Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.5 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

1. Use a suitable identity to get each of the following products:

Question (i)

(x + 3) (x + 3)

Solution:

= (x + 3)^{2}

= (x)^{2} + 2(x)(3) + (3)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= x^{2} + 6x + 9

Question (ii)

(2y + 5) (2y + 5)

Solution:

= (2y + 5)^{2}

= (2y)^{2} + 2 (2y)(5) + (5)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 4y^{2} + 20y + 25

Question (iii)

(2a – 7) (2a – 7)

Solution:

= (2a – 7)^{2}

= (2a)^{2} – 2(2a)(7) + (7)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 4a^{2} – 28a + 49

Question (iv)

(3a – \(\frac {1}{2}\))(3a – \(\frac {1}{2}\))

Solution:

= (3a – \(\frac {1}{2}\))^{2}

= (3a)^{2} – 2(3a)(\(\frac {1}{2}\)) + (\(\frac {1}{2}\))^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 9a^{2} – 3a + \(\frac {1}{4}\)

Question (v)

(1.1m – 0.4) (1.1m + 0.4)

Solution:

= (1.1m)^{2} – (0.4)^{2}

[∵ (a + b) (a – b) = a^{2} – b^{2}]

= 1.21m^{2} – 0.16

Question (vi)

(a^{2} + b^{2}) (-a^{2} + b^{2})

Solution:

= (b^{2} + a^{2}) (b^{2} – a^{2})

= (b^{2})^{2} – (a^{2})^{2}

[∵ (a + b) (a – b) = a^{2} – b^{2}]

= b^{4} – a^{4}

Question (vii)

(6x – 7) (6x + 7)

Solution:

= (6x)^{2} – (7)^{2}

[∵ (a + b) (a – b) = a^{2} – b^{2}]

= 36x^{2} – 49

Question (viii)

(-a + c) (-a + c)

Solution:

= (-a + c)^{2}

= (-a)^{2} + 2 (-a) (c) + (c)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= a^{2} – 2ac + c^{2}

Question (ix)

(\(\frac{x}{2}+\frac{3 y}{4}\))

Solution:

= (\(\frac{x}{2}+\frac{3 y}{4}\))^{2}

= (\(\frac {x}{2}\))^{2} + 2(\(\frac {x}{2}\))(\(\frac {3y}{4}\)) + (\(\frac {3y}{4}\))^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= \(\frac{x^{2}}{4}+\frac{3 x y}{4}+\frac{9 y^{2}}{16}\)

Question (x)

(7a – 9b) (7a – 9b)

Solution:

= (7a – 9b)^{2}

= (7a)^{2} – 2(7a)(9b) + (9b)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + bsup>2]

= 49a^{2} – 126ab + 81b^{2}

2. Use the identity (x + a) (x + b) = x^{2} + (a + b) x + ab to find the following products:

Question (i)

(x + 3) (x + 7)

Solution:

Identity : (x + a) (x + b) = x^{2} + (a + b) x + ab

= (x)^{2} + (3 + 7)x + (3) (7)

= x^{2} + (10) x + 21

= x^{2} + 10x + 21

Question (ii)

(4x + 5) (4x + 1)

Solution:

= (4x)^{2} + (5 + 1) 4x + (5)(1)

= 16x^{2} + (6) 4x + 5

= 16x^{2} + 24x + 5

Question (iii)

(4x – 5) (4x – 1)

Solution:

= (4x)^{2} + (- 5 – 1) 4x + (- 5) (- 1)

= 16x^{2} + (- 6) 4x + 5

= 16x^{2} – 24x + 5

Question (iv)

(4x + 5) (4x- 1)

Solution:

= (4x)^{2} + (5 – 1) 4x + (5) (- 1)

= 16x^{2} + (4) 4x – 5

= 16x^{2} + 16x – 5

Question (v)

(2x + 5y) (2x + 3y)

Solution:

= (2x)^{2} + (5y + 3y) 2x + (5y) (3y)

= 4x^{2} + (8y) 2x + 15y^{2}

= 4x^{2} + 16xy + 15y^{2}

Question (vi)

(2a^{2} + 9) (2a^{2} + 5)

Solution:

= (2a^{2})^{2} + (9 + 5) 2a^{2} + (9)(5)

= 4a^{4} + (14)2a^{2} + 45

= 4a^{4} + 28a^{2} + 45

Question (vii)

(xyz – 4) (xyz – 2)

Solution:

= (xyz)^{2} + (- 4 – 2) xyz + (- 4)(- 2)

= x^{2}y^{2}z^{2} + (- 6) xyz + 8

= x^{2}y^{2}z^{2} – 6xyz + 8

3. Find the following squares by using the identities:

Question (i)

(b – 7)^{2}

Solution:

= (b)^{2} – 2 (b)(7) + (7)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= b^{2} – 14 b + 49

Question (ii)

(xy + 3z)^{2}

Solution:

= (xy)^{2} + 2 (xy)(3z) + (3z)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= x^{2}y^{2} + 6xyz + 9z^{2}

Question (iii)

(6x^{2} – 5y)^{2}

Solution:

= (6x^{2})^{2} – 2 (6x^{2}) (5y) + (5y)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 36x^{4} – 60x^{2}y + 25y^{2}

Question (iv)

(\(\frac {2}{3}\)m + \(\frac {3}{2}\)n)^{2}

Solution:

= (\(\frac {2}{3}\)m)^{2} + 2(\(\frac {2}{3}\)m)(\(\frac {3}{2}\)n) + (\(\frac {3}{2}\)n)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= \(\frac {4}{9}\)m^{2} + 2mn + \(\frac {9}{4}\)n^{2}

Question (v)

(0.4p – 0.5q)^{2}

Solution:

= (0.4p)^{2} – 2 (0.4p)(0.5q) + (0.5q)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 0.16p^{2} – 0.4pq + 0.25q^{2}

Question (vi)

(2xy + 5y)^{2}

Solution:

= (2xy)^{2} + 2 (2xy)(5y) + (5y)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 4x^{2}y^{2} + 20xy^{2} + 25 y^{2}

4. Simplify:

Question (i)

(a^{2} – b^{2})^{2}

Solution:

= (a^{2})^{2} – 2(a^{2})(b^{2}) + (b^{2})^{2}

= a^{4} – 2a^{2}b^{2} + b^{4}

Question (ii)

(2x + 5)^{2} – (2x – 5)^{2}

Solution:

= [(2x)^{2} + 2(2x)(5) + (5)^{2}] – [(2x)^{2} – 2 (2x)(5) + (5)^{2}]

= [4x^{2} + 20x + 25] – [4x^{2} – 20x + 25]

= 4x^{2} + 20x + 25 – 4x^{2} + 20x – 25

= 4x^{2} – 4x^{2} + 20x + 20x + 25 – 25

= 40x

Question (iii)

(7m – 8n)^{2} + (7m + 8n)^{2}

Solution:

= [(7m)^{2} – 2(7m)(8n) + (8n)^{2}] + [(7m)^{2} + 2 (7m)(8n) + (8n)^{2}]

= [49m^{2} – 112mn + 64n^{2}] + [49m^{2} + 112mn + 64n^{2}]

= 49m^{2} – 112mn + 64n^{2} + 49m^{2} + 112mn + 64n^{2}

= 49m^{2} + 49m^{2} – 112mn + 112mn + 64n^{2} + 64n^{2}

= 98m^{2} + 128n^{2}

Question (iv)

(4m + 5n)^{2} + (5m + 4n)^{2}

Solution:

= [(4m)^{2} + 2 (4m)(5n) + (5n)^{2}] + [(5m)^{2} + 2 (5m)(4n) + (4n)^{2}]

= 16m^{2} + 40mn + 25n^{2} + 25m^{2} + 40mn + 16n^{2}

= 16m^{2} + 25m^{2} + 40mn + 40 mn + 25n^{2} + 16n^{2}

= 41m^{2} + 80mn + 41n^{2}

Question (v)

(2.5p – 1.5q)^{2} – (1.5p – 2.5q)^{2}

Solution:

= [(2.5p)^{2} – 2 (2.5p)(1.5q) + (1.5q)^{2}] – [(1.5p)^{2} – 2 (1.5p)(2.5q) + (2.5q)^{2}]

= [6.25p^{2} – 7.5pq + 2.25q^{2}] – [2.25p^{2} – 7.5pq + 6.25q^{2}]

= 6.25p^{2} – 7.5pq + 2.25q^{2} – 2.25p^{2} + 7.5pq – 6.25q^{2}

= 6.25p^{2} – 2.25p^{2} – 7.5pq + 7.5pq + 2.25q^{2} – 6.25q^{2}

= 4p^{2} – 4q^{2}

Question (vi)

(ab + bc)^{2} – 2ab^{2}c

Solution:

= [(ab)^{2} + 2 (ab)(bc) + (bc)^{2}] – 2ab^{2}c

= a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2} – 2ab^{2}c

= a^{2}b^{2} + 2ab^{2}c – 2ab^{2}c + b^{2}c^{2}

= a^{2}b^{2} + b^{2}c^{2}

Question (vii)

(m^{2} – n^{2}m)^{2} + 2m^{3}n^{2}

Solution:

= [(m^{2})^{2} – 2 (m^{2})(n^{2}m)^{2} + (n^{2}m)^{2}] + 2m^{3}n^{2}

= m^{4} – 2 m^{3}n^{2} + n^{4}m^{2} + 2 m^{3 }n^{2}

= m^{4} – 2 m^{3}n^{2} + 2 m^{3}n^{2} + n^{4}m^{2}

= m^{4} + m^{2}n^{4}

5. Show that:

Question (i)

(3x + 7)^{2} – 84x = (3x – 7)^{2}

Solution:

LHS = (3x + 7)^{2} – 84x

= (3x)^{2} + 2(3x)(7) + (7)^{2} – 84x

= 9x^{2} + 42x + 49 – 84x

= 9x^{2} + 42x – 84x + 49

= 9x^{2} – 42x + 49

RHS = (3x – 7)^{2}

= (3x)^{2} – 2(3x)(7) + (7)^{2}

= 9x^{2} – 42x + 49

Thus, LHS = RHS

∴ (3x + 7)^{2} – 84x = (3x – 7)^{2}

Question (ii)

(9p – 5q)^{2} + 180pq = (9p + 5q)^{2}

Solution:

LHS = (9p – 5q)^{2} + 180pq

= (9p)^{2} – 2(9p)(5q) + (5q)^{2} + 180pq

= 81p^{2} – 90pq + 25q^{2} + 180pq

= 81p^{2} – 90pq + 180pq + 25q^{2}

= 81p^{2} + 90pq + 25q^{2}

RHS = (9p + 5q)^{2}

= (9p)^{2} + 2(9p)(5q) + (5q)^{2}

= 81p^{2} + 90pq + 25q^{2}

Thus, LHS = RHS

∴ (9p – 5q)^{2} + 180pq = (9p + 5q)^{2}

Question (iii)

(\(\frac {4}{3}\)m – \(\frac {3}{4}\)n)^{2} + 2mn = \(\frac {16}{9}\)m^{2} + \(\frac {9}{16}\)n^{2}

Solution:

LHS = [\(\frac {4}{3}\)m – \(\frac {3}{4}\)n]^{2} + 2mn

= [(\(\frac {4}{3}\)m)^{2} – 2(\(\frac {4}{3}\)m)(\(\frac {3}{4}\)n) + (\(\frac {3}{4}\)n)^{2}] + 2mn

= \(\frac {16}{9}\)m^{2} – 2mn + \(\frac {9}{16}\)n^{2} + 2mn

= \(\frac {16}{9}\)m^{2} – 2mn + 2mn + \(\frac {9}{16}\)n^{2}

= \(\frac {16}{9}\)m^{2} + \(\frac {9}{16}\)n^{2} = RHS

Thus, LHS = RHS

∴ (\(\frac {4}{3}\)m – \(\frac {3}{4}\)n)^{2} + 2mn = \(\frac {16}{9}\)m^{2} + \(\frac {9}{16}\)n^{2}

Question (iv)

(4pq + 3q)^{2} – (4pq – 3q)^{2} = 48pq^{2}

Solution:

LHS = (4pq + 3q)^{2} – (4pq – 3q)^{2}

= [(4pq)^{2} + 2 (4pq)(3q) + (3q)^{2}] – [(4pq)^{2} – 2 (4pq)(3q) + (3q)^{2}]

= [16p^{2}q^{2} + 24pq^{2} + 9q^{2}] – [16p^{2}q^{2} – 24pq^{2} + 9q^{2}]

= 16p^{2}q^{2} + 24pq^{2} + 9q^{2} – 16p^{2}q^{2} + 24pq^{2} – 9q^{2}

= 16p^{2}q^{2} – 16p^{2}q^{2} + 24pq^{2} + 24pq^{2} + 9q^{2} – 9q^{2}

= 48pq^{2} = RHS

Thus, LHS = RHS

∴ (4pq + 3q)^{2} – (4pq – 3q)^{2} = 48pq^{2}

Question (v)

(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Solution:

LHS = (a – b)(a + b) + (b – c)(b + c) + (c – a) (c + a)

= (a^{2} – b^{2}) + (b^{2} – c^{2}) + (c^{2} – a^{2})

= a^{2} – b^{2} + b^{2} – c^{2} + c^{2} – a^{2}

= a^{2} – a^{2} + b^{2} – b^{2} + c^{2} – c^{2}

= 0 = RHS

Thus, LHS = RHS

∴ (a -b)(a + b) + (b- c) (b + c) + (c – a) (c + a) = 0

6. Using identities, evaluate:

Question (i)

71^{2}

Solution:

= (70 + 1)^{2}

= (70)^{2} + 2 (70)(1) + (1)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 4900 + 140 + 1

= 5041

Question (ii)

99^{2}

Solution:

= (100 – 1)2

= (100)^{2} – 2(100)(1) + (1)^{2}

[∵ (a – b)^{2} – a^{2} – 2ab + b^{2}]

= 10000 – 200 + 1

= 9801

Question (iii)

102^{2}

Solution:

= (100 + 2)^{2}

= (100)^{2} + 2 (100)(2) + (2)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 10000 + 400 + 4

= 10404

Question (iv)

998^{2}

Solution:

= (1000 – 2)^{2}

= (1000)^{2} – 2 (1000)(2) + (2)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 1000000 – 4000 + 4

= 996004

Question (v)

5.2^{2}

Solution:

= (5 + 0.2)^{2}

= (5)^{2} + 2 (5)(0.2) + (0.2)^{2}

[∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 25 + 2 + 0.04

= 27 + 0.04

= 27.04

Question (vi)

297 × 303

Solution:

= (300 – 3) × (300 + 3)

= (300)^{2} – (3)^{2}

[∵ (a – b)(a + b) = a^{2} – b^{2}]

= 90000 – 9

= 89991

Question (vii)

78 × 82

Solution:

= (80 – 2) × (80 + 2)

= (80)^{2} – (2)^{2}

[∵ (a – b)(a + b) = a^{2} – b^{2}]

= 6400 – 4

= 6396

Question (viii)

8.9^{2}

Solution:

= (9 – 0.1)^{2}

= (9)^{2} – 2(9)(0.1) + (0.1)^{2}

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 81 – 1.8 + 0.01

= 81.01 – 1.8

= 79.21

Question (ix)

10.5 × 9.5

Solution:

= (10 + 0.5) × (10 – 0.5)

= (10)^{2} – (0.5)^{2}

[∵ (a + b)(a – b) = a^{2} – b^{2}]

= 100 – 0.25

= 99.75

7. Using a^{2} – b^{2} = (a + b) (a – b), find:

Question (i)

51^{2} – 49^{2}

Solution:

= (51 + 49) (51 – 49)

= (100) × (2)

= 200

Question (ii)

(1.02)^{2} – (0.98)^{2}

Solution:

= (1.02 + 0.98) (1.02 – 0.98)

= (2.0) × (0.04)

= 0.08

Question (iii)

153^{2} – 147^{2}

Solution:

= (153 + 147) (153 – 147)

= (300) × (6)

= 1800

Question (iv)

12.1^{2} – 7.9^{2}

Solution:

= (12.1 + 7.9) (12.1 – 7.9)

= 20 × 4.2

= 84

8. Using (x + a)(x + b) = x^{2} + (a + b) x + ab, find:

Question (i)

103 × 104

Solution:

= (100 + 3) × (100 + 4)

= (100)^{2} + (3 + 4) × 100 + (3)(4)

= 10000 + 700 + 12

=10712

Question (ii)

5.1 × 5.2

Solution:

= (5 + 0.1) (5 + 0.2)

= (5)^{2} + (0.1 + 0.2) × 5 + (0.1)(0.2)

= 25 + (0.3) × 5 + 0.02

= 25 + 1.5 + 0.02

= 26.52

Question (iii)

103 × 98

Solution:

= (100 + 3) (100-2)

= (100)^{2} + (3 – 2) 100 + (3)(-2)

= 10000 + 100 – 6

= 10094

Question (iv)

9.7 × 9.8

Solution:

= (10 – 0.3) (10 – 0.2)

= (10)^{2} + [(-0.3) + (-0.2)] 10 + (-0.3) (-0.2)

= 100 + [-0.5] × 10 + 0.06

= 100 – 5 + 0.06

= 95.06