Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Q1: Find the following squares using one of the identities. Determine which identity makes these calculations easier.
(i) (117)²
Use (a + b)² = a² + 2ab + b²
117 = 100 + 17
(117)² = (100 + 17)²
= 100² + 2(100)(17) + 17²
= 10000 + 3400 + 289
= 13689
Answer: 13689
(ii) (78)²
Use (a − b)² = a² − 2ab + b²
78 = 80 − 2
(78)² = (80 − 2)²
= 6400 − 320 + 4
= 6084
Answer: 6084
(iii) (198)²
198 = 200 − 2
(198)² = (200 − 2)²
= 40000 − 800 + 4
= 39204
Answer: 39204
(iv) (214)²
214 = 200 + 14
(214)² = (200 + 14)²
= 40000 + 5600 + 196
= 45796
Answer: 45796
(v) (1104)²
1104 = 1100 + 4
(1104)² = (1100 + 4)²
= 1210000 + 8800 + 16
= 1218816
Answer: 1218816
(vi) (1120)²
1120 = 1100 + 20
(1120)² = (1100 + 20)²
= 1210000 + 44000 + 400
= 1254400
Answer: 1254400
Q2: Factor using suitable identities.
(i) 16y² − 24y + 9
= (4y)² − 2(4y)(3) + 3²
= (4y − 3)²
Answer: (4y − 3)²
(ii) (9/4)s² + 6st + 4t²
= (3s/2)² + 2(3s/2)(2t) + (2t)²
= (3s/2 + 2t)²
Answer: (3s/2 + 2t)²
(iii)
m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²
Observe:
(m/3 + k/2 + 3n)²
= m²/9 + k²/4 + 9n²
+ mk/3 + 2mn + 3nk
which matches the given expression.
Answer:
(m/3 + k/2 + 3n)²
(iv) p²/16 − 2 + 16/p²
= (p/4)² − 2(p/4)(4/p) + (4/p)²
= (p/4 − 4/p)²
Answer:
(p/4 − 4/p)²
(v) 9a² + 4b² + c² − 12ab + 6ac − 4bc
= (3a)² + (-2b)² + c²
+ 2(3a)(-2b)
+ 2(3a)(c)
+ 2(-2b)(c)
= (3a − 2b + c)²
Answer:
(3a − 2b + c)²
Q3: Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(i) (p + 3q + 7r)²
= p² + (3q)² + (7r)²
+ 2(p)(3q)
+ 2(3q)(7r)
+ 2(p)(7r)
= p² + 9q² + 49r² + 6pq + 42qr + 14pr
Answer:
p² + 9q² + 49r² + 6pq + 42qr + 14pr
(ii) (3x − 2y + 4z)²
= (3x)² + (-2y)² + (4z)²
+ 2(3x)(-2y)
+ 2(-2y)(4z)
+ 2(3x)(4z)
= 9x² + 4y² + 16z²
− 12xy − 16yz + 24xz
Answer:
9x² + 4y² + 16z² − 12xy − 16yz + 24xz
Q4: Is this an identity?
(a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c²
Solution:
Expand each square:
(a + b − c)²
= a² + b² + c² + 2ab − 2ac − 2bc
(a − b + c)²
= a² + b² + c² − 2ab + 2ac − 2bc
(a − b − c)²
= a² + b² + c² − 2ab − 2ac + 2bc
Adding all three:
= 3a² + 3b² + 3c²
− 2ab − 2ac − 2bc
This is not equal to
2a² + 2b² + 2c²
for all values of a, b and c.
Verification:
Take a = b = c = 1
LHS = 1² + 1² + (-1)² = 3
RHS = 2 + 2 + 2 = 6
3 ≠ 6
Answer: No, this is not an identity.