Chapter 5 Squares and Square Roots solutions

Exercise 5.1

Q1. What will be the unit digit of the squares of the following numbers?
(i) 81 
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Ans: (i) Since, 1 × 1 = 1
∴ The unit’s digit of (81)² will be 1.
(ii) Since, 2 × 2 = 4
∴ The unit’s digits of (272)² will be 4.
(iii) Since, 9 × 9 = 81
∴ The unit’s digit of (799)² will be 1.
(iv) Since, 3 × 3 = 9
∴ The unit’s digit of (3853)² will be 9.
(v) Since, 4 × 4 = 16
∴ The unit’s digit of (1234)² will be 6.
(vi) Since 7 × 7 = 49
∴ The unit’s digit of (26387)² will be 9.
(vii) Since, 8 × 8 = 64
∴ The unit’s digit of (52698)² will be 4.
(viii) Since 0 × 0 = 0
∴ The unit’s digit of (99880)² will be 0.
(ix) Since 6 × 6 = 36
∴ The unit’s digit of (12796)² will be 6.
(x) Since, 5 × 5 = 25
∴ The unit’s digit of (55555)² will be 5.

Q2. The following numbers are obviously not perfect squares. Give reason.
(i) 1057 
(ii) 23453 
(iii) 7928 
(iv) 222222
(v) 64000 
(vi) 89722 
(vii) 222000 
(viii) 505050
Ans: 
(i) 1057
Since, the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9)
∴ 1057 is not a perfect square.
(ii) 23453
Since, the ending digit is 3 (which is not one of 0, 1, 4, 5, 6 or 9).
∴ 23453 is not a perfect square.
(iii) 7928
Since, the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9).
∴ 7928 is not a perfect square.
(iv) 222222
Since, the ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).
∴ 222222 is not a perfect square.
(v) 64000
Since the number of zeros is odd.
∴ 64000 is not a perfect square.
(vi) 89722
Since, the ending digits is 2 (which is not one of 0, 1, 4, 5, 6 or 9).
∴ 89722 is not a perfect square.
(viii) 222000
Since the number of zeros is odd.
∴  222000 is not a perfect square.
(viii) 505050
The unit’s digit is odd zero.
∴ 505050 can not be a perfect square.

Q3. The squares of which of the following would be odd numbers?
(i) 431 
(ii) 2826 
(iii) 7779 
(iv) 82004
Ans: Since the square of an odd natural number is odd and that of an even number is an even number.
(i) The square of 431 is an odd number.
     [∵ 431 is an odd number.]
(ii) The square of 2826 is an even number.
      [∵ 2826 is an even number.]
(iii) The square of 7779 is an odd number.
       [∵ 7779 is an odd number.]
(iv) The square of 82004 is an even number.
       [∵ 82004 is an even number.]

Q4. Observe the following pattern and find the missing digits.
11² = 1 2 1
101² = 10201
1001² = 1002001
100001² = 1 … 2 …1
10000001² = …
Ans: We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.
∴ (100001)² = 10000200001
(10000001)² = 100000020000001

Q5. Observe the following pattern and supply the missing number.
11² = 1 2 1
101² = 1 0 2 0 1
10101² = 102030201
1010101² = ………….
………….² = 10203040504030201
Ans: Observing the above, we have:
(i) (1010101)² = 1020304030201
(ii) 10203040504030201 = (101010101)²

Q6. Using the given pattern, find the missing numbers.
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + __² = 21²
5² + __² + 30² = 31²
6² + 7² + __² = __²

Note: To find pattern:
Third number is related to first and second number. How?
Fourth number is related to third number. How?

Ans: Given, 1² + 2² + 2² = 3²
i.e 1² + 2² + (1×2 )² = ( 1² + 2² -1 × 2 )²
2² + 3² + 6² =7²
∴ 2² + 3² + (2×3 )² = (2² + 3² -2 × 3)²
3² + 4² + 12² = 13²
∴ 3² + 4² + (3×4 )² = (3² + 4² – 3 × 4)²
4² + 5² + (4×5 )² = (4² + 5² – 4 × 5)²
∴ 4² + 5² + 20² = 21²
5² + 6² + (5×6 )² = (5²+ 6² – 5 × 6)²
∴ 5² + 6² + 30² = 31²
6² + 7² + (6×7 )² = (6² + 7² – 6 × 7)²
∴ 6² + 7² + 42² = 43²

Q7. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Ans:
(i) The sum of first 5 odd numbers = 5²  = 25
(ii) The sum of first 10 odd numbers = 10²  = 100
(iii) The sum of first 12 odd numbers = 12² = 144

Q8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Ans:
(i) 49 =7² = Sum of first 7 odd numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = 11² = Sum of first 11 odd numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Q9. How many numbers lie between squares of the following numbers?
(i) 12 and 13    
(ii) 25 and 26     
(iii) 99 and 100
Solution: Since between n² and (n + 1)², there are 2n non-square numbers.
(i) Between 12² and 13², there are 2 *12, i.e. 24 numbers.
(ii) Between 25² and 26², there are 2 * 25, i.e. 50 numbers. 
(iii) Between 99² and 100², there are 2 * 99, i.e. 198 numbers.

Deleted Questions from NCERT

Q.1. Find the perfect square numbers between 
(i) 30 and 40
(ii) 50 and 60.
Solution.
(i) Since,
⇒ 1 *1 = 1  
⇒ 2 * 2 = 4  
⇒ 3 * 3 = 9
⇒ 4 * 4 = 16  
⇒ 5 * 5 = 25  
⇒ 6 * 6 = 36
⇒ 7 * 7 = 49
Thus, 36 is a perfect square number between 30 and 40.

(ii) Since, 7 * 7 = 49 and 8 * 8 = 64.
It means there is no perfect number between 49 and 64, and thus there is no perfect number between 50 and 60.

Q.2. Can we say whether the following numbers are perfect squares? How do we know?

(i) 1057 
(ii) 23453 
(iii) 7928 
(iv) 222222 
(v) 1069 
(vi) 2061

Write five numbers which you can decide by looking at their one’s digit that they are not square numbers.
Solution.
(i) 1057

• The ending digit is 7 (which is not one of 0, 1, 4, 5, 6, or 9)
• 1057 cannot be a square number.

(ii) 23453

•  The ending digit is 3 (which is not one of 0, 1, 4, 5, 6, and 9).
•  23453 cannot be a square number.

(iii) 7928

• The ending digit is 8 (which is not one of 0, 1, 4, 5, 6 and 9).
• 7928 cannot be a square number.

(iv) 222222

• The ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).
• 222222 cannot be a square number.

(v) 1069

• The ending digit is 9.
• It may or may not be a square number.

Also,

• 30 * 30 = 900
• 31 * 31 = 961
• 32 * 32 = 1024
• 33 * 33 = 1089

i.e. No natural number between 1024 and 1089 is a square number.

∴ 1069 cannot be a square number.

(vi) 2061

• The ending digit is 1
• It may or may not be a square number.

 45 * 45 = 2025

and 46 * 46 = 2116

i.e. No natural number between 2025 and 2116 is a square number.

∴ 2061 is not a square number.

We can write many numbers which do not end with 0, 1, 4, 5, 6 or 9. (i.e. which are not square number).
Five such numbers can be:

1234, 4312, 5678, 87543, 1002007.

Q.3. Write five numbers that you cannot decide just by looking at their unit’s digit (or one’s place) whether they are square numbers or not.
Solution. Any natural number ending in 0, 1, 4, 5, 6 or 9 can be or cannot be a square-number.
Five such numbers are:

56790, 3671, 2454, 76555, 69209

Property 1. If a number has 1 or 9 in the unit’s place, then its square ends in 1.
Example: (1)² = 1, (9)² = 81, (11)² = 121, (19)² = 361, (21)² = 441.

Q.4. Which of 123², 77², 82², 161², 109² would end with digit 1?
Solution. The squares of those numbers end in 1 which ends in either 1 or 9.

∴ The squares of 161 and 109 would end in 1.

Property 2. When a square number ends in 6, then the number whose square it will have 4 or 6 in its unit place.

Q.5. Which of the following numbers would have digit 6 at unit place.
(i) 19² 
(ii) 24² 
(iii) 26² 
(iv) 36² 
(v) 34²
Solution.
(i) 19²: Unit’s place digit = 9

∴ 19² would not have unit’s digit as 6.

(ii) 24²: Unit’s place digit = 4

∴ 24² would have unit’s digit as 6.

(iii) 26²: Unit’s place digit = 6

∴ 26² would have 6 as unit’s place.

(iv) 36²: Unit place digit = 6

∴ 36² would end in 6.

(v) 34²: Since the unit place digit is 4

∴ 34² would have unit place digit as 6.

Q.6. What will be the “one’s digit” in the square of the following numbers?

(i) 1234 
(ii) 26387 
(iii) 52698 
(iv) 99880 
(v) 21222 
(vi) 9106
Solution.
(i) Since Ending digit = 4 and 4² = 16

∴ (1234)² will have 6 as the one’s digit.

(ii) Since Ending digit is 7 and 7² = 49

∴  (26387)² will have 9 as the one’s digit.

(iii) Since Ending digit is 8, and 8² = 64

∴ (52692)² will end in 4.

(iv) Since Ending digit is 0.

∴ (99880)² will end in 0.

(v) Since Ending digit = 2 and 2² = 4

∴ Ending digit of (21222)² is 4.

(vi) ∵ 6² = 36

∴ Ending digit of (9106)² is 6.

Property 3. A square number can only have even number of zeros at the end.

Property 4. The squares of odd numbers are odd and the squares of even numbers are even.

Q.7. The square of which of the following numbers would be an odd number/an even number? Why?

(i) 727 
(ii) 158 
(iii) 269 
(iv) 1980

Solution.

(i) 727 

Since 727 is an odd number.

∴ Its square is also an odd number.

(ii) 158

Since 158 is an even number.

∴ Its square is also an even number.

(iii) 269

Since 269 is an odd number.

∴ Its square is also an odd number.

(iv) 1980

Since 1980 is an even number.

∴ Its square is also an even number.

Q.8. What will be the number of zeros in the square of the following numbers?

(i) 60 
(ii) 400

Solution. 

(i) In 60, the number of zero is 1.

∴ Its square will have 2 zeros.

(ii) Since there are 2 zeroes in 400.

∴ Its square will have 4 zeros.

Property 5. The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.

Property 6. There are 2n non-perfect square numbers between the squares of the numbers n and n + 1.

Q.9. How many natural numbers lie between 9² and 10²? Between 11² and 12²?

Solution.

(a) Between 9² and 10²

Here, n = 9 and n + 1 = 10

∴ Natural number between 9² and 10² are (2 * n) or 2 * 9, i.e. 18.

(b) Between 11² and 12²

Here, n = 11 and n + 1 = 12

∴ Natural numbers between 11² and 12² are (2 * n) or 2 *11, i.e. 22.

Q.10. How many non-square numbers lie between the following pairs of numbers:

(i) 100² and 101²       
(ii) 90² and 91²      
(iii) 1000² and 1001²

Solution. 
(i) Between 100² and 101²

Here, n = 100
∴ n * 2 = 100 * 2 = 200

∴ 200 non-square numbers lie between 1002 and 1012.

(ii) Between 90² and 91²

Here, n = 90
∴ 2 * n = 2 * 90 or 180

∴ 180 non-square numbers lie between 90 and 91.

(iii) Between 1000² and 1001²

Here, n = 1000
∴ 2 * n = 2 * 1000 or 2000

∴  2000 non-square numbers lie between 1000² and 1001².

Property 7. The sum of first n odd natural numbers is n².
OR
If there is a square number, it has to be the sum of the successive odd numbers starting from 1.

Q.11. Find whether each of the following numbers is a perfect square or not?

(i) 121  
(ii) 55  
(iii) 81  
(iv) 49  
(v)  69

Solution. Remember: If a natural number cannot be expressed as a sum of successive odd natural numbers starting from 1, then it is not a perfect square.

(i) 121

Since, 

   121 – 1 = 120
   120 – 3 = 117
   117 – 5 = 112
   112 – 7 = 105
   105 – 9 = 96
    96 – 11 = 85

    85 – 13 = 72
    72 – 15 = 57
    57 – 17 = 40
    40 – 19 = 21
    21 – 21 = 0

i.e. 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21. Thus, 121 is a perfect square.

(ii) 55

Since, 

    55 – 1 = 54     

    54 – 3 = 51

    51 – 5 = 46

    46 – 7 = 39

    39 – 9 = 30

    30 – 11 = 19

    19 – 13 = 6        
     6 – 15 = –9    

Since, 55 cannot be expressed as the sum of successive odd numbers starting from 1. Thus, 55 is not a perfect square.

(iii) 81

Since,
 81 – 1 = 80
 80 – 3 = 77
 77 – 5 = 72
 72 – 7 = 65
 65 – 9 = 56
 56 – 11 = 45
 45 – 13 = 32
 32 – 15 = 17
 17 – 17 = 0

∴ 81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17. Thus, 81 is a perfect square.

(iv) 49

Since, 

   49 – 1 = 48

   48 – 3 = 45

   45 – 5 = 40

   40 – 7 = 33

   33 – 9 = 24

   24 – 11 = 13

   13 – 13 = 0

∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13. Thus, 49 is a perfect square.

(v) 69

Since,
    69 – 1 = 68
    68 – 3 = 65
    65 – 5 = 60
    60 – 7 = 53
    53 – 9 = 44
    44 – 11 = 33
    33 – 13 = 20
    20 – 15 = 5
     5 – 17 = –12

∴ 69 cannot be expressed as the sum of consecutive odd numbers starting from 1. Thus, 69 is not a perfect square.

Property 8. The square of an odd number can be expressed as the sum of two consecutive natural numbers.

Try These
Q1. Express the following as the sum of two consecutive integers.
(i) 21² 
(ii) 13² 
(iii) 11² 
(iv) 19²
Solution: 
(i) n = 21

or  21² = 220 + 221 = 441

(ii) n = 13

∴ 132 = 85 + 84 = 169

Similarly,

(iii) 11² = 60 + 61 = 121

(iv) 19² = 180 + 181 = 361

Q2. Do you think the reverse is also true, i.e. is the sum of any two consecutive positive integers is perfect square of a number? Give example to support your answer.
Solution: No, it is not always true.
Example:
(i) 5 + 6 = 11, 11 is not a perfect square.
(ii) 21 + 22 = 43, 43 is not a perfect square.

Property 10. The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.

Examples:

9² – 8² = 81 – 64 = 17 = 9 + 8

10² – 9² = 100 – 81 = 19 = 10 + 9

15² – 14² = 225 – 196 = 29 = 15 + 14

1012 – 1002 = 10201 – 10000 = 201 = 101 + 100

Property 11. If (n + 1) and (n – 1) are two consecutive even or odd natural numbers, then (n + 1) X  (n – 1) = n² – 1.

Example:

10 * 12 = (11 – 1) * (11 + 1) = 11² – 1

11 * 13 = (12 – 1) * (12 + 1) = 12² – 1

25 * 27 = (26 – 1) * (26 + 1) = 26² – 1

Q3. Write the square, making use of the above pattern.

(i) 111111² 
(ii) 1111111²

Solution: Using above pattern we can write:

(i) (111111)² = 12345654321

(ii) (1111111)² = 1234567654321

Q4. Can you find the square of the following numbers using the above pattern?
(i) 6666667² 
(ii) 66666667²
Solution: Using the above pattern, we can have:
(i) 6666667² = 44444448888889
(ii) 66666667² = 4444444488888889

Exercise 5.2
Que 1. Find the square of the following numbers.
(i) 32 
(ii) 35 
(iii) 86 
(iv) 93 
(v) 71 
(vi) 46
Ans:

(i) (32)² = (30 + 2)²

= 30² + 2(30)(2) + (2)²
= 900 + 120 + 4 = 1024

(ii) (35)² = (30 + 5)²
= (30)² + 2(30)(5) + (5)²
= 900 + 300 + 25
= 1200 + 25 = 1225
Second method
35² = 3 × (3 + 1) × 100 + 25
= 3 × 4 × 100 + 25
= 1200 + 25 = 1225

(iii) (86)² = (80 + 6)²
= (80)² + 2(80)(6) + (6)²
= 6400 + 960 + 36 = 7396

(iv) (93)² = (90 + 3)²
= (90)² + 2(90)(3) + (3)²
= 8100 + 540 + 9 = 8649

(v) (71)² = (70 + 1)²
= (70)² + 2(70)(1) + (1)²
= 4900 + 140 + 1 = 5041

(vi) (46)² = (40 + 6)²
= (40)² + 2(40)(6) + (6)²
= 1600 + 480 + 36 = 2116

Q2. Write a Pythagorean triplet whose one member is
(i) 6 
(ii) 14 
(iii) 16 
(iv) 18
Ans: 
(i) Let 2n = 6       ∴n = 3
Now, n² – 1 = 3² – 1 = 8
and n² + 1 = 3² + 1 = 10
Thus, the required Pythagorean triplet is 6, 8, 10.

(ii) Let 2n = 14        ∴ n = 7
Now, n² – 1 = 7² – 1 = 48
and n² + 1 = 7² + 1 = 50
Thus, the required Pythagorean triplet is 14, 48, 50.

(iii) Let 2n = 16      ∴n = 8
Now, n² – 1 = 8² – 1
= 64 – 1 = 63
and n² + 1 = 8² + 1
= 64 + 1 = 65
Thus, the required Pythagorean triplet is 16, 63, 65.

(iv) Let 2n = 18          ∴n = 9
Now, n² – 1 = 9² – 1
= 81 – 1 = 80
and n² + 1 = 9² + 1
= 81 + 1 = 82
Thus, the required Pythagorean triple is 18, 80, 82.

Deleted Questions from NCERT

Finding the Square of a number

Example . Find the square of 27.

Solution: (27)² = (20 + 7)²

Using the formula (a + b)² = a² + 2ab + b², we have

(20 + 7)² = (20)² + 2 * (20) * (7) + (7)²

= 400 + 280 + 49

= 729

Thus, (27)² = 729

Note: For any number ending with 5, the square is a(a + 1) hundred + 25.

For example,

(25)² = 2(2 + 1) * 100 + 25 = 625

(35)² = 3(3 + 1) * 100 + 25 = 1225

(65)² = 6(6 + 1) *100 + 25 = 4225

(125)² = 12(12 + 1) * 100 + 25 = 15625

Pythagorean Triplets

If three numbers a, b and c are such that a² + b² = c², then they are called Pythagorean Triplets and they represent the sides of a right triangle.

Example: 

(i) 3, 4, 5 form a Pythagorean triplet.

     [∵ 3² + 4² = 5²]

(ii) 8, 15, 17 form a Pythagorean triplet.

     [∵ 8² + 15² = 17²]

Note: For any natural number n, (n > 1), we have

         (2n)² + (n² – 1)² = (n² + 1)²

such that 2n, n² – 1 and n² + 1 are a Pythagorean triplet.

Example. Write a Pythagorean triplet whose one member is 15.

Solution: Since, a Pythagorean triplet is given by 2n, n² – 1 and n² + 1.

∴ 2n = 15 or n = 15/2 is not an integer.

So, let us assume that

n² – 1 = 15

or n² = 15 + 1 = 16

or n² = 4², i.e. n = 4

Now, the required Pythagorean triplet is

2n, n² – 1 and n² + 1

or   2(4), 4² – 1 and 4² + 1

or   8, 15 and 17

Remember

All Pythagorean triplets may not be obtained using the above form.

Question: Find the square of the following numbers containing 5 in unit’s place.

(i) 15 (ii) 95 (iii) 105 (iv) 205

Solution:

(i) (15)² = 1 * (1 + 1) * 100 + 25

= 1 * 2 * 100 + 25

= 200 + 25 = 225

(ii) (95)² = 9(9 + 1) * 100 + 25

= 9 * 10 * 100 + 25

= 9000 + 25 = 9025

(iii) (105)² = 10 * (10 + 1) * 100 + 25

= 10 *11 * 100 + 25

= 11000 + 25 = 11025

(iv) (205)² = 20 * (20 + 1) * 100 = 25

= 20 * 21 * 100 + 25

= 42000 + 25 = 42025

Question: Write all the square numbers between 100 and 300.

Solution: 121, 144, 169, 196, 225, 256 and 289 

 Question: Which of the following would end with digit 9:
123², 77², 84², 161², and 10².
Solution: (123)² and (77)²

Question: Write all the non-square numbers between 42 and 52.

Solution:  17, 18, 19, 20, 21, 22, 23 and 24
 

Question: Fill in the blanks:
(i) 1 + 3 + 5 + 7 + 9 + … = 5²
(ii) 1 + 3 + 5 + 7 + 9 + … = 6²

Solution: (i) 11 (ii) 11 + 13
 

Question: Fill in the blanks:
(i) 11² = …… + ……
(ii) (…..)² = 112 + 113
Solution: (i) 60 + 61 (ii) (15)²

Question: Find the square of the following numbers actual multiplication:
 (i) 39 (ii) 42

Ans: (i) 1521 (ii) 1764
Question: Find the square of:
 (i) 35 (ii) 105
Solution: (i) 1225 (ii) 11025

Question: Write the Pythagorean triplet whose smallest number is 8.

Solution: 8, 15, 17

Question: Find a Pythagorean triplet in which one member is 12.

Solution: 12, 35, 37

Square Roots

Finding the square root of a number is just the opposite operation of squaring it. For example, the square of 5 is 25.

∴ Square root of 25 is 5.

Note: We kwon that (–2) * (–2) = 4, then we say that (–2) is also the square root of 4. Similarly, square roots of 100 are 10 and (–10).

But in this chapter, we shall be studying about positive square roots only.

Finding Square Root Through Prime Factorisation

We know that a factor that occurs once in the prime factorisation of a number, occurs twice in the prime factorisation of its square. Thus, we can use this fact of prime factorisation of a number to find the square root of a perfect square.

Note: A perfect square has complete pairs of its prime factors.

Example. Is 1008 a perfect square? If not, find the smallest multiply of 1008 which is a perfect square end then find the square root of the new number.

Solution: We have 

1008 = 2 * 2 * 2 * 2 * 3 * 3 * 7

As the prime factor 7 has no pair.

∴1008 is not a perfect square. Obviously, if 7 gets a pair, then the number will become a perfect square. 

∴1008 * 7 = [2 * 2 * 2 * 2 * 3 * 3 * 7] * 7

or  7056 = 2 * 2 * 2 * 2 * 3 * 3 * 7 

Thus, 7056 is the required multiple of 1008 which is a perfect square.

Now,    = 2 * 2 * 3 * 7 = 84

Question: (i) 11² = 121. What is the square root of 121?
(ii) 14² = 196. What is the square root of 196?

Solution: (i) The square root of 121 is 11.
(ii) The square root of 196 is 14.

Think, Discuss and Write

Question:

(–1)² = 1. Is –1, a square root of 1? 

(–2)² = 4. Is –2, a square root of 4? 

(–9)² = 81. Is –9, a square root of 81?

Solution:

(i) Since (–1) * (–1) = 1

i.e. (–1)² = 1

∴ Square root of 1 can also be –1.

Similarly,

(ii) Yes (–2) is a square root of 4.

(iii) Yes (–9) is a square root of 81.

Since, we have to consider the positive square roots only and the symbol for a positive square
root is √

∴  = 4 [and not (–4)]

Similarly,    means, the positive square root of 25, i.e. 5.

Note: We can also find the square root by subtracting successive odd numbers starting from 1.

Question: By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square, then find its square root.

(i) 121 (ii) 55 (iii) 36 (iv) 49 (v) 90

Solution: 

(i) Subtracting the successive odd numbers from 121, we have

121 – 1 = 120          120 – 3 = 117

117 – 5 = 112          112 – 7 = 105

105 – 9 = 96            96 – 11 = 85

85 – 13 = 72            72 – 15 = 57

57 – 17 = 40            40 – 19 = 21

21 – 21 = 0

∴   = 11. [∵ We had to subtract the first 11 odd numbers.]

(ii) 

∵ 55 – 1 = 54    54 – 3 = 51
   51 – 5 = 46    46 – 7 = 39
   39 – 9 = 30    30 – 11 = 19
   19 – 13 = 6    6 – 15 = –9

and we do not reach to 0. ∴ 55 is not a perfect square.

(iii) 

∵ 36 – 1 = 35    35 – 3 = 32
   32 – 5 = 27    27 – 7 = 20
   20 – 9 = 11    11 – 11 = 0

and we have obtained 0 after subtracting 6 successive odd numbers.

∴ 36 is a perfect square.

Thus,    = 6.

(iv) We have

49 – 1 = 48       48 – 3 = 45
45 – 5 = 40       40 – 7 = 33
33 – 9 = 24       24 – 11 = 13
13 – 13 = 0

∵ We have obtained 0 after successive subtraction of 7 odd numbers.

∴ 49 is a perfect square,

Thus,   = 7.

(v) We have:

90 – 1 = 89       89 – 3 = 86
86 – 5 = 81       81 – 7 = 74
74 – 9 = 65       65 – 11 = 54
54 – 13 = 41     41 – 15 = 26
26 – 17 = 9        9 – 19 = –10

Since, we can not reach to 0 after subtracting successive odd numbers.

∴ 90 is not a perfect square.