Exercise 6.2 — Prime Factorisation Method

Finding the square root by prime factorisation method.

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Exercise 6.2 – Finding Square Roots by Prime Factorisation

Exercise 6.2 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) is the core exercise on square roots. You will learn what a square root means, how to find it using the prime factorisation method, how to find the smallest number by which a given number must be multiplied or divided to make it a perfect square, and how to solve real-life word problems that depend on square roots. The exercise also revises the important concept of Pythagorean triplets.

This exercise builds directly on Exercise 6.1 (perfect squares and their properties) and is essential preparation for the long-division method of finding square roots covered in Exercise 6.3.

Pythagorean Triplets — Concept & Examples

Before jumping into square roots, the chapter introduces a beautiful relationship between squares. Three positive integers a, b, c form a Pythagorean triplet when the sum of the squares of the two smaller numbers equals the square of the largest:

a² + b² = c²

This is simply the Pythagorean theorem applied to whole numbers. The most famous triplet is (3, 4, 5) — you can verify: 9 + 16 = 25 = 5². Another common one is (5, 12, 13): 25 + 144 = 169 = 13².

Primitive vs Non-Primitive Triplets

When the three numbers in a triplet share no common factor other than 1, the triplet is called a primitive triplet — for example (3, 4, 5). Multiplying every number in a primitive triplet by the same integer gives another valid (but non-primitive) triplet.

Starting from (3, 4, 5):
× 2 → (6, 8, 10) | × 3 → (9, 12, 15) | × 10 → (30, 40, 50)

💡 Key Insight: If (a, b, c) is a Pythagorean triplet, then (ka, kb, kc) is also a Pythagorean triplet for any positive integer k. This is because squaring distributes over multiplication: (ka)² + (kb)² = k²(a² + b²) = k²c² = (kc)².

What is a Square Root?

When a number is expressed as the product of two equal factors, that factor is called its square root. The square root of a number x is written as √x.

16 = 4 × 4, so √16 = 4
121 = 11 × 11, so √121 = 11
144 = 12 × 12, so √144 = 12

Square and Square Root Reference Table (1² to 30²)

Memorising these values up to at least 20² is very useful for board exams in Telangana and Andhra Pradesh:

Number (n)	Square (n²)	Square Root (√n²)	Number (n)	Square (n²)	Square Root (√n²)
1	1	√1 = 1	16	256	√256 = 16
2	4	√4 = 2	17	289	√289 = 17
3	9	√9 = 3	18	324	√324 = 18
4	16	√16 = 4	19	361	√361 = 19
5	25	√25 = 5	20	400	√400 = 20
6	36	√36 = 6	21	441	√441 = 21
7	49	√49 = 7	22	484	√484 = 22
8	64	√64 = 8	23	529	√529 = 23
9	81	√81 = 9	24	576	√576 = 24
10	100	√100 = 10	25	625	√625 = 25
11	121	√121 = 11	26	676	√676 = 26
12	144	√144 = 12	27	729	√729 = 27
13	169	√169 = 13	28	784	√784 = 28
14	196	√196 = 14	29	841	√841 = 29
15	225	√225 = 15	30	900	√900 = 30

Two Introductory Methods for Finding Square Roots

Method 1 – Repeated Subtraction of Successive Odd Numbers

Every perfect square n² is the sum of the first n odd numbers. This means you can find √36 by repeatedly subtracting consecutive odd numbers (1, 3, 5, 7, …) until you reach zero — the number of subtractions needed equals the square root.

Step

Calculation

Result

36 − 1

35 − 3

32 − 5

27 − 7

20 − 9

11 − 11

0 ✓

We reached zero after 6 steps, therefore √36 = 6.

⚠️ This method works only for perfect squares and becomes impractical for large numbers. The prime factorisation method (used in Exercise 6.2) is far more efficient.

Method 2 – Prime Factorisation (the main method for Exercise 6.2)

To find the square root by prime factorisation: (1) break the number into its prime factors, (2) pair up identical prime factors, and (3) take one factor from each pair and multiply them together.

√36 = √(2 × 2 × 3 × 3) = √(2²× 3²) = 2 × 3 = 6

The logic: since √(a²) = a, pulling one factor out of each squared pair and multiplying gives the root.

Exercise 6.2 — Complete Solutions with Step-by-Step Explanation

Question 1 (i)

Find √441 by Prime Factorisation

Break 441 into prime factors using successive division:

3	441
3	147
7	49
7	7
	1

441 = 3 × 3 × 7 × 7 = (3 × 7)²
√441 = 3 × 7 = 21

Both prime factors 3 and 7 appear in perfect pairs, confirming that 441 is a perfect square. Taking one factor from each pair: 3 × 7 = 21.

Question 1 (ii)

Find √784 by Prime Factorisation

2	784
2	392
2	196
2	98
7	49
7	7
	1

784 = 2 × 2 × 2 × 2 × 7 × 7 = (2 × 2 × 7)²
√784 = 2 × 2 × 7 = 28

We have three pairs: (2, 2), (2, 2), (7, 7). Taking one from each: 2 × 2 × 7 = 28.

Question 1 (iii)

Find √4096 by Prime Factorisation

4096 is a power of 2. Keep dividing by 2:

2	4096
2	2048
2	1024
2	512
2	256
2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

4096 = 2¹² = (2⁶)²
√4096 = 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

Twelve 2's make six pairs. Taking one 2 from each of the six pairs and multiplying: 2⁶ = 64.

Question 1 (iv)

Find √7056 by Prime Factorisation

2	7056
2	3528
2	1764
2	882
3	441
3	147
7	49
7	7
	1

7056 = 2⁴ × 3² × 7² = (2² × 3 × 7)²
√7056 = 2 × 2 × 3 × 7 = 84

Four pairs in total: (2,2), (2,2), (3,3), (7,7). One from each: 2 × 2 × 3 × 7 = 84.

Question 2

Smallest number to multiply 3645 to make it a perfect square

Multiplication to get perfect square

To find out which number is "missing" a pair, factorise 3645 completely and look for any unpaired prime factor.

3	3645
3	1215
3	405
3	135
3	45
3	15
5	5
	1

3645 = 3⁶ × 5¹

The six 3's form three perfect pairs. But the single 5 has no pair — it is the unpaired factor. To complete its pair, multiply 3645 by 5.

3645 × 5 = 18225 = (3³ × 5)² = 135²

∴ The smallest number to multiply is 5.

Question 3

Smallest number to multiply 2400 to make it a perfect square, and find the resulting square root

Multiplication to get perfect square + find √

2	2400
2	1200
2	600
2	300
2	150
3	75
5	25
5	5
	1

2400 = 2⁵ × 3¹ × 5²

The 5's are paired. But 2 appears 5 times (one 2 has no pair) and 3 appears once (also unpaired). To pair both the lone 2 and the lone 3, multiply by 2 × 3 = 6.

2400 × 6 = 14400 = (2³ × 3 × 5)² = 120²

Smallest multiplier = 6
√14400 = 2 × 2 × 2 × 3 × 5 = 120

Question 4

Smallest number to divide 7776 to make it a perfect square

Division to get perfect square

2	7776
2	3888
2	1944
2	972
2	486
3	243
3	81
3	27
3	9
3	3
	1

7776 = 2⁵ × 3⁵

Both 2 and 3 appear 5 times each — in each case, four of them form two perfect pairs but one is left over. Dividing by the product of these unpaired factors removes them: divide by 2 × 3 = 6.

7776 ÷ 6 = 1296 = (2² × 3²)² = 36²

∴ The smallest divisor is 6.

Question 5

1521 trees planted with equal rows and trees per row — find the number of rows

Word Problem – Square root application

Let the number of rows = x. Since the number of trees per row is also x, the total number of trees = x × x = x².

x² = 1521
x = √1521

3	1521
3	507
13	169
13	13
	1

1521 = 3² × 13² → √1521 = 3 × 13 = 39

∴ Number of rows = Number of trees per row = 39.

Question 6

School collected ₹2601 as fees — number of students (fee per student = number of students)

Word Problem – Square root application

Let the number of students = x. Fee paid by each student is also x (in rupees). So total fees = x × x = x².

x² = 2601 → x = √2601

3	2601
3	867
17	289
17	17
	1

2601 = 3² × 17² → √2601 = 3 × 17 = 51

∴ Number of students in the school = 51.

Question 7

Product of two numbers is 1296; one number is 16 times the other — find both numbers

Word Problem – Algebra + Square root

Let the smaller number = x. Then the larger number = 16x. Their product: 16x × x = 16x².

16x² = 1296
x² = 1296 ÷ 16 = 81
x = √81 = 9

3	81
3	27
3	9
3	3
	1

√81 = 9, so x = 9. The two numbers are x = 9 and 16x = 16 × 9 = 144.

Smaller number = 9
Larger number = 144
Verification: 144 × 9 = 1296 ✓ | 144 = 16 × 9 ✓

Question 8

7921 soldiers seated with equal rows and soldiers per row — find number of rows

Word Problem – Square root application

Let the number of rows = x. Soldiers per row also = x. Total = x².

x² = 7921 → x = √7921

89	7921
89	89
	1

7921 = 89 × 89 = 89² → √7921 = 89

∴ Number of rows in the auditorium = 89.

💡 89 is a prime number, so division by all primes up to √89 (≈ 9.4) — that is 2, 3, 5, 7 — fails, and 89 divides 7921 exactly twice. This is a useful technique for large numbers without obvious factors.

Question 9

Area of square field = 5184 m². Rectangular field with same perimeter and length = 2 × breadth — find area of rectangle

Word Problem – Square root + Geometry

Step 1: Find the side of the square.

Let the side of the square = x m. Area = x² = 5184.

2	5184
2	2592
2	1296
2	648
2	324
2	162
3	81
3	27
3	9
3	3
	1

5184 = 2⁶ × 3⁴ = (2³ × 3²)² → √5184 = 8 × 9 = 72

Side of square = x = 72 m. Perimeter of square = 4 × 72 = 288 m.

Step 2: Find the dimensions of the rectangle.

Let breadth = y. Then length = 2y. Perimeter of rectangle = 2(2y + y) = 2(3y) = 6y. Since the perimeters are equal: 6y = 288 → y = 48.

Breadth = y = 48 m
Length = 2y = 96 m

Step 3: Calculate the area of the rectangle.

Area = length × breadth = 96 × 48 = 4608 sq.m

Square field (area = 5184 m²)

Rectangular field (area = 4608 m²)

∴ Area of the rectangle = 4608 sq.m.

Exercise 6.2 — Quick Answer Summary

Question	Type	Answer
1 (i) √441	Prime factorisation	21
1 (ii) √784	Prime factorisation	28
1 (iii) √4096	Prime factorisation	64
1 (iv) √7056	Prime factorisation	84
2. Multiply 3645 by ?	Make perfect square	5 (→ 18225 = 135²)
3. Multiply 2400 by ?	Make perfect square + √	6 (→ 14400, √ = 120)
4. Divide 7776 by ?	Make perfect square	6 (→ 1296 = 36²)
5. 1521 trees	Word problem	39 rows, 39 trees/row
6. ₹2601 fees	Word problem	51 students
7. Product 1296, one = 16×other	Algebra + √	9 and 144
8. 7921 soldiers	Word problem	89 rows
9. 5184 m² square + rectangle	Geometry + √	4608 sq.m

Common Mistakes to Avoid in Exercise 6.2

Incomplete prime factorisation: Always continue dividing until the quotient is 1. Stopping early gives wrong pairs.
Multiplying instead of dividing (or vice versa) for perfect square questions: To make a perfect square by multiplication, find the unpaired factor and multiply by it. To make one by division, find the unpaired factor and divide by it.
Forgetting to multiply all unpaired factors: In Q3 (2400), both 2 and 3 are unpaired — you must multiply by both (2 × 3 = 6), not just one of them.
Taking all factors, not one from each pair: √(3² × 7²) = 3 × 7 = 21, not 3 × 3 × 7 × 7 = 441.
Word problems — setting up x²: When rows = trees/row (or students = fee), the product is always x². Identify this structure before computing.

📝 Board exam tip (Telangana & AP): Always write the prime factorisation step in full, show the pairing of factors explicitly, and write the final answer with the √ symbol resolved. Examiners award marks for each step — a correct answer without working may receive partial or zero credit.

What Exercise 6.2 Prepares You For

The prime factorisation method practised here is the foundation for the long-division method of finding square roots (Exercise 6.3), which handles non-perfect-square numbers and large values that factorisation alone cannot address efficiently.

The Pythagorean triplet concept introduced at the start of this chapter connects directly to the Theorem of Pythagoras in Class 9 Triangles, where you prove and apply it in various geometric problems. The idea that x² = total (rows × trees or students × fee) is also the basis for solving quadratic equations of the form x² = k, encountered formally in Class 9 and 10.