## MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 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)

(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)

(x) (7a – 9b)(7a – 9b)

Solution:

(i) (x + 3) (x + 3) = (x + 3)2

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

[Using identity : (a + b)^{2} = a^{2} + 2ab + b^{2}] = x^{2} + 6x + 9

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

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

[Using identity : (a + b)^{2} = a^{2} + 2ab + b^{2}] = 4a^{2} + 20y + 25

(iii) (2a – 7)(2a – 7) = (2a – 7)^{2}

= (2a)^{2} – 2(2a)(7) + (7)^{2}

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}] = 4a^{2} – 28a + 49

(v) (1.1m – 0.4) (1.1m + 0.4) = (1.1m)^{2} – (0.4)^{2}

[Using identity : a^{2} – b^{2} = (a + b) (a – b)]

= 1.21m^{2} – 0.16

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

= (b^{2})^{2} – (a^{2})^{2}

[Using identity : (a^{2} – b^{2}) = (a + b) (a – b)] = b^{4} – a^{4}

(vii) (6x – 7) (6x + 7) = (6x)^{2} – (7)^{2} = 36x^{2} – 49

[Using identity : (a^{2} – b^{2}) = (a + b) (a – b)]

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

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}]

= c^{2} – 2ac + a^{2}

(x) (7a – 9b)(7a – 9b) = (7a – 9b)^{2} = (7a)^{2} – 2(7a)(9b) + (9b)^{2}

[Using identity : (a – b)^{2} = a^{2}– 2ab + b^{2}]

= 49a^{2} – 126ab + 81b^{2}

Question 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:

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

= x^{2} + 10x + 21

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

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

(iii) (4x – 5) (4x – 1) – [4x + (-5)] [4x +(-1)]

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

= 16x^{2} – 24x +5

(iv) (4x + 5) (4x – 1) (4x + 5) (4x + (-1)]

= (4x + (5 – 1)(4x) + (5) × (-1)

= 16x^{2} + 16x – 5

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

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

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

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

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

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

(vii) (xyz – 4) (xyz – 2)

= [xyz + (-4)] [xyz + (-2)]

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

= x^{2}y^{2}z^{2} – 6xyz + 8

Question 3.

Find the following squares by using the identities.

(i) (b – 7)^{2}

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

(iii) (6x^{2} – 5y)^{2}

(v) (0.4p – 0.5q)^{2}

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

Solution:

(i) (b – 7)^{2} = (b)^{2} – 2(b)(7) + (7)^{2}

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}]

= b^{2} – 14b + 49

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

[Using identity : (a + b)^{2} = a^{2} + b^{2} + 2ab]

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

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

[Using identity : (a – b)^{2} = a^{2} + b^{2} – 2ab]

= 36x^{4} + 25y^{2} – 60x^{2}y

(v) (0.4p – 0.5p)^{2}

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

[Using identity : (a – b)^{2} = a^{2} + b^{2} – 2ab]

= 0.16p^{2} + 0.25q^{2} – 0.4pq

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

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

[Using identity : (a + b)^{2} = a^{2} + b^{2} + 2ab] = 4x^{2}y^{2} + 25 y^{2} + 20xy^{2}

Question 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^{2}c

(vii) (m^{2} – n^{2}m)^{2} + 2m^{3}n^{2}

Solution:

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

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}]

= a^{4} – 2a^{2}b^{2} + b^{4}

(ii) (2x + 5)^{2} – (2x – 5)^{2}

=[2x + 5 – (2x – 5)][2x + 5 + 2x – 5]

[Using identity : a^{2} – b^{2} – (a – b) (a + h)]

= (10)(4x) = 40x

(iii) (7m – 8n)^{2} + (7m + 8n)^{2}

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

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}]

and (a + b)^{2} = a^{2} + 2ab + b^{2}]

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

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

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

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

(Using identity : (a + b)^{2} = a^{2} + 2ab + b^{2})

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

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

(v) (2.5p – 1.5q)^{2} – (1.5p – 2.5q)^{2}

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

[Using identity: (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 6.25p^{2} – 7.5pq + 2.25q^{2} – [225p^{2} – 7.5pq + 6.25q^{2}]

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

= 4p^{2} – 4q^{2}

(vi) (ab + bc)^{2} – 2ab^{2}c

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

[Using identity : (a + b)^{2} = a^{2} + 2ab + b^{2}]

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

(vii) (m^{2} – n^{2}n)^{2} + 2m^{3}n^{2}

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

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}]

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

Question 5.

Show that

(i) (3x + 7)^{2} – 84x = (3x – 7)^{2}

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

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

(v) (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = O

Solution:

(i) To prove: (3x + 7)^{2} – 84x = (3x – 7)^{2}

LH.S. = (3x + 7)^{2} – 84x

= (3x)^{2} + 2(3x)(7) + (7)^{2} – 84x

= 9x^{2} + 42x +49 – 84x = 9x^{2} – 42x + 49

R.H.S. = (3x – 7)^{2} = (3x)^{2} – 2(3x)(7) + (7)^{2} = 9x^{2} – 42x + 49

∴ L.H.S. = R.H.S.

(ii) To prove : (9p – 5q)^{2} + 180pq = (9p + 5q)^{2}

L.H.S. = (9p – 5q)^{2} + 180pq

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

= 81p^{2} – 90pq + 25q^{2} + 180pq

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

R.H.S. = (9p + 5q)^{2}

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

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

∴ L.H.S. = R.H.S.

(iii) To prove :

(iv) To prove : (4pq + 3q)^{2} – (4pq – 3q)^{2} = 48pq^{2}

L.H.S. = (4pq + 3q)^{2} – (4pq – 3q)^{2}

= (4 pq)^{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}

= 48pq^{2} = R.H.S.

(v) To prove : (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

L.H.S. = (a – b) (a + b) + (b – c) (b + c) + (c – a)(c + a)

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

[Using identity : (x – y)(x + y) = x^{2} – y^{2}]

= 0 = R.H.S.

Question 6.

Using identities, evaluate,

(i) 71^{2}

(ii) 99^{2}

(iii) 102^{2}

(iv) 998^{2}

(v) 5.2^{2}

(vi) 297 × 303

(vii) 78 × 82

(viii) 8.9^{2}

(ix) 1.05 × 9.5

Solution:

(i) 71^{2} = (70 + 1)^{2}

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

[Using identity : (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 4900 + 140 + 1 = 5041

(ii) 99^{2} = (100 – 1)^{2}

= (100)^{2} – 2(100) (1) + 1^{2}

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 10000 – 200 + 1 = 9801

(iii) (102)^{2} = (100 + 2)^{2}

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

[Using identity : (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 10000 + 400 + 4 = 10404

(iv) (998)^{2} = (1000 – 2)^{2}

= (1000)^{2} – 2(1000)(2) + (2)^{2}

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 1000000 – 4000 + 4 = 996004

(v) (5.2)^{2} = (5 + 0.2)^{2}

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

[Using identity : (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 15 + 1 + 0.04 = 27.04

(vi) 297 ^{2} × 303 = (300 – 3) (300 + 3)

= (300)^{2} – 3^{2}

[Using identity : (a – b) (a + b) = (a^{2} – b^{2})

= 90000 – 9 = 89991

(vii) 78 × 82 = (80 – 2) (80 + 2)

= (80)^{2} – (2)^{2}

[Using identity : (a -b) (a + b) = a^{2} – b^{2}]

= 6400 – 4 = 6396

(viii) (8.9)^{2} = (9 – 0.1)^{2}

= (9)^{2} – 2(9)(0.1) + (0.1)^{2}

[Using identity : (a – b)^{2} = a^{2} – 2ab + b^{2}] = 81 – 1.8 + 0.01 = 79.21

(ix) 1.05 × 9.5

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}

Solution:

(i) 51^{2} – 49^{2} = (51 – 49) (51 + 49)

= (2)(100) = 200

(ii) (1.02)^{2} – (0.98)^{2} = (1.02 – 0.98) (1.02 + 0.98)

= (0.04)(2) = 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)

= (20.0) (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

Solution:

(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 + 3) [100+ (- 2)]

= (100)^{2} + (3 – 2) × 100 + (3) (- 2) = 10000 + 100 – 6 = 10094

(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2)

= [10 + (- 0.3)[10 + (- 0.2)]

= (10)^{2} + (- 0 3 – 0 2) × 10 + (- 0.3) (- 0.2)

= 100 – 5 + 0.06 = 95.06