# Square Roots

## Exercise 6.3 Part 1

Question 1: What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801

Answer: 1 and 9.
Explanation: Since 12 and 92 give 1 at unit’s place, so these are the possible values of unit digit of the square root.

(ii) 99856

Answer: 4 and 6
Explanation: Since, 42 = 16 and 62 = 36, hence, 4 and 6 are possible digits

(iii) 998001

Answer: 1 and 9

(iv) 657666025

Explanation: Since, 52 = 25, hence 5 is possible.

Question 2: Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153 (ii) 257 (iii) 408 (iv) 441

Answer: (i) 153 (ii) 257 (iii) 408
Explanation: Since, (i), (ii) and (iii) are surely not be perfect square as these numbers end with 3, 7 and 8. A number can be a perfect square if it ends with 0, 1, 4, 5, 6, 9 only

Question 3: Find the square roots of 100 and 169 by the method of repeated subtraction.

Answer: Square root of 100 by Repeated subtraction:

1. 100 - 1 = 99
2. 99 - 3 = 96
3. 96 - 5 = 91
4. 91 -7 = 84
5. 84 - 9 = 75
6. 75 - 11 = 64
7. 64 - 13 = 51
8. 51 - 15 = 3610. 19 - 19 = 0
We get 0 at 10th step. Thus, 10 is the square root of 100.

So, sqrt(100)=100

Square root of 169 by Repeated subtraction:

1. 169 - 1 = 168
2. 168 - 3 = 165
3. 165 - 5 = 160
4. 160 - 7 = 153
5. 153 - 9 = 144
6. 144 - 11 = 133
7. 133 - 13 = 120
8. 120 - 15 = 105
9. 105 - 17 = 88
10. 88 - 19 = 69
11. 69 - 21 = 48
12. 48 - 23 = 25
13. 25 - 25 = 0
We get 0 at 13th step. Thus 13 is the square root of 169

So, sqrt(169)=13

Question 4: Find the square roots of the following numbers by the Prime Factorisation Method.

(i) 729

 3 729 3 243 3 81 3 27 3 9 3

So, 729=3^6

Or, sqrt(729)=3^3=27

(ii) 400

 2 400 2 200 2 100 2 50 5 25 5

So, 400=2^4×5^2

Or, sqrt(400)=2^2×5=20

(iii) 1764

 2 1764 2 882 3 441 3 147 7 49 7

1764=2^2xx3^2xx7^2

sqrt(1764)=2xx3xx7=42

(iv) 4096

 2 4096 2 2048 2 1024 2 512 2 256 2 128 2 64 2 32 2 16 2 8 2 4 2

4096=2^(12)

sqrt(4096)=2^6=64

(v) 7744

 2 7744 2 3872 2 1936 2 968 2 484 2 242 11 121 11

7744=2^6xx11^2

sqrt(7744)=2^3xx11=88

(vi) 9604

 2 9604 2 4802 7 2401 7 343 7 49 7

9604=2^2xx7^4

sqrt(9604)=2xx7^2=98

(vii) 5929

 11 5929 11 539 7 49 7

5929=7^2xx11^2

sqrt(5929)=7xx11=77

(viii) 9216

 2 9216 2 4608 2 2304 2 1152 2 576 2 288 2 144 2 72 2 36 2 18 3 9 3

9216=2^(10)xx3^2

sqrt(9216)=2^5xx3=96

(ix) 529

Answer: 529=23xx23

sqrt(529)=23

(x) 8100

8100=2^2xx3^4xx5^2
sqrt(8100)=2xx3^2xx5=90