FINDING SQUARE ROOT BY PRIME FACTORIZATION METHOD

The following steps will be useful to find square root of a number by prime factorization.  

(i) Decompose the number inside the square root into prime factors. 

(ii) Inside the square root, for every two same numbers multiplied, one number can be taken out of the square root.  

(iii) Combine the like square root terms using mathematical operations. 

Example : 

√3 - √12 + √27 = √3 + √(2 ⋅ 2 ⋅ 3) + √(3 ⋅ 3 ⋅ 3)

√3 - 2√3 + 3√3

= 2√3

Practice Questions

Find the square root each of the following numbers prime factorization.

Question 1 : 

36

Answer :

Decompose 36 into prime factors using synthetic division. 

√36 = √(2 ⋅ 2 ⋅ 3 ⋅ 3)

(2 ⋅ 3)

= 6

Question 2 : 

144

Answer :

Decompose 144 into prime factors using synthetic division. 

√144 = √(2 ⋅ ⋅ ⋅ ⋅ 3 ⋅ 3)

(2 ⋅ ⋅ 3)

= 12

Question 3 : 

256

Answer :

Decompose 256 into prime factors using synthetic division. 

√256 = √(2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 2)

(2 ⋅ ⋅ ⋅ 2)

= 16

Question 4 : 

324

Answer :

Decompose 324 into prime factors using synthetic division. 

√324 = √(2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3)

= 2 ⋅ 3 ⋅ 3

= 18

Question 5 : 

400

Answer :

Decompose 400 into prime factors using synthetic division. 

√400 = √(2 ⋅ ⋅ ⋅ ⋅ 5 ⋅ 5)

(2 ⋅ ⋅ 5)

= 20

Question 6 : 

625

Answer :

Decompose 625 into prime factors using synthetic division. 

√625 = √(5 ⋅ 5 ⋅ 5 ⋅ 5)

= 5 ⋅ 5

= 25

Question 7 : 

1024

Answer :

Decompose 1024 into prime factors using synthetic division. 

√1024 = √(2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 2)

(2 ⋅ ⋅ ⋅ ⋅ 2)

= 32

Question 8 : 

2025

Answer :

Decompose 2025 into prime factors using synthetic division. 

√2025 = √(5 ⋅ 5 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3)

(5 ⋅ 3 ⋅ 3)

= 45

Question 9 : 

3136

Answer :

Decompose 3136 into prime factors using synthetic division. 

√3136 = √(2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 7 ⋅ 7)

(2 ⋅ ⋅ ⋅ 7)

= 56

Question 10 : 

4096

Answer :

Decompose 4096 into prime factors using synthetic division. 

√4096 = √(2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 2)

(2 ⋅ ⋅ ⋅ ⋅ ⋅ 2)

= 64

Question 11 : 

√8 + √125

Answer : 

√8 = √(2 ⋅ 2 ⋅ 2)

√8 = 2

√125 = √(5 ⋅ 5 ⋅ 5)

√125 = 5

√8 + √125 = 2 + 5

= 7

Question 12 : 

(√34)(√136)

Answer : 

(√34)(√136) = √(34 ⋅ 136)

√(2 ⋅ 17 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 17)

= 2 ⋅ 2 ⋅ 17

= 68

Question 13 : 

(14√117) ÷ (7√52)

Answer : 

Decompose 117 and 52 into prime factors using synthetic division.

√117 = √(3 ⋅ 3 ⋅ 13)

√117 = 3√13

√52 = √(2 ⋅ 2 ⋅ 13)

√52 = 2√13

(14√117) ÷ (7√52) :

= 14(3√13) ÷ 7(2√13)

= 42√13 ÷ 14√13

= 42√13/14√13

= 3

Question 14 : 

(4√7)2

Answer :

(4√7)= 4√7 ⋅ 4√7

= (4 ⋅ 4)(√7 ⋅ √7)

= (16)(7)

= 112

Question 15 : 

(√5)3(√125)  

Answer :

(√5)3 + √125 = (√5 ⋅ √5  √5)√(5 ⋅ 5 ⋅ 5)

= (5√5)(5√5)

= (5 ⋅ 5)(√5 ⋅ √5)

= (25)(5)

= 125

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