Exercise 6.3 — Division Method
Finding square root by division method, square roots of decimal numbers.
Exercise 6.3 – Square Root by Long Division Method
Exercise 6.3 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) introduces the long division method for finding square roots — a powerful technique that works for any number, including large numbers and decimals, where prime factorisation becomes slow or impossible. This exercise has 8 questions covering four types of problems: finding square roots of integers, finding square roots of decimals, adjusting a number by subtraction or addition to make it a perfect square, and estimating square roots to the nearest whole number.
Unlike the prime factorisation method from Exercise 6.2, the long division method works even when a number has large or awkward prime factors. It is a key technique tested in board exams across Telangana, Andhra Pradesh, and CBSE Class 8.
How the Long Division Method Works — Step-by-Step
The long division method for square roots is built on a simple idea: group the digits in pairs starting from the decimal point (or from the right for whole numbers), then find how many times a carefully chosen divisor fits into each successive group. Here is the complete procedure:
Question 1 — Find Square Roots by Long Division Method
Group the digits: 10 | 89
| Quotient digit 1 | 3 (since 3² = 9 ≤ 10) |
| Subtract 9 from 10 | Remainder = 1 |
| Bring down 89 | New dividend = 189 |
| Trial divisor | 2 × 3 = 6 → try 6_ |
| Find d | 63 × 3 = 189 ≤ 189 → d = 3 |
| Subtract 189 | Remainder = 0 |
| √1089 | = 33 |
√1089 = 33 Verify: 33 × 33 = 1089 ✓
Group the digits: 23 | 04
| Quotient digit 1 | 4 (since 4² = 16 ≤ 23) |
| Subtract 16 from 23 | Remainder = 7 |
| Bring down 04 | New dividend = 704 |
| Trial divisor | 2 × 4 = 8 → try 8_ |
| Find d | 88 × 8 = 704 ≤ 704 → d = 8 |
| Subtract 704 | Remainder = 0 |
| √2304 | = 48 |
√2304 = 48 Verify: 48 × 48 = 2304 ✓
Group the digits: 77 | 44
| Quotient digit 1 | 8 (since 8² = 64 ≤ 77) |
| Subtract 64 from 77 | Remainder = 13 |
| Bring down 44 | New dividend = 1344 |
| Trial divisor | 2 × 8 = 16 → try 16_ |
| Find d | 168 × 8 = 1344 ≤ 1344 → d = 8 |
| Subtract 1344 | Remainder = 0 |
| √7744 | = 88 |
√7744 = 88 Verify: 88 × 88 = 7744 ✓
Group the digits: 60 | 84
| Quotient digit 1 | 7 (since 7² = 49 ≤ 60) |
| Subtract 49 from 60 | Remainder = 11 |
| Bring down 84 | New dividend = 1184 |
| Trial divisor | 2 × 7 = 14 → try 14_ |
| Find d | 148 × 8 = 1184 ≤ 1184 → d = 8 |
| Subtract 1184 | Remainder = 0 |
| √6084 | = 78 |
√6084 = 78 Verify: 78 × 78 = 6084 ✓
Group the digits: 90 | 25
| Quotient digit 1 | 9 (since 9² = 81 ≤ 90) |
| Subtract 81 from 90 | Remainder = 9 |
| Bring down 25 | New dividend = 925 |
| Trial divisor | 2 × 9 = 18 → try 18_ |
| Find d | 185 × 5 = 925 ≤ 925 → d = 5 |
| Subtract 925 | Remainder = 0 |
| √9025 | = 95 |
√9025 = 95 Verify: 95 × 95 = 9025 ✓
Question 2 — Square Roots of Decimal Numbers
For decimal numbers, pair the digits on each side of the decimal point separately — going left for the integer part and going right for the decimal part. Place the decimal point in the square root exactly when you bring down the first decimal pair. The method is otherwise identical to the integer case.
Group as: 2 . 56 → integer part: "2" (one group), decimal part: "56" (one pair)
| Quotient digit 1 | 1 (since 1² = 1 ≤ 2) |
| Subtract 1 from 2 | Remainder = 1 |
| Bring down .56 | New dividend = 156 (place decimal point in quotient) |
| Trial divisor | 2 × 1 = 2 → try 2_ |
| Find d | 26 × 6 = 156 ≤ 156 → d = 6 |
| Subtract 156 | Remainder = 0 |
| √2.56 | = 1.6 |
√2.56 = 1.6 Verify: 1.6 × 1.6 = 2.56 ✓Group as: 18 . 49 → integer part: "18", decimal part: "49"
| Quotient digit 1 | 4 (since 4² = 16 ≤ 18) |
| Subtract 16 from 18 | Remainder = 2 |
| Bring down .49 | New dividend = 249 |
| Trial divisor | 2 × 4 = 8 → try 8_ |
| Find d | 83 × 3 = 249 ≤ 249 → d = 3 |
| Subtract 249 | Remainder = 0 |
| √18.49 | = 4.3 |
√18.49 = 4.3 Verify: 4.3 × 4.3 = 18.49 ✓Group as: 68 . 89
| Quotient digit 1 | 8 (since 8² = 64 ≤ 68) |
| Subtract 64 from 68 | Remainder = 4 |
| Bring down .89 | New dividend = 489 |
| Trial divisor | 2 × 8 = 16 → try 16_ |
| Find d | 163 × 3 = 489 ≤ 489 → d = 3 |
| Subtract 489 | Remainder = 0 |
| √68.89 | = 8.3 |
√68.89 = 8.3 Verify: 8.3 × 8.3 = 68.89 ✓Group as: 84 . 64
| Quotient digit 1 | 9 (since 9² = 81 ≤ 84) |
| Subtract 81 from 84 | Remainder = 3 |
| Bring down .64 | New dividend = 364 |
| Trial divisor | 2 × 9 = 18 → try 18_ |
| Find d | 182 × 2 = 364 ≤ 364 → d = 2 |
| Subtract 364 | Remainder = 0 |
| √84.64 | = 9.2 |
√84.64 = 9.2 Verify: 9.2 × 9.2 = 84.64 ✓Questions 3 to 7 — Applying the Division Method
The strategy: find √4000 by long division and note the remainder at the end. That remainder is the number which, when subtracted, leaves the largest perfect square ≤ 4000.
Group 4000 as: 40 | 00
| Quotient digit 1 | 6 (since 6² = 36 ≤ 40) |
| Subtract 36 from 40 | Remainder = 4 |
| Bring down 00 | New dividend = 400 |
| Trial divisor | 2 × 6 = 12 → try 12_ |
| Find d | 123 × 3 = 369 ≤ 400; 124 × 4 = 496 > 400 → d = 3 |
| Subtract 369 | Remainder = 31 |
| Quotient = 63 | Remainder = 31 |
4000 − 31 = 3969 = 63²
The remainder 31 is the excess. Subtracting it gives 3969, which is a perfect square.
If the side of the square is x cm, then area = x² = 4489. So x = √4489. Group as: 44 | 89
| Quotient digit 1 | 6 (since 6² = 36 ≤ 44) |
| Subtract 36 from 44 | Remainder = 8 |
| Bring down 89 | New dividend = 889 |
| Trial divisor | 2 × 6 = 12 → try 12_ |
| Find d | 127 × 7 = 889 ≤ 889 → d = 7 |
| Subtract 889 | Remainder = 0 |
| √4489 | = 67 |
Side = √4489 = 67 cmSince 8 plants could not be arranged in the square, the plants actually arranged form a perfect square: 8289 − 8 = 8281. If each row has x plants and there are x rows, then x² = 8281.
Group 8281 as: 82 | 81
| Quotient digit 1 | 9 (since 9² = 81 ≤ 82) |
| Subtract 81 from 82 | Remainder = 1 |
| Bring down 81 | New dividend = 181 |
| Trial divisor | 2 × 9 = 18 → try 18_ |
| Find d | 181 × 1 = 181 ≤ 181 → d = 1 |
| Subtract 181 | Remainder = 0 |
| √8281 | = 91 |
The smallest four-digit number is 1000. Find √1000 by long division and check what the next quotient would be — squaring it gives the least four-digit perfect square.
Group 1000 as: 10 | 00
| Quotient digit 1 | 3 (since 3² = 9 ≤ 10) |
| Subtract 9 from 10 | Remainder = 1 |
| Bring down 00 | New dividend = 100 |
| Trial divisor | 2 × 3 = 6 → try 6_ |
| Find d | 61 × 1 = 61 ≤ 100; 62 × 2 = 124 > 100 → d = 1 |
| Subtract 61 | Remainder = 39 (not zero — 1000 is NOT a perfect square) |
| Floor quotient | = 31 |
Since √1000 lies between 31 and 32, the largest three-digit perfect square is 31² = 961 and the least four-digit perfect square is the next one: 32² = 1024.
32² = 32 × 32 = 1024
The strategy: find √6412 by long division. The quotient gives the floor square root (call it n). The next perfect square above 6412 is (n+1)². The number to add is (n+1)² − 6412.
Group 6412 as: 64 | 12
| Quotient digit 1 | 8 (since 8² = 64 ≤ 64) |
| Subtract 64 from 64 | Remainder = 0 |
| Bring down 12 | New dividend = 12 |
| Trial divisor | 2 × 8 = 16 → try 16_ |
| Find d | 160 × 0 = 0 ≤ 12; 161 × 1 = 161 > 12 → d = 0 |
| Subtract 0 | Remainder = 12 (6412 is NOT a perfect square) |
| Floor quotient | = 80 |
√6412 lies between 80 and 81. The next perfect square above 6412 is 81² = 6561.
6561 − 6412 = 149
Question 8 — Estimate Square Roots to the Nearest Whole Number
When a number is not a perfect square, you can estimate its square root by identifying which two consecutive perfect squares it falls between, then choosing the nearer one. This is a quick mental technique useful in CBSE and state board MCQs and estimation questions.
Method: Find n such that n² ≤ given number < (n+1)². If the number is closer to n², √ ≈ n; if closer to (n+1)², √ ≈ n+1.
9² = 81 < 97 < 100 = 10²97 − 81 = 16 | 100 − 97 = 397 is closer to 100 → √97 ≈ 10
15² = 225 < 250 < 256 = 16²250 − 225 = 25 | 256 − 250 = 6250 is closer to 256 → √250 ≈ 16
27² = 729 < 780 < 784 = 28²780 − 729 = 51 | 784 − 780 = 4780 is closer to 784 → √780 ≈ 28
Exercise 6.3 — Quick Answer Summary
| Question | Type | Answer |
|---|---|---|
| 1(i) √1089 | Long division — integer | 33 |
| 1(ii) √2304 | Long division — integer | 48 |
| 1(iii) √7744 | Long division — integer | 88 |
| 1(iv) √6084 | Long division — integer | 78 |
| 1(v) √9025 | Long division — integer | 95 |
| 2(i) √2.56 | Long division — decimal | 1.6 |
| 2(ii) √18.49 | Long division — decimal | 4.3 |
| 2(iii) √68.89 | Long division — decimal | 8.3 |
| 2(iv) √84.64 | Long division — decimal | 9.2 |
| 3. Subtract from 4000 | Make perfect square | 31 (→ 3969 = 63²) |
| 4. Side of square (area 4489) | Word problem | 67 cm |
| 5. Plants per row (8281 planted) | Word problem | 91 |
| 6. Least 4-digit perfect square | Conceptual | 1024 = 32² |
| 7. Add to 6412 | Make perfect square | 149 (→ 6561 = 81²) |
| 8(i) √97 ≈ | Estimation | 10 |
| 8(ii) √250 ≈ | Estimation | 16 |
| 8(iii) √780 ≈ | Estimation | 28 |
Common Mistakes to Avoid in Exercise 6.3
- Wrong digit grouping: Always pair digits from the decimal point outward. "1089" groups as "10|89", not "1|08|9". One wrong group ruins the entire calculation.
- Forgetting to double the quotient for the trial divisor: The trial divisor is always 2 × (current quotient), not the quotient itself. Missing this step is the most common error in board exams.
- Decimal square roots — wrong placement of decimal point: The decimal point goes in the answer exactly when you cross the decimal point in the number. Practice this with 2(i)–(iv) until it is automatic.
- Subtract vs Add questions — confusing the two: For Q3 (subtract), the answer is the remainder in the division. For Q7 (add), the answer is next perfect square minus the given number. These are opposite operations.
- Estimation — choosing the wrong square: Always compute both differences (number − lower square) and (upper square − number) and pick the smaller one. Do not guess by eye.
What Exercise 6.3 Prepares You For
The long division method is the last major technique in Chapter 6. Together with the prime factorisation method from Exercise 6.2, it gives you a complete toolkit for handling square roots at Class 8 level.
The estimation skill from Question 8 directly connects to irrational numbers in Class 9, where you locate surds like √2 and √3 on the number line by identifying which consecutive integers they fall between — the exact same reasoning used here. The subtraction and addition technique (Questions 3 and 7) also reappears in completing the square for quadratic equations in Class 9 and 10.