 29 + 12, Q = 2 and R = 5, write quotient(Q) and remain
at the proper place.
New dividend = 53 and Corrected dividend = 53 - 22 = '
For 49+12, Q=4andR= 1, writeQandRatitsproper
So far we have got three digits in the quotient column, her
next digit in the quotient column will be the digit placer
decimal.
Next dividend = 43
Net dividend = 43 - Duplex of 361
=43-(2x3x1+62)
=43-42=l
Divide 1 by
Since there are three groups in the above square root, hen
square root consists of 3 digits.
.
r
I
New dividend = 17 and remainder = l7 - 2 x 4 x 2 =
New dividend = 16 and the last remainder = 16 - 42 = C
Square Root = 624
Example: Find the square root of 41254929.
1
Solution:
16
Q=0and R=1
Next dividend = t2
Net dividend = 12 -
t2l4tl
Duplex of
=12-2x6xI=0
134
al6l
61
s2 45 s4 29 t2 09
423.000
The step by step illustration is done here for your convenience.
r Number of digits in the square root = 4
o The first digit of square root = 6
r Remainder = 4l - 62 - 5, place it below 2.
o Next dividend = 52 and divisor = 2 x 6 = 12
o For 52 + 12, Q = 4 and R = 4, place it at proper place.
o Next dividend = 4-5 tmd corrected dividend = 45 - 42 = 29
o For29+12,Q=2andR=5
o Nextdividend= 54 and corrected dividend = 54-2x2x4=
Solution: Perfect square approaching 47 is 49'
Deviation=4"1 -49=-2
2
Jn = J4e- 2xr/49
-
1
I
-' 1 ="/'A.1428= 6.8512
Exarnple: Find the square rctot of 1'74.
Solution: Perfect square approaching 114 is 169
Deviation
=
Jll+
= Vl69*-
l'74
-
169
=
5
38
o For38+12,Q=3andR=2
o
o
e
o
Since we have so far got 4 digits in the quotient column, hence
the perfect square root is obtained. The next digit in the quotient
column will give the remainder, if any.
Remainder
= 29 - 4 x 3 x 2 +
22
=
I
= 72 andrernainder = 12 - 2 x 3 x 2 = O
Next dividend = 9 and the last remainder = 9 - 32 = 0
Hence square root = 6423
Next dividend
Square Root
=
13+
L=13.192
Exarnple: Find the square root of 147.
Solution: Perf'ect square approaching 141 is 144.
Deviation
=147
[4i
-
-I44=3
.1144*]*lA
I
Many a times, while solving the problems of Mensuration, you must
have encountered the need to find the squal'e root of inational number. This
method presented below will help yon to find the square root of any
irrational number in a few moments. and once you are in running practice.
you can find the square root of such irrational number mentally.
=
Deviation fmm nearrcst Perfect square
Z x ./Nearrcst
Example: Find the square root of 47.
i_:
2x.l169
of Irrational Nurnber
Square Root of Irrational Number
-5
fertect squ:re
= l2+!=12.125
8
Example: Find the square root of 187.
Solution: Perfect square approachirrg 187 is
-196=-9
Deviation
=181
Jl8?
= Jlt6 zxJrx
A-2*=14-
0.32
28
13.6"1
136
t37
196.
Exnnctsr non PnacrlcE
Exercise
7.1,
FindthesquareRootofthefollowingbyAryabhatta'sMethod.
(e) 4225
(e) s9049
(b) 10329796
(d) 2116
(f) 688e
(h) t2s3r6
(i)
c)
(a)
4413225
(c)
14047504
t69744
CueE
r26s62s
Exercise 7.2
Introduction
Find the Square Root by Duplex Method.
(b) e3r22s
(a) 99920016
(c)
4s369
(d)
(e)
(e)
78978769
3249
(f) 453t9824
(h) ss2049
6990632t
0)
(i)
number
When a number is multiplied by itself three times' the
obtained is called the cube of that number'
79032r
1xlx1=13=1
2xZx2=23=8
3x3x3=33=2J
4x4x4=43=64
5x5x5=53=125
6x6x6=63=216
7 x7 x] ='13 =343
8x8x8=83=512
9x9x9=93=729
4t2s4929
Exercise 7.3
Find the Square Root of following krationals'
(b) t64
(a) 37
(c)
(d)
r75
(e) 23t
(e) 6e
(i)
43
(o
1e8
(h)
287
so
Ingeneral,axaXl=o3
63t
Cubesoflargernumbersarerarelyusedinourdailycalculations.We
sometimesneedthecubevalueoftwo-digitnumbersandgenerallymultiply
tr3tr
plesents several very interesting
the number three times. Vedic Mathematics
methodstofindthecubesofanygivennumberinafewseconds.Allthe
to practice'
methods discussed here are easy to understand and easy
139
138
,l