RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.1

RD Sharma Class 9 Solutions Chapter 4 Ex 4.1 Free Download

RD Sharma class 9 mathematics chapter 4 exercise 4.1 Algebraic Identities is provided here. BYJU’S experts have prepared RD Sharma Class 9 chapter 4 solutions in accordance with CBSE syllabus for class 9. Studying this chapter will help you to gain extra knowledge of algebraic identities. Students can download RD Sharma exercise 4.1 by clicking on the link below.

Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.1

 

RD Sharma Solution class 9 Maths Chapter 4 Algebraic Identities 01
RD Sharma Solution class 9 Maths Chapter 4 Algebraic Identities 02
RD Sharma Solution class 9 Maths Chapter 4 Algebraic Identities 03
RD Sharma Solution class 9 Maths Chapter 4 Algebraic Identities 04
RD Sharma Solution class 9 Maths Chapter 4 Algebraic Identities 05

 

Access Answers to Maths RD Sharma Class 9 Chapter 4 Algebraic Identities Exercise 4.1 Page number 4.6

Exercise 4.1 Page No: 4.6

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

[Use identity: (a – b)2 = a2 + b2 – 2ab ]

(2x – 1/x)2 = (2x)2 + (1/x)2 – 2 (2x)(1/x)

= 4x2 + 1/x2 – 4

(ii) (2x + y) (2x – y)

[Use identity: (a – b)(a + b) = a2 – b2 ]

(2x + y) (2x – y) = (2x )2 – (y)2

= 4x2 y2

(iii) (a2b – b2a)2

[Use identity: (a – b)2 = a2 + b2 – 2ab ]

(a2b – b2a)2 = (a2b) 2 + (b2a)2 – 2 (a2b)( b2a)

= a4b 2 + b4a2 – 2 a3b3

(iv) (a – 0.1) (a + 0.1)

[Use identity: (a – b)(a + b) = a2 – b2 ]

(a – 0.1) (a + 0.1) = (a)2 – (0.1)2

= (a)2 – 0.01

(v) (1.5 x2 – 0.3y2) (1.5 x2 + 0.3y2)

[Use identity: (a – b)(a + b) = a2 – b2 ]

(1.5 x2 – 0.3y2) (1.5x2 + 0.3y2) = (1.5 x2 ) 2 – (0.3y2)2

= 2.25 x4 – 0.09y4

Question 2: Evaluate each of the following using identities:

(i)(399)2

(ii)(0.98)2

(iii)991 x 1009

(iv) 117 x 83

Solution:

(i)

RD sharma class 9 maths chapter 4 ex 4.1 question 2 part 1 Solution

(ii)

RD sharma class 9 maths chapter 4 ex 4.1 question 2 part 2 Solution

(iii)

RD sharma class 9 maths chapter 4 ex 4.1 question 2 part 3 Solution

(iv)

RD sharma class 9 maths chapter 4 ex 4.1 question 2 part 4 Solution

Question 3: Simplify each of the following:

(i) 175 x 175 +2 x 175 x 25 + 25 x 25

(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22

(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

(iv)

RD sharma class 9 maths chapter 4 ex 4.1 question 3

Solution:

(i) 175 x 175 +2 x 175 x 25 + 25 x 25 = (175)2 + 2 (175) (25) + (25)2

= (175 + 25)2

[Because a2+ b2+2ab = (a+b)2 ]

= (200)2

= 40000

So, 175 x 175 +2 x 175 x 25 + 25 x 25 = 40000.

(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22

= (322)2 – 2 x 322 x 22 + (22)2

= (322 – 22)2

[Because a2+ b2-2ab = (a-b)2]

= (300)2

= 90000

So, 322 x 322 – 2 x 322 x 22 + 22 x 22= 90000.

(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

= (0.76) 2 + 2 x 0.76 x 0.24 + (0.24) 2

= (0.76+0.24) 2

[ Because a2+ b2+2ab = (a+b)2]

= (1.00)2

= 1

So, 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.

(iv)

RD sharma class 9 maths chapter 4 ex 4.1 question 3 part 4 solution

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

Solution:

RD sharma class 9 maths chapter 4 ex 4.1 question 4

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

Solution:

RD sharma class 9 maths chapter 4 ex 4.1 question 5 solution


Access other exercise solutions of Class 9 Maths Chapter 4 Algebraic Identities

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Exercise 4.4 Solutions

Exercise 4.5 Solutions

Exercise VSAQs Solutions

RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.1

Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.1 is based on the following algebraic Identities:

  • (a + b)2 = a2 + 2ab + b2
  • (a – b)2 = a2 – 2ab + b2
  • (a – b)(a + b) = a2 – b2
  • (x + a)(x + b) = x2 + (a + b)x + ab

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