Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.1 | Set 1

Chapter 4 of RD Sharma's Class 9 mathematics textbook delves into the crucial concept of Algebraic Identities, with Exercise 4.1 Set 1 specifically focusing on fundamental algebraic identities and their applications in various mathematical contexts. This section forms the cornerstone of algebraic manipulation, introducing students to powerful tools for simplifying expressions, factoring polynomials, and solving equations efficiently. The exercise covers key identities such as $(a + b)^2$, $(a - b)^2$, $a^2 - b^2$, and $(x + a)(x + b)$, among others, providing a solid foundation for more advanced algebraic concepts in higher mathematics.

Question 1. Evaluate each of the following using identities:

(i) (2x – 1/x)²
(ii) (2x + y) (2x – y)
(iii) (a²b – b²a)²
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x² – 0.3y²) (1.5x² + 0.3y²)

Solution:

i) (2x – 1/x)² 

We know that, (a -b)² = a² – 2ab + b²

So, (2x – 1/x)² = (2x)² – (2 × 2x × 1/x) + (1/x)²

= 4x² + 1/x² – 4

ii) (2x + y) (2x – y)

We know that, (a + b)(a – b) = a² – b²

So, (2x + y) (2x – y) = (2x)² – (y)²

= 4x² – y²

iii) (a²b – b²a)²

We know that, (a -b)² = a² – 2ab + b²

So, (a²b – b²a)² = (a²b)² – (2 × a²b × b²a) + (b²a)²

= a⁴b² + b⁴a² – 2a³b³

iv) (a – 0.1) (a + 0.1)

We know that, (a + b)(a – b) = a² – b²

So, (a – 0.1) (a + 0.1) = (a)² – (0.1)²

= a² – 0.01

v) (1.5.x² – 0.3y²) (1.5x² + 0.3y²)

We know that, (a + b)(a – b) = a² – b²

So, (1.5.x² – 0.3y²) (1.5.x² + 0.3y²) = (1.5.x²)² – (0.3y²)²

= 2.225x⁴ – 0.09y⁴

Question 2. Evaluate each of the following using identities:

(i)(399)²
(ii)(0.98)²
(iii)991 × 1009
(iv) 117 × 83

Solution:

i) (399)²

We can write (399)² as (400 – 1)²

Also, (a -b)² = a² – 2ab + b²

= (400)² + (1)² – 2 × 400 ×1

= 160000 + 1 – 800 = 159201

Hence, (399)² = 159201

ii) (0.98)²

We can write (0.98)² as (1 – 0.02)²

Also, (a -b)² = a² – 2ab + b²

= (1)² + (0.02)² – 2 × 0.02 ×1

= 1 + 0.0004 – 0.04 = 0.9604

Hence, (0.98)² = 0.9604

iii) 991 × 1009

We can write 991 × 1009 as (1000 – 9)(1000 + 9)

Also, (a + b)(a – b) = a² – b²

= (1000)² – (9)²

= 1000000 – 81 = 999919

Hence, 991 × 1009 = 999919

iv) 117 × 83

We can write 117 × 83 as (100 + 17)(100 – 17)

Also, (a + b)(a – b) = a² – b²

= (100)² – (17)²

= 10000 – 289 = 9711

Hence, 117 × 83 = 9711

Question 3. Simplify each of the following:

(i) 175 × 175 +2 × 175 × 25 + 25 × 25
(ii) 322 × 322 – 2 × 322 × 22 + 22 × 22
(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24
(iv) (7.83 × 7.83 – 1.17 × 1.17)/6.66

Solution:

i) 175 × 175 +2 × 175 × 25 + 25 × 25

It can be written as (175)² + 2(175)(25) + (25)²

And we also know that, (a + b)² = a² + b² + 2ab

So, we can conclude (175)² + 2(175)(25) + (25)² as (175 + 25)²

= (200)² = 40000

Hence, 175 × 175 +2 × 175 × 25 + 25 × 25 = 40000

ii) 322 × 322 – 2 × 322 × 22 + 22 × 22

It can be written as (322)² – 2(322)(22) + (22)²

And we also know that, (a – b)² = a² + b² – 2ab

So, we can conclude (322)² – 2(322)(22) + (22)² as (322 – 22)²

= (300)² = 90000

Hence, 322 × 322 – 2 × 322 × 22 + 22 × 22 = 90000

iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24

It can be written as (0.76)² + 2(0.76)(0.24) + (0.24)²

And we also know that, (a + b)² = a² + b² + 2ab

So, we can conclude (0.76)² + 2(0.76)(0.24) + (0.24)² as (0.76 + 0.24)²

= (1.0)² = 1

Hence, 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24 = 1

iv) (7.83 × 7.83 – 1.17 × 1.17)/6.66

It can be written as (7.83² – 1.17²)/6.66

And we also know that, (a + b)(a – b) = a² – b²

So, we can conclude (7.83² – 1.17²)/6.66 as [(7.83 + 1.17)(7.83 – 1.17)]/6.66

= (9 × 6.66)/6.66 = 9

Hence, (7.83 × 7.83 – 1.17 × 1.17)/6.66 = 9

Question 4. If x + 1/x = 11, find the value of x² +1/x²

Solution:

Given, x + 1/x = 11

So, (x + 1/x)² = (x)² + (1/x)² + 2 × (x) × (1/x)

We also know that, (a + b)² = a² + b² + 2ab

So,

(11)² = x² + 1/x² + 2

121 – 2 = x² + 1/x²

x² + 1/x² = 119

Hence, value of x² + 1/x² is 119

Question 5. If x – 1/x = -1, find the value of x² +1/x²

Solution:

Given, x – 1/x = -1

So, (x – 1/x)² = (x)² + (1/x)² – 2 × (x) × (1/x)

We also know that, (a – b)² = a² + b² – 2ab

So,

(-1)² = x² + 1/x² – 2

1 + 2 = x² + 1/x²

x² + 1/x² = 3

Hence, value of x² – 1/x² is 3

Question 6. If x + 1/x = √5, find the value of x² +1/x² and x⁴ +1/x⁴

Solution:

Given, x + 1/x = √5

So, (x + 1/x)² = (x)² + (1/x)² + 2 × (x) × (1/x)

We also know that, (a + b)² = a² + b² + 2ab

So,

(√5)² = x² + 1/x² + 2

5 – 2 = x² + 1/x²

x² + 1/x² = 3

Now, taking the square of x² + 1/x²

(x² + 1/x²)² = x⁴ + 1/x⁴ + 2 × (x)² × (1/x²)

(3)² = x⁴ + 1/x⁴ + 2

x⁴ + 1/x⁴ = 7

Hence, value of x² + 1/x² is 3 and that of x⁴ +1/x⁴ is 7

Question 7. If x² +1/x² = 66, find the value of x – 1/x

Solution:

Given, x² +1/x² = 66

Let us take the square of x – 1/x

So, (x – 1/x)² = (x)² + (1/x)² – 2 × (x) × (1/x)

Since, (a – b)² = a² + b² – 2ab

So,

(x – 1/x)² = 66 – 2

(x – 1/x)² = 64

x – 1/x = ±8

Hence, the value of x – 1/x is 8

Summary

Chapter 4 of RD Sharma's Class 9 mathematics textbook, focusing on Algebraic Identities, introduces students to fundamental algebraic tools essential for simplifying expressions, factoring polynomials, and solving equations. Exercise 4.1 Set 1 specifically covers key identities such as $(a + b)^2$, $(a - b)^2$, and $a^2 - b^2$, among others. The practice questions in this set are designed to reinforce understanding and application of these identities, progressing from simple expansions to more complex problems involving multiple identities or real-world applications. Students learn to recognize, apply, and manipulate these identities in various contexts, building a strong foundation for advanced algebraic concepts. The exercise emphasizes the importance of these identities in mathematical problem-solving, highlighting their relevance in higher-level mathematics and real-world scenarios in fields like physics, engineering, and computer science. Through consistent practice and application, students develop crucial skills in algebraic manipulation, pattern recognition, and logical reasoning, preparing them for more advanced mathematical studies.