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NCERT Solutions Ganita Prakash (Part 1) Chapter 6 Investigating Patterns — Figure it Out

Book page 149 Updated on2026-09-05

Q1.
Which is greater: (a – b)² or (b – a)²? Justify your answer.
Answer

Neither — they are always equal.

b – a = –(a – b)
so (b – a)² = (–(a – b))² = (a – b)²

Or expand both and compare:

(a – b)² = a² – 2ab + b²
(b – a)² = b² – 2ba + a² = a² – 2ab + b²

The two expansions are the same expression, so the two are equal for every a and b.

Test at a = 3, b = 7: (3 – 7)² = 16 and (7 – 3)² = 16 ✓

Why it happens: squaring destroys sign information — (–x)² = x². Two numbers that differ only in sign always have the same square, and a – b and b – a differ only in sign.
Q2.
Express 100 as the difference of two squares.
Answer

Use Identity 1C backwards: a² – b² = (a + b)(a – b), so we need two factors of 100 whose product is 100.

a + b = first factor, a – b = second factor
a and b are whole numbers only if the two factors have the same parity
(their sum 2a and difference 2b must be even)
Factor pair of 100Same parity?a, bResult
50 × 2both even ✓a = 26, b = 2426² – 24² = 676 – 576 = 100
10 × 10both even ✓a = 10, b = 010² – 0² = 100
100 × 1, 20 × 5, 25 × 4mixed ✗—no whole-number answer
100 = 26² – 24² (and trivially 100 = 10² – 0²)
Why it happens: from a + b = m and a – b = n we get a = (m + n)⁄2 and b = (m – n)⁄2. Those are whole numbers only when m and n are both even or both odd. Since 100 is even but not a multiple of 4 in an odd × odd way, only the 50 × 2 and 10 × 10 splits survive.
Q3.
Find 406², 72², 145², 1097², and 124² using the identities you have learnt so far.
Answer
406² = (400 + 6)² — Identity 1A
= 160000 + 2 × 400 × 6 + 36
= 160000 + 4800 + 36 = 164836
72² = (70 + 2)² — Identity 1A
= 4900 + 280 + 4 = 5184
145² = (150 – 5)² — Identity 1B
= 22500 + 25 – 1500 = 21025
1097² = (1100 – 3)² — Identity 1B
= 1210000 + 9 – 6600 = 1203409
124² = (120 + 4)² — Identity 1A
= 14400 + 960 + 16 = 15376
Check it yourself: 145² can also be done Sridharacharya's way: 145² = 150 × 140 + 5² = 21000 + 25 = 21025 ✓
Q4.
Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.
Answer

They hold for all of them. Both patterns were proved by expanding brackets, and expanding uses only distributivity and commutativity — rules that integers and fractions obey just as counting numbers do.

Pattern 1: 2(a² + b²) = (a + b)² + (a – b)²

a, b2(a² + b²)(a + b)² + (a – b)²
–3, –52(9 + 25) = 68(–8)² + 2² = 68 ✓
–4, 62(16 + 36) = 1042² + (–10)² = 104 ✓
1⁄2, 1⁄32(1⁄4 + 1⁄9) = 13⁄18(5⁄6)² + (1⁄6)² = 13⁄18 ✓

Pattern 2: a² – b² = (a + b)(a – b)

a, ba² – b²(a + b)(a – b)
–7, 349 – 9 = 40(–4)(–10) = 40 ✓
2⁄3, 1⁄34⁄9 – 1⁄9 = 1⁄31 × 1⁄3 = 1⁄3 ✓
Why it happens: nowhere in the proofs did we say “a is a counting number”. Once a statement is proved from the rules of arithmetic alone, it is true in every number system where those rules hold — which is what makes an identity so much stronger than a list of checked examples.
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