Squares and Square Roots

1. Perfect Squares

A perfect square is the square of an integer.

0² = 0, 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100, 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225, 16² = 256, 17² = 289, 18² = 324, 19² = 361, 20² = 400, 25² = 625, 30² = 900, 40² = 1600, 50² = 2500.

'Memorise squares up to 25 — they appear in EVERY chapter of mathematics.'


2. Properties of Square Numbers

PropertyExample
A number ending in 2, 3, 7, or 8 is NEVER a perfect square22, 33, 47, 58 are NOT perfect squares
A number ending in an ODD number of zeros is NEVER a perfect square1000, 250, 40 are NOT perfect squares
Square of an EVEN number is EVEN12² = 144 (even)
Square of an ODD number is ODD13² = 169 (odd)
Square of a number ending in 5 ends in 2535² = 1225, 45² = 2025
Square of a number ending in 1 or 9 ends in 111² = 121, 19² = 361
Square of a number ending in 4 or 6 ends in 614² = 196, 16² = 256

Pattern: (n + 1)² = n² + n + (n + 1)

Example: If 15² = 225, then 16² = 225 + 15 + 16 = 256. ✓


3. Square Roots by Prime Factorisation

To find the square root of a perfect square:

  1. Write the number as a product of its prime factors.
  2. GROUP the factors into PAIRS of identical factors.
  3. Take ONE factor from each pair.
  4. MULTIPLY those factors.

Worked Example: Find √144

144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3² Pair: (2×2), (2×2), (3×3) √144 = 2 × 2 × 3 = 12

Worked Example: Find √1764

1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7² √1764 = 2 × 3 × 7 = 42


4. Square Roots by Long Division Method

Used for LARGER numbers where prime factorisation is difficult.

Worked Example: Find √4489 by long division.

Step 1: Place bars on the digits in PAIRS from the RIGHT. 44 | 89

Step 2: Find the largest number whose square ≤ 44. 6² = 36. Write 6 as quotient and divisor.

Step 3: Subtract 36 from 44. Remainder = 8. Bring down 89. New dividend = 889.

Step 4: DOUBLE the quotient (6 × 2 = 12). Find digit d such that 12d × d ≤ 889. d = 7 because 127 × 7 = 889.

Step 5: Subtract. Remainder = 0.

Therefore √4489 = 67.

Worked Example: Find √5579.56 (decimal).

Follow the same steps. Place bars on INTEGER part from RIGHT and DECIMAL part from LEFT.


5. Pythagorean Triplets

Three natural numbers (a, b, c) such that a² + b² = c².

Generating Triplets

For any natural number m > 1: (2m, m² — 1, m² + 1)

mTripletCheck
2(4, 3, 5)4² + 3² = 16 + 9 = 25 = 5²
3(6, 8, 10)6² + 8² = 36 + 64 = 100 = 10²
4(8, 15, 17)8² + 15² = 64 + 225 = 289 = 17²
5(10, 24, 26)10² + 24² = 100 + 576 = 676 = 26²
6(12, 35, 37)12² + 35² = 144 + 1225 = 1369 = 37²

'Pythagorean triplets are useful in geometry problems involving right triangles.'


Common Mistakes and Fixes

MistakeFix
'√(a + b) = √a + √b'FALSE! √(9 + 16) = √25 = 5, but √9 + √16 = 3 + 4 = 7
'625 is not a perfect square'625 = 25²! Check the last two digits: 25 → could be a perfect square
'A number ending in 4 cannot be a perfect square'It CAN: 4, 64, 144, 324 are all perfect squares ending in 4
'Leaving out a factor when pairing'Ensure EVERY prime factor is paired. Unpaired factors mean the number is NOT a perfect square

ICSE Exam Focus (4–6 marks)

  • 2-mark questions: Identifying perfect squares, finding square roots of small numbers
  • 3-mark questions: Prime factorisation method for square roots
  • 4-mark questions: Long division method or Pythagorean triplet problems
  • 6-mark questions: Application — area problems using square roots

Self-Test

Q1. Is 1568 a perfect square? Justify. A1. 1568 = 2⁵ × 7². The factors are 2² × 2² × 7² × 2. One '2' is unpaired. So 1568 is NOT a perfect square.

Q2. Find √3136 by prime factorisation. A2. 3136 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 = 2⁶ × 7². √3136 = 2 × 2 × 2 × 7 = 56.

Q3. Find the smallest number by which 720 must be MULTIPLIED to get a perfect square. A3. 720 = 2⁴ × 3² × 5. '5' is unpaired. Multiply by 5: 720 × 5 = 3600 = 60².

Q4. Find √53,824 by long division. A4. 5,38,24. Step-by-step: √53824 = 232. (Check: 232² = 53824)

Q5. Write a Pythagorean triplet whose smallest member is 8. A5. If 2m = 8, then m = 4. Triplet: (2×4, 4²—1, 4²+1) = (8, 15, 17). Check: 8² + 15² = 64 + 225 = 289 = 17² ✓.

Q6. The area of a square field is 8281 m². Find its side. A6. Side = √8281 = 91 m (by long division: 91² = 8281).

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