Class 8 maths ch 9 ex 9.5

NCERT Solutions for class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4 NCERT Solutions have been structured in logical and easy language for quick revisions. Well-illustrated solutions for CBSE Class 8 Maths Exercise 9.4 are extremely accurate and solved using a step-by-step problem-solving approach. This exercise helps students to learn how to perform the multiplication of two polynomials. Download free Download PDF carouselExampleControls112 Previous Next Access Answers to NCERT Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4 Page number 148  Exercise 9.4 Page No: 148 1. Multiply the binomials. (i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4) (iii) (2.5l – 0.5m) and (2.5l + 0.5m) (iv) (a + 3b) and (x + 5) (v) (2pq + 3q 2) and (3pq – 2q 2) (vi) (3/4 a 2 + 3b 2) and 4( a 2 – 2/3 b 2 ) Solution : (i) (2x + 5)(4x – 3) = 2x x 4x – 2x x 3 + 5 x 4x – 5 x 3 = 8x² – 6x + 20x -15 = 8x² + 14x -15 (ii) ( y – 8)(3y – 4) = y x 3y – 4y – 8 x 3y + 32 = 3y 2 – 4y – 24y + 32 = 3y 2 – 28y + 32 (iii) (2.5l  – 0.5m)(2.5l + 0.5m) = 2.5l x 2.5 l + 2.5l x 0.5m – 0.5m x 2.5l – 0.5m x 0.5m = 6.25l 2 + 1.25 lm – 1.25 lm – 0.25 m 2 = 6.25l 2– 0.25 m 2 (iv) (a + 3b) (x + 5) = ax + 5a + 3bx + 15b (v) (2pq + 3q 2) (3pq – 2q 2) = 2pq x 3pq – 2pq x 2q 2 + 3q 2 x 3pq – 3q 2 x 2q 2 = 6p 2q 2 – 4pq 3 + 9pq 3 – 6q 4 = 6p 2q 2 + 5pq 3 – 6q 4 (vi) (3/4 a ² + 3b...

Table of Contents • • • • • • • • • • • • • NCERT Solutions for Class 8 Maths Exercise 9.5 solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 8 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free. Class 8 Maths Algebraic Expressions and Identities NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 1. Use a suitable identity to get each of the following products: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) Ans. (i) [Using identity ] = (ii) = [Using identity ] = (iii) = [Using identity ] = (iv) = [Using identity ] = (v) [Using identity ] = (vi) = [Using identity ] = (vii) [Using identity ] = (viii) = [Using identity ] = (ix) = [Using identity ] = (x) = [Using identity ] = 2. Use the identity to find the following products: (i) (ii) (iii) (iv) (v) (vi) (vii) Ans. (i) [Using identity ] = (ii) [Using identity ] = (iii) = [Using identity ] = (iv) [Using identity ] = = = (v) [Using identity ] = = (vi) [Using identity ] = = (vii) [Using identity ] = NCERT Solutions for Class 8 Maths Exercise 9.5 3. Find the following squares by using identities: (i) (ii) (iii) (iv) (v) (vi) Ans. (i) [Using identity ] = (ii) [Using identity ] = (iii) [Using identity ] = (iv) [Using iden...

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5, are part of NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Question 1. Use a suitable identity to get each of the following products: Solution. Question 2. Use the identity \((x+a)(x+b)=+(a+b)x+ab\), find (i) 103 x 104 (ii) 5.1 x 5.2 (iii) 103 x 98 (iv) 9.7 x 9.8 Solution. We hope the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5, help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5, drop a comment below and we will get back to you at the earliest. Filed Under: Tagged With:

September 26, 2021bySuman NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 • • • • • • NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Ex 9.5 Class 8 Maths Question 1. Use a suitable identity to get each of the following products: (i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a – 7) (2a – 7) (iv) (3a –1/2) (3a –1/2) (v) (1.1m – 0.4) (1.1m + 0.4) (vi) (a 2+ b 2) (-a 2+ b 2) (vii) (6x – 7) (6x + 7) (viii) (-a + c) (-a + c) (ix) (x/2+3y/4) (x/2+3y/4) (x) (7a – 9b) (7a – 9b) Solution: Ex 9.5 Class 8 MathsQuestion 2. Use the identity (x + a)(x + b) = x 2+ (a + b)x + ab to find the following products. (i) (x + 3) (x + 7) (ii) (4x + 5)(4x + 1) (iii) (4x – 5) (4x – 1) (iv) (4x + 5) (4x – 1) (v) (2x + 5y) (2x + 3y) (vi) (2a 2+ 9) (2a 2+ 5) (vii) (xyz – 4) (xyz – 2) Solution: Ex 9.5 Class 8 MathsQuestion 3. Find the following squares by using the identities. (i) (b – 7) 2 (ii) (xy + 3z) 2 (iii) (6x 2– 5y) 2 (iv) (2/3m +3/2n) 2 (v) (0.4p – 0.5q) 2 (vi) (2xy + 5y) 2 Solution: Ex 9.5 Class 8 MathsQuestion 4. Simplify: (i) (a 2– b 2) 2 (ii) (2x + 5) 2– (2x – 5) 2 (iii) (7m – 8n) 2+ (7m + 8n) 2 (iv) (4m + 5n) 2+ (5m + 4n) 2 (v) (2.5p – 1.5q) 2– (1.5p – 2.5q) 2 (vi) (ab + bc) 2– 2ab 2c (vii) (m 2– n 2m) 2+ 2m 3n 2 Solution: Ex 9.5 Class 8 MathsQuestion 5. Show that: (i) (3x + 7) 2– 84x = (3x – 7) 2 (ii) (9p – 5q) 2+ 180pq = (9p + 5q) 2 (iii) (4/3m –3/4n) 2+ 2mn =16/9m 2+9/16n 2 (iv) (4pq + 3q) 2– (4p...

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 • • • • • • NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3 Ex 9.3 Class 8 Maths Question 1. Carry out the multiplication of the expressions in each of the following pairs: (i) 4p, q + r (ii) ab, a – b (iii) a + b, 7a 2b 2 (iv) a 2– 9, 4a (v) pq + qr + rp, 0 Solution: (i) 4p × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr (ii) ab, a – b = ab × (a – b) = (ab × a) – (ab × b) = a 2b – ab 2 (iii) (a + b) × 7a 2b 2 = (a × 7a 2b 2) + (b × 7a 2b 2) = 7a 3b 2 + 7a 2b 3 (iv) (a 2– 9) × 4a = (a 2 × 4a) – (9 × 4a) = 4a 3– 36a (v) (pq + qr + rp) × 0 = 0 [∵ Any number multiplied by 0 is = 0] Ex 9.3 Class 8 Maths Question 2. Complete the table. S.No. First Expression Second Expression Product (i) a b + c + d – (ii) x + y – 5 5xy – (iii) p 6p 2– 7p + 5 – (iv) 4p 2q 2 p 2– q 2 – (v) a + b + c abc – Solution: (i) a × (b + c + d) = (a × b) + (a × c) + (a × d) = ab + ac + ad (ii) (x + y – 5) (5xy) = (x × 5xy) + (y × 5xy) – (5 × 5xy) = 5x 2y + 5xy 2– 25xy (iii) p × (6p 2– 7p + 5) = (p × 6p 2) – (p × 7p) + (p × 5) = 6p 3– 7p 2 + 5p (iv) 4p 2q 2 × (p 2– q 2) = 4p 2q 2 × p 2– 4p 2q 2 × q 2 = 4p 4q 2– 4p 2q 4 (v) (a + b + c) × (abc) = (a × abc) + (b × abc) + (c × abc) = a 2bc + ab 2c + abc 2 Completed Table: S.No. First Expression Second Expression Product (i) a b + c + d ab + ac + ad (ii) x + y – 5 5xy 5x 2y + 5xy 2– 25xy (iii) p 6p 2– 7p + 5 6p 3– 7p 2 + 5p (iv) 4p 2...

NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.5 Algebraic Expressions & Identities ncert solutions for class 8 maths chapter 9 exercise 9.5 contains 8 questions and each question has explained in detail and stepwise. If you are class 8th student and currently preparing chapter 9 exercise 9.1 then you must be looking for the class 8 maths chapter 9 exercise 9.5 solution for your exams preparation. Here we are providing complete solutions for class 8 maths chapter 9 exercise 9.5. Table of Content Category Subject Maths Chapter Chapter 9 Exercise Exercise 9.5 Chapter Name Algebraic Expressions & Identities NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.5 – Algebraic Expressions & Identities, has been designed by the NCERT to test the knowledge of the student on the following topics:- • What is an Identity? • Standard Identities • Applying Identities NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 1. Use a suitable identity to get each of the following products. i. (x + 3) (x + 3) Sol.: – (x + 3) (x + 3) = (x + 3) 2 = x 2 + 2 × x × 3 + 3 2 … = 1 + 8.55 + 0.425 = 9.975 7. Using a 2 – b 2 = (a + b) (a – b), find i. 51 2 – 49 2 Sol.: – 51 2 – 49 2 = (51 + 49) (51 – 49) = 100 × 2 = 200 ii. (1.02) 2 – (0.98) 2 Sol.: – (1.02) 2 – (0.98) 2 = (1.02 + 0.98) (1.02 – 0.98) = 2 × 0.04 = 0.08 iii. 153 2 – 147 2 Sol.: – 153 2 – 147 2 = (153 + 147) (153 – 147) = 300 × 6 = 1800 iv. 12.1 2 – 7.9 2 Sol.: – 12.1 2 – 7.9 2 = (12.1 + 7.9) (12.1 – 7.9) = 20 × 4.2 = 84 8. ...

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Question 1. Use a suitable identity to get each of the following products. (i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a – 7) (2a – 7) (iv) (3a – \(\frac\right)\) (x) (7a – 9b) (7a – 9b) Answer: (i) (x + 3) (x + 3) = (x + 3) 2 = x 2 + 2 × x × 3 + 3 2 = x 2 + 6x +9 [Using the identity (a + b) 2 = a 2+ 2ab + b 2] (ii) (2y + 5) (2y + 5) = (2y + 5) 2 = (2y) 2 + 2 × 2y × 5 + 5 2 = 4y 2 + 20y + 25 [Using the identity (a + b) 2 = a 2 + 2ab + b 2 ] (iii) (2a – 7) (2a – 7) = (2a – 7) 2 = (2a) 2– 2 × 2a × 7 + (7) 2 [using the identity (a – b) 2 = a 2 -2ab + b 2] = 4a 2– 28a + 49 (iv) (3a – \(\frac \) × 99.75 = 9.975 Question 7. Using a 2– b 2 = (a + b) (a – b) find (i) 51 2– 49 2 (ii) (1.02) 2– (0.98) 2 (iii) 153 2– 147 2 (iv) 12.1 2– 7.9 2 Answer: (i) 51 2– 49 2 = (51 + 49)(51 – 49) = (100) × 2 = 200 (ii) (1.02) 2– (0.98) 2 = (1.02 + 0.98) (1.02 – 0.98) = 2 × 0.04 = 0.08 (iii) 153 2– 147 2 = (153 + 147) (153 – 147) = 300 × 6 = 1800 (iv) (12.1) 2– (7.9) 2 = (12.1 + 7.9) (12.1 – 7.9) = 20 × 4.2 = 84 Question 8. Using (x + a) (x + b) = x 2 + (a + b) x + ab, find (i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8 Answer: (i) 103 × 104 = (100+ 3) (100 + 4) = 100 2 + (3 + 4) 100 + 3 × 4 = 10000 + 700 + 12 = 10712 (ii) 5.1 × 5.2= (5 + 0.1) (5 + 0.2) = 5 2 + (0.1 +0.2) 5+ 0.1 × 0.2 = 25 + 1.5 + 0.02 = 26.52 (iii) 103 × 98 = (100 + 3) (100 – 2) = 100 2 + (3 – 2) 100 + (3) (- 2)...