SQUARE ROOTS AND CUBE ROOTS
IMPORTANT FACTS AND FORMULAE
Square Root: If x2 = y, we say
that the square root of y is x and we write, √y = x.
Thus, √4 = 2, √9 = 3, √196 = 14.
Cube Root: The cube root of a given number x is the
number whose cube is x. We denote the cube root of x by 3√x.
Thus, 3√8
= 3√2 x 2 x 2 = 2, 3√343 = 3√7 x 7 x 7
= 7 etc.
Note:
1.√xy = √x * √y 2. √(x/y) = √x / √y = (√x / √y) * (√y / √y) = √xy / y
SOLVED EXAMPLES
Ex. 1. Evaluate √6084 by
factorization method .
Sol.
Method: Express the given number as the product of prime
factors. 2 6084
Now, take the product of these
prime factors choosing one out of 2 3042
every pair of the same primes. This
product gives the square root
3 1521
of the given number.
3 507
Thus, resolving 6084 into prime
factors, we get: 13 1696084 = 22 x 32
x 132 13 \ √6084
= (2 x 3 x 13) = 78. 
Ex.
2. Find the square root of 1471369.
Sol. Explanation: In the given number, mark off
the digits 1 1471369 (1213
in pairs starting from the unit's
digit. Each pair and
1
the remaining one digit is called
a period.
22 47
Now, 12 = 1. On
subtracting, we get 0 as remainder. 44
Now, bring down the next period i.e.,
47.
241 313
Now, trial divisor is 1 x 2 = 2
and trial dividend is 47.
241
So, we take 22 as divisor and put
2 as quotient.
2423 7269
The remainder is 3.
7269
Next, we bring down the next
period which is 13. x
Now, trial divisor is 12 x 2 = 24
and trial dividend is
313. So, we take 241 as dividend
and 1 as quotient.
The remainder is 72.
Bring down the next period i.e.,
69.
Now, the trial divisor is 121 x 2
= 242 and the trial
dividend is 7269. So, we take 3as
quotient and 2423
as divisor. The remainder is then
zero.
Hence, √1471369 = 1213.
Ex. 3. Evaluate: √248 + √51 + √ 169 . 
Sol. Given expression = √248 + √51 + 13 = √248
+ √64 = √ 248 + 8 = √256 =
16.Ex. 4. If a * b * c = √(a + 2)(b + 3)
/ (c + 1), find the value of 6 * 15 * 3.
Sol. 6 * 15 * 3 = √(6 + 2)(15 + 3) / (3 + 1) = √8 * 18 / 4 = √144 /
4 = 12 / 4 = 3. Ex. 5. Find the value of √25/16.
Sol. √ 25 / 16
= √ 25 / √ 16 = 5 / 4
Ex. 6. What is the square root of 0.0009?
Sol. √0.0009=
√ 9 / 1000 = 3 / 100 = 0.03.
Ex.
7. Evaluate √175.2976.
Sol. Method:
We make even number of decimal places 1 175.2976 (13.24
by affixing a zero, if necessary.
Now, we mark off 1
periods and extract the square
root as shown.
23 75
69
\√175.2976 = 13.24
262 629
524
2644 10576
10576
x
Ex. 8. What will come in
place of question mark in each of the following questions?
(i) √32.4 / ? = 2 (ii)
√86.49 + √ 5 + ( ? )2 = 12.3. Sol. (i) Let √32.4 / x = 2. Then,
32.4/x = 4 <=> 4x = 32.4 <=> x = 8.1.(ii) Let √86.49 + √5 + x2
= 12.3.
Then, 9.3 + √5+x2 = 12.3 <=> √5+x2 = 12.3 - 9.3 = 3
<=> 5 + x2 = 9
<=> x2 = 9 - 5= 4
<=> x = √4 = 2.
Ex.9. Find the value of √ 0.289
/ 0.00121. |
 |
 |
Sol. √0.289 / 0.00121 = √0.28900/0.00121 = √28900/121
= 170 / 11.
 |
Ex.10.
If √1 + (x / 144) = 13 / 12, the find the value of x.
Sol. √1 + (x / 144) = 13 / 12 Þ ( 1 + (x / 144)) = (13 /
12 )2 = 169 / 144
Þx /
144 = (169 / 144) - 1
Þx /
144 = 25/144 Þ
x = 25.
Ex. 11. Find the value of √3 up to three places of
decimal.
Sol.
1
3.000000 (1.732
1
27 200
189
343 1100
1029
3462 7100
6924 \√3 = 1.732.
Ex. 12. If √3 = 1.732, find the value of √192
- 1 √48 - √75 correct to 3 places
2
of decimal. (S.S.C.
2004)
Sol.
√192 - (1 / 2)√48 - √75
= √64 * 3 - (1/2) √ 16 * 3 - √ 25 * 3
=8√3
- (1/2) * 4√3 - 5√3
=3√3
- 2√3 = √3 = 1.732
 |
Ex. 13. Evaluate: √(9.5
* 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)
Sol. Given exp. = √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)
Now,
since the sum of decimal places in the numerator and denominator under the radical sign is the same, we remove
the decimal.
\ Given exp = √(95 * 85 * 18900) / (17 *
19 * 21) = √ 5 * 5 * 900 = 5 * 30 = 150.Ex. 14. Simplify: √ [(
12.1 )2 - (8.1)2] / [(0.25)2 +
(0.25)(19.95)]
Sol. Given exp. = √ [(12.1 + 8.1)(12.1 - 8.1)] / [(0.25)(0.25 +
19.95)]
=√
(20.2 * 4) /( 0.25 * 20.2) = √ 4 / 0.25
= √400 / 25 = √16 = 4.
Ex. 15. If x = 1 + √2 and y = 1 - √2,
find the value of (x2 + y2).
Sol. x2
+ y2 = (1 + √2)2 + (1 - √2)2 =
2[(1)2 + (√2)2] = 2 * 3 = 6.
Ex. 16. Evaluate: √0.9 up to 3 places of decimal.
Sol.
9 0.900000(0.948
81
184 900
736
1888 16400
15104 \√0.9
= 0.948
Ex.17. If √15 = 3.88, find the value of √ (5/3).
Sol.
√
(5/3) = √(5 * 3) / (3 * 3) = √15 / 3 =
3.88 / 3 = 1.2933…. = 1.293.
√
(5/3) = √(5 * 3) / (3 * 3) = √15 / 3 =
3.88 / 3 = 1.2933…. = 1.293.
Ex. 18. Find
the least square number which is exactly divisible by 10,12,15 and 18.
Sol. L.C.M.
of 10, 12, 15, 18 = 180. Now, 180 = 2 * 2 * 3 * 3 *5 = 22 * 32 *
5.
To make
it a perfect square, it must be multiplied by 5.
\ Required
number = (22 * 32 * 52) = 900.
Ex. 19. Find the greatest number of five digits which is a
perfect square.
(R.R.B. 1998)
Sol. Greatest
number of 5 digits is 99999.
3
99999(316
9
61 99
61
626 3899
3756
143
\
Required number == (99999 - 143) = 99856.
Ex. 20. Find the smallest number
that must be added to 1780 to make it a perfect
square.
Sol.
4 1780 (42
16
82 180
164
16 |
\ Number to be added = (43)2
- 1780 = 1849 - 1780 = 69.
Ex. 21. √2 =
1.4142, find the value of √2 / (2 + √2).
Sol. √2 /
(2 + √2) = √2 / (2 + √2) * (2 - √2) / (2 - √2) = (2√2 – 2) / (4 – 2)
= 2(√2 – 1) / 2 = √2 – 1 = 0.4142.
22. If x = (√5 + √3) / (√5 - √3) and y = (√5 - √3) /
(√5 + √3), find the value of (x2 +
y2).
Sol.
x = [(√5 + √3) / (√5 - √3)] * [(√5
+ √3) / (√5 + √3)] = (√5 + √3)2 / (5 - 3)
=(5 + 3 + 2√15) / 2 = 4 + √15.
y =
[(√5 - √3) / (√5 + √3)] * [(√5 - √3) / (√5 - √3)] = (√5 - √3)2 / (5
- 3)
=(5 + 3 - 2√15) / 2 = 4 - √15.
\ x2 + y2 = (4 + √15)2
+ (4 - √15)2 = 2[(4)2 + (√15)2] = 2 * 31
= 62.
Ex. 23. Find the cube root of 2744.
Sol. Method: Resolve the given number as the
product 2 2744 of prime factors and take the product
of prime 2 1372
factors, choosing one out of three of the same 2 686
prime
factors. Resolving 2744 as the product of 7 343 prime factors, we get:
7 49
7
2744 = 23 x 73.
\
3√2744= 2 x 7 = 14.
Ex. 24. By what least number 4320 be multiplied to obtain
a number which is a perfect cube?
Sol. Clearly, 4320 = 23 * 33 * 22 *
5.
To
make it a perfect cube, it must be multiplied by 2 * 52 i.e,50.
e
�7o s �� P� es'> =
32000 : y Þ y =
15 * 32000 =30000
16
4.(b):
Population of village Y in 1997 = 30000(given) .
Let the population village V in 1997
be v.
Then, 15:10 = 30000:v Þ v =
30000 * 10 = 20000.
15
Now population of village V in 1998
= 20000 + (10% 0f 20000) = 20000.
\Population of village V
below poverty line in 1998 = 58% of 22000 = 12760.
5.(a) :
Let the total population of all the seven villages in 1997 be N.
Then,population of village R in 1997
= 16% of N = 16/ 100 N
And population of village Z in 1997
= 11% of N =11/100 N
\Population of village R in
1999 = {16/100N+(10%
of 16/100 N)}=1760/10000 N
and population of village Z in 1999
= {11/100 N-(5% of 11/100 N)} = 1045/10000 N.
Now,population of village R below
poverty line for 1999 = 51% of(1760/10000 N)
And population of village Z below
poverty line 1999 = 42% of (1045/10000 n)
51% of (1760/10000 N) 51 *
1760 2
Required ratio = -------------------------- = ------------- =
------
42%
of (1045/10000 N) 42 * 1045 1.
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