RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.3

RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.3 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Exercise 9.3.

 Board RBSE Textbook SIERT, Rajasthan Class Class 8 Subject Maths Chapter Chapter 9 Chapter Name Algebraic Expressions Exercise Exercise 9.3 Number of Questions 8 Category RBSE Solutions

Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.3

Question 1.
Find the products of the following using suitable identity
(i) (x + 5) (x + 5)
(ii) (3x + 2) (3x + 2)
(iii) (5a – 7) (5a – 7)
(iv) (3p – $$\frac { 1 }{ 2 }$$) (3p – $$\frac { 1 }{ 2 }$$)
(v) (1.2m – 0.3) (1.2m – 0.3)
(vi) (x² + y²) (x² – y²)
(vii) (6y + 7) (- 6y + 7)
(viii) (7a + 9b) (7a – 9b)
Solution
(i) (x + 5) (x + 5)
= (x + 5)²
= (x)² + 2 (x) (5) + (5)² (Using identity I)
= x² + 10x + 25

(ii) (3x + 2) (3x + 2)
= (3x + 2)²
= (3x)² + 2 (3x) (2) + (2)² (Using identity I)
= 9x² + 12x + 4

(iii) (5a – 7) (5a – 7)
= (5a – 7)²
= (5a)² – 2 (5a) (7) + (7)² (Using identity II)
= 25 a² – 70a + 49

(iv) (3p – $$\frac { 1 }{ 2 }$$) (3p – $$\frac { 1 }{ 2 }$$)

(v) (1.2m – 0.3) (1.2m – 0.3)
= (1.2m – 0.3)²
= (1.2 m)² – 2 (1.2 m) (0.3) + (0.3)² (Using identity II)
= 1.44 m² – 0.72 m + 0.09

(vi) (x² + y²) (x² – y²)
= (x²)² – (y²)²(Using identity II)
= x4 – y4

(vii) (6y + 7) (- 6y + 7)
= (7 + 6y)(7 – 6y)
= (7)² – (6y)² (Using identity II)
= 49 – 36y²

(viii) (7a + 9b) (7a – 9b)
= (7a)² – (9b)² (Using identity III)
= 49a² – 81b²

Question 2.
Use the identity (x + a) (x + b) = x² + (a + b) x + ab to find the following products:
(i) (x + 1) (x + 2)
(ii) (3x + 5) (3x + 1)
(iii) (4x – 5) (4x – 1)
(iv) (3a + 5) (3a – 8)
(v) (xyz – 1) (xyz – 2)
Solution
(i) (x + 1) (x + 2)
= x² + (1 + 2)x + 1 × 2 (Using given identity)
= x² + 3x + 2

(ii) (3x + 5) (3x + 1)
= (3x)² + (5 + 1) 3x + 5 x 1 (Using(RBSESolutions.com)given identity)
= 9x² + 18x + 5

(iii) (4x – 5) (4x – 1)
= {4x + (- 5)} {4x + {- 1)}
= (4x)² + {(- 5) + (- 1)} 4x + (- 5) (- 1) (Using given identity)
= 16x² + (- 6) 4x + (5)
= 16x² – 24x + 5

(iv) (3a + 5) (3a – 8)
= (3a + 5) {3a + (- 8)}
= (3a)² + {5 + (- 8)} 3a + (5) (- 8) (Using given identity)
= 9a² + (- 3) 3a – 40
= 9a² – 9a – 40

(v) {xyz – 1) (xyz – 2)
= {xyz + {- 1)} {xyz + {- 2)}
= {xyz)² + {(- 1) + (- 2)} xyz + (- 1) (- 2)
= x²y²z² – 3xyz + 2

Question 3.
Find the following squares by using the identities.
(i) (b – 7)²
(ii) (xy + 3z)²
(iii) (6m² – 5n)²
(iv) $${ \left( \frac { 3 }{ 2 } x+\frac { 2 }{ 3 } y \right) }^{ 2 }$$
Solution
(i) (b – 7)²
= (b)² – 2 (b) (7) + (7)² (Using identity II)
= b² – 14b + 49

(ii) (xy + 3z)²
= (xy)² + 2 (xy) (3z) + (3z)² (Using identity I)
= x²y² + 6xyz + 9z²

(iii) (6m² – 5n)²
= (6m²)² – 2(6m²) (5n) (5n)² (Using identity II)
= 36m4 – 60m²n + 25n²

(iv) $${ \left( \frac { 3 }{ 2 } x+\frac { 2 }{ 3 } y \right) }^{ 2 }$$

Question 4.
Simplify
(i) (a² – b²)²
(ii) (2n + 5)² – (2n – 5)²
(iii) (7m – 8n)² + (7m + 8n)²
(iv) (m² – n²m)² + 2m3
Solution
(i) (a² – b²)²
= (a²)² – 2(a²) (b²) + (b²)² (Using identity II)
= a4 – 2a²b² + b4

(ii) (2n + 5)² – (2n – 5)²
= {(2n)² + 2 (2n) (5) + (5)²} – {(2n)² – 2(2n) (5) + (5)²}
(Using identity I and II)
= (4n² + 20n + 25) – (4n² – 20n + 25)
= 4n² + 20n + 25 – 4n² + 20n – 25
= 4n² – 4n² + 20n + 20n + 25 – 25
= 0 + 40n + 0
= 40n
Alternative Method-
(2n + 5)² – (2n – 5)²
= {(2n + 5) + (2n – 5)} {(2n + 5) – (2n – 5)}
(Using identity II)
= (4n) (10)
= 40n

(iii) (7m – 8n)² + (7m + 8n)²
= {(7m)² – 2(7m) (8n) + (8n)²} + {(7m)² + 2(7m) (8n) + (8n)²} (Using identity I and II)
= (49m² – 112mn + 64n²) + (49m² + 112mn + 64n²)
= 49m² – 112mn + 64n² + 49m² + 112 mn + 64 n²
= 49m² + 49m² – 112mn + 112mn + 64m² + 64n²
= 98m² + 128n²

(iv) (m² – n²m)² + 2m3
= [(m²)² – 2(m²) (n²m) + (n²m)²] + 2m3n² (Using identity II)
= m4 – 2m3n² + n4m² + 2m3
= m4 – 2m3n² + 2m3n² + n4
= m + n4

Question 5.
Show that
(i) (2a + 3b)² – (2a – 3b)² = 24ab
(ii) (4x + 5)² – 80x = (4x – 5)²
(iii) (3x – 2y)² + 24xy = (3x + 2y)²
(iv) (a – b) (a + b) + (b – c)(b + c) + (c – a) (c + a) = 0
Solution
(i) LHS
= (2a + 3b)² – (2a – 3b)²
= [(2a)² + 2(2a) (3b) + (3b)²] – [(2a)² – 2(2a) (3b) + (3b)²] (Using identity I and II)
= (4a² + 12ab + 9b²) – (4a² – 12ab + 9b²)
= 4a² + 12ab + 9b² – 4a² + 12ab – 9b²
= (4a² – 4a²) + (12ab + 12ab) + (9b² – 9b²)
= 0 + 24ab + 0
= 24 ab
= RHS

(ii) LHS
(4x + 5)² – 80x
= [(4x)² + 2(4x) (5) + (5)²] – 80x (Using identity II)
= (16x² + 40x + 25) – 80x
= 16x² + 40x – 80x + 25
= 16x² – 40x + 25
= (4x)² – 2(4x) (5) + (5)²
= (4x – 5)² (Using identity II)
= RHS

(iii) LHS
(3x – 2y)² + 24xy
= [(3x)² – 2(3x) (2y) + (2y)²] + 24xy
= (9x² – 12xy + 4y²) + 24xy
= 9x² – 12xy + 24xy + 4y²
= 9x² + 12xy + 4y²
= (3x)² + 2 (3x) (2y) + (2y)²
= (3x + 2y)² (Using identity I)
= RHS

(iv) LHS
(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= (a² – b²) + (b² – c²) + (c² – a²)
= a² – b² + b² – c² + c² – a²
= (a² – a²) + (b² – b²) + (c² – c²)
= 0 + 0 + 0 + 0
= RHS

Question 6.
Using identities, evaluate the following
(i) 99²
(ii) 103²
(iii) 297 x 303
(iv) 78 x 82
Solution
(i) 99²
= (100 – 1)²
= (100)² – 2 (100) (1) + (1)²
(Using identity II)
= 10000 – 200 + 1
= 10000 + 1 – 200
= 10001 – 200
= 9801

(ii) 103²
= (100 + 3)²
= (100)² + 2 (100) (3) + (3)²
(Using identity I)
= 10000 + 600 + 9
= 10609

(iii) 297 x 303
= (300 – 3) x (300 + 3)
= (300)² – (3)² (Using identity III)
= 90000 – 9
= 89991

(iv) 78 x 82
= (80 – 2) x (80 + 2)
= (80)² – (2)² (Using identity III)
= 6400 – 4
= 6396

Question 7.
Using a² – b² = (a + b) (a – b), find
(i) 101² – 99²
(ii) (10.3)² – (9.7)²
(iii) 153² – 147²
Solution
(i) 101² – 99²
= (101 + 99) (101 – 99)
(Using given identity)
= (200) (2)
= 400

(ii) (10.3)² – (9.7)²
= (10.3 + 9.7) (10.3 – 9.7)
(Using given identity)
= (20) (0.6)
= 12

(iii) 153² – 147²
= 1 (153 + 147) (153 – 147)
(Using given identity)
= (300) (6)
= 1800

Question 8.
Using a² – b² = (a + b) (a – b), find
(i) 103 x 102
(ii) 7.1 x 7.3
(iii) 102 x 99
(iv) 9.8 x 9.6
Solution
(i) 103 x 102
= (100 + 3) x (100 + 2)
= (100)2 + (3 + 2) 100 + 3 x 2
(Using given identity)
= 10000 + 500 + 6
= 10506

(ii) 7.1 x 7.3
= (7 + 0.1) x (7 + 0.3)
= (7)² + (0.1 + 0.3) 7 + (0.1) x (0.3)
(Using given identity)
= 49 + 2.8 + 0.03
= 51.83

(iii) 102 x 99
= (100 + 2) x (100 – 1)
= (100 + 2) x {100 + (-1)}
= (100)² + {2 + (- 1)} 100 + (2) (- 1)
(Using given identity)
= 10000 + 100 – 2
= 10098

(iv) 9.8 x 9.6
= (10 – 0.2) x (10 – 0.4) .
= {10 + (-0.2)} x {10 +(-0.4)}
= (10)² + {(- 0.2) + (- 0.4)} 10 + (- 0.2) (- 0.4) (Using given identity)
= 100 – 6 + 0.08
= 94.08

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