Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3 Solutions: Exploring Algebraic Identities

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3 Solutions cover square identities, factorisation using identities and expansion of three-term algebraic expressions.

Exercise 4.3 brings together the identities students have learnt so far in Chapter 4. In this exercise, students use (a + b)², (a − b)² and (a + b + c)² to find numerical squares, factor algebraic expressions and expand expressions with three terms.

In Ganita Manjari Class 9 Chapter 4 Exercise 4.3, the focus is on choosing the right identity for each question. Some questions are easier with (a − b)², such as 198², while others need the identity a plus b plus c whole square, such as (p + 3q + 7r)². These Class 9 Maths Chapter 4 Exercise 4.3 Solutions from Exploring Algebraic Identities Exercise 4.3 give clear steps for Class 9 Maths algebraic identities answers, factorisation and expansion.

Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3 Solutions

Exercise 4.3 Question 1

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

(i) 117²

Solution:

Write 117 as:

117 = 100 + 10 + 7

Use:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

So,

117² = (100 + 10 + 7)²

= 100² + 10² + 7² + 2(100)(10) + 2(10)(7) + 2(7)(100)

= 10000 + 100 + 49 + 2000 + 140 + 1400

= 13689

Answer: 117² = 13689

(ii) 78²

Solution:

Write 78 as:

78 = 80 − 2

Use:

(a − b)² = a² − 2ab + b²

So,

78² = (80 − 2)²

= 80² − 2(80)(2) + 2²

= 6400 − 320 + 4

= 6084

Answer: 78² = 6084

(iii) 198²

Solution:

Write 198 as:

198 = 200 − 2

So,

198² = (200 − 2)²

= 200² − 2(200)(2) + 2²

= 40000 − 800 + 4

= 39204

Answer: 198² = 39204

(iv) 214²

Solution:

Write 214 as:

214 = 200 + 10 + 4

Use the identity for the square of three terms.

214² = (200 + 10 + 4)²

= 200² + 10² + 4² + 2(200)(10) + 2(10)(4) + 2(4)(200)

= 40000 + 100 + 16 + 4000 + 80 + 1600

= 45796

Answer: 214² = 45796

(v) 1104²

Solution:

Write 1104 as:

1104 = 1100 + 4

Use:

(a + b)² = a² + 2ab + b²

So,

1104² = (1100 + 4)²

= 1100² + 2(1100)(4) + 4²

= 1210000 + 8800 + 16

= 1218816

Answer: 1104² = 1218816

(vi) 1120²

Solution:

Write 1120 as:

1120 = 1100 + 20

So,

1120² = (1100 + 20)²

= 1100² + 2(1100)(20) + 20²

= 1210000 + 44000 + 400

= 1254400

Answer: 1120² = 1254400

Exercise 4.3 Question 2

Factor using suitable identities:

(i) 16y² − 24y + 9

Solution:

16y² = (4y)²

9 = 3²

−24y = −2(4y)(3)

So,

16y² − 24y + 9 = (4y)² − 2(4y)(3) + 3²

Using:

a² − 2ab + b² = (a − b)²

we get:

16y² − 24y + 9 = (4y − 3)²

Answer: (4y − 3)²

(ii) 9/4 s² + 6st + 4t²

Solution:

9/4 s² = (3s/2)²

4t² = (2t)²

6st = 2(3s/2)(2t)

So,

9/4 s² + 6st + 4t² = (3s/2)² + 2(3s/2)(2t) + (2t)²

Therefore:

9/4 s² + 6st + 4t² = (3s/2 + 2t)²

Answer: (3s/2 + 2t)²

(iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²

Solution:

Write the expression as the square of three terms.

m²/9 = (m/3)²

k²/4 = (k/2)²

9n² = (3n)²

Now check the middle terms.

2(m/3)(k/2) = mk/3

2(k/2)(3n) = 3kn

2(m/3)(3n) = 2mn

So,

m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²

= (m/3)² + (k/2)² + (3n)² + 2(m/3)(k/2) + 2(k/2)(3n) + 2(m/3)(3n)

= (m/3 + k/2 + 3n)²

Answer: (m/3 + k/2 + 3n)²

(iv) p²/16 − 2 + 16/p²

Solution:

p²/16 = (p/4)²

16/p² = (4/p)²

−2 = −2(p/4)(4/p)

So,

p²/16 − 2 + 16/p² = (p/4)² − 2(p/4)(4/p) + (4/p)²

Therefore:

p²/16 − 2 + 16/p² = (p/4 − 4/p)²

Answer: (p/4 − 4/p)²

(v) 9a² + 4b² + c² − 12ab + 6ac − 4bc

Solution:

This expression can be compared with:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here, take the three terms as:

3a, −2b and c

Now:

(3a)² = 9a²

(−2b)² = 4b²

c² = c²

2(3a)(−2b) = −12ab

2(3a)(c) = 6ac

2(−2b)(c) = −4bc

Therefore:

9a² + 4b² + c² − 12ab + 6ac − 4bc = (3a − 2b + c)²

Answer: (3a − 2b + c)²

Exercise 4.3 Question 3

Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:

(i) (p + 3q + 7r)²

Solution:

Here:

a = p
b = 3q
c = 7r

Using:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

we get:

(p + 3q + 7r)²

= p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)

= p² + 9q² + 49r² + 6pq + 42qr + 14pr

Answer: p² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii) (3x − 2y + 4z)²

Solution:

Here:

a = 3x
b = −2y
c = 4z

Using the identity:

(3x − 2y + 4z)²

= (3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x)

= 9x² + 4y² + 16z² − 12xy − 16yz + 24xz

Answer: 9x² + 4y² + 16z² − 12xy − 16yz + 24xz

Exercise 4.3 Question 4

Is this an identity?

(a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c²

Solution:

To check whether it is an identity, test it with simple values.

Take:

a = 1, b = 1, c = 1

Now calculate the left-hand side.

(a + b − c)² = (1 + 1 − 1)² = 1² = 1

(a − b + c)² = (1 − 1 + 1)² = 1² = 1

(a − b − c)² = (1 − 1 − 1)² = (−1)² = 1

So,

LHS = 1 + 1 + 1 = 3

Now calculate the right-hand side.

RHS = 2a² + 2b² + 2c²

= 2(1)² + 2(1)² + 2(1)²

= 2 + 2 + 2

= 6

Since:

3 ≠ 6

the given statement is not true for all values of a, b and c.

Answer: No, this is not an identity.

NCERT Solutions for Class 9 Maths Chapter 4 - Related Links

• NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1
• NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.2
• NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.4
• NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.5
• NCERT Solutions for Class 9 Maths Chapter 4 End-of-Chapter Exercises