NCERT Solutions Ganita Prakash (Part 1) Chapter 6 .2 Special Cases of the Distributive Property — In-text Questions

Book page 1476 Updated on2026-09-05

Q1.
We have seen what (a + b)² gives when expanded. What is the expansion of (a – b)²?
Answer

Multiply the bracket by itself, using the distributive property.

(a – b)² = (a – b) × (a – b)
= a² – ba – ab + b²
= a² – 2ab + b²
Identity 1B: (a – b)² = a² + b² – 2ab

Test at a = 9, b = 4: LHS = 5² = 25; RHS = 81 + 16 – 72 = 25 ✓

Why it happens: the four products are a·a, a·(–b), (–b)·a and (–b)(–b). Two of them are –ab, giving –2ab; the last is +b² because a negative times a negative is positive. This is the algebraic twin of the picture on page 146: subtract two strips, add the corner back.
Q2.
We can also use the expansion of (a + b)² to find the expansion of (a – b)². Think how. Hint: (a – b)² = (a + (–b))².
Answer

A subtraction is an addition of the opposite, so Identity 1A already covers it.

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

The only two things used are (–b)² = b² and a × (–b) = –ab.

Why it happens: Identity 1A was proved for any numbers a and b, so we are free to put –b in place of b. Getting a second identity out of the first by substitution — rather than by starting again — is one of the real economies algebra offers.
Q3.
Find the general expansion of (a – b)² using geometry, as we did for 55².
Answer

Draw a square of side a. Mark off a square of side (a – b) inside it, in one corner.

Area of the big square = a²
Remove the strip along the right: a × b
Remove the strip along the bottom: b × a
The corner b × b has now gone twice, so add it back:
(a – b)² = a² – ab – ba + b² = a² – 2ab + b²
(a – b)² b(a–b) b(a – b) a b a
The square of side a splits into (a – b)², two strips of b(a – b) and the corner b². Since 2b(a – b) + b² = 2ab – b², the shaded square is a² – 2ab + b².

Reading the picture the other way round: a² = (a – b)² + 2b(a – b) + b², and expanding 2b(a – b) = 2ab – 2b² gives (a – b)² = a² – 2ab + b² again.

Q4.
Use the identity (a – b)² to find the values of (a) 99² and (b) 58².
Answer
(a) 99² = (100 – 1)²
= 100² + 1² – 2 × 100 × 1
= 10000 + 1 – 200
= 9801
(b) 58² = (60 – 2)²
= 60² + 2² – 2 × 60 × 2
= 3600 + 4 – 240
= 3364
Tip: the split to look for is “a round number, minus a small number”. Squaring 99 or 58 straight out takes a full multiplication; this way it is two easy squares and one doubling.
Q5.
Expand the following using both Identity 1B and by applying the distributive property (i) (b – 6)² (ii) (–2a + 3)² (iii) (7y – 3⁄4 z)²
Answer

(i) (b – 6)²

Identity 1B: b² + 6² – 2 × b × 6 = b² – 12b + 36
Distributive: (b – 6)(b – 6) = b² – 6b – 6b + 36 = b² – 12b + 36 ✓

(ii) (–2a + 3)² — read it as (3 – 2a)²

Identity 1B: 3² + (2a)² – 2 × 3 × 2a = 4a² – 12a + 9
Distributive: (–2a + 3)(–2a + 3) = 4a² – 6a – 6a + 9 = 4a² – 12a + 9 ✓

(iii) (7y – 3⁄4 z)²

Identity 1B: (7y)² + (3⁄4 z)² – 2 × 7y × 3⁄4 z
= 49y² – 21⁄2 yz + 9⁄16 z²
Distributive: 7y(7y – 3⁄4 z) – 3⁄4 z(7y – 3⁄4 z)
= 49y² – 21⁄4 yz – 21⁄4 yz + 9⁄16 z² = 49y² – 21⁄2 yz + 9⁄16 z² ✓

Test (ii) at a = 1: (–2 + 3)² = 1 and 4 – 12 + 9 = 1 ✓

Tip: in (ii) the square makes the sign of the whole bracket irrelevant — (–2a + 3)² and (2a – 3)² are equal, because (–x)² = x².
Was this helpful? Report an error