Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Q1: Using the identity (a + b)² = a² + 2ab + b², expand the following:
(i) (7x + 4y)²
Using (a + b)² = a² + 2ab + b²
= (7x)² + 2(7x)(4y) + (4y)²
= 49x² + 56xy + 16y²
Answer: 49x² + 56xy + 16y²
(ii) (7x/5 + 3y/2)²
= (7x/5)² + 2(7x/5)(3y/2) + (3y/2)²
= 49x²/25 + 21xy/5 + 9y²/4
Answer:
49x²/25 + 21xy/5 + 9y²/4
(iii) (2.5p + 1.5q)²
= (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²
= 6.25p² + 7.5pq + 2.25q²
Answer:
6.25p² + 7.5pq + 2.25q²
(iv) (3s/4 + 8t)²
= (3s/4)² + 2(3s/4)(8t) + (8t)²
= 9s²/16 + 12st + 64t²
Answer:
9s²/16 + 12st + 64t²
(v) (x + 1/2y)²
= x² + 2(x)(1/2y) + (1/2y)²
= x² + x/y + 1/(4y²)
Answer:
x² + x/y + 1/(4y²)
(vi) (1/x + 1/y)²
= (1/x)² + 2(1/x)(1/y) + (1/y)²
= 1/x² + 2/(xy) + 1/y²
Answer:
1/x² + 2/(xy) + 1/y²
Q2: Using the same identity, find the values of the following:
(i) (64)²
64 = 60 + 4
(64)² = (60 + 4)²
= 60² + 2(60)(4) + 4²
= 3600 + 480 + 16
= 4096
Answer: 4096
(ii) (105)²
105 = 100 + 5
(105)² = (100 + 5)²
= 100² + 2(100)(5) + 5²
= 10000 + 1000 + 25
= 11025
Answer: 11025
(iii) (205)²
205 = 200 + 5
(205)² = (200 + 5)²
= 200² + 2(200)(5) + 5²
= 40000 + 2000 + 25
= 42025
Answer: 42025