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Chapter 9 Algebraic Expressions and Identities Ex 9.5 Solutions

Question - 1 : -

Use a suitableidentity 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) (a2+ b2) (- a2+ b2)

(vii) (6x – 7) (6x + 7)

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

(ix) (1/2x + 3/4y) (1/2x + 3/4y)

(x) (7a – 9b) (7a – 9b)

Answer - 1 : -






Question - 2 : -
Use the identity (x + a)(x + b) = x2 + (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) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2)

Answer - 2 : -



Question - 3 : -

Find the followingsquares by using the identities.

(i) (b – 7)2

(ii) (xy + 3z)2

(iii) (6x2 –5y)2

(iv) [(2m/3) + (3n/2)]2

(v) (0.4p – 0.5q)2

(vi) (2xy + 5y)2

Answer - 3 : -




Question - 4 : -

Simplify:
(i) (a2 – b2)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 – 2ab2c
(vii) (m2 – n2m)2 + 2m3n2

Answer - 4 : -





Question - 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 = 16/9 m2 +9/16 n2

(iv) (4pq + 3q)2– (4pq– 3q)2 = 48pq2

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

Answer - 5 : -





Question - 6 : -

Using identities, evaluate:
(i) 712
(ii) 992
(iii) 1022
(iv) 9982
(v) 5.22
(vi) 297 × 303
(vii) 78 × 82
(viii) 8.92
(ix) 1.05 × 9.5

Answer - 6 : -





Question - 7 : -

Using a2 – b2 = (a + b) (a – b), find
(i) 512 – 492
(ii) (1.02)2 – (0.98)2
(iii) 1532 – 1472
(iv) 12.12 – 7.92

Answer - 7 : -

(i)512 – 492 = (51 + 49) (51 – 49) = 100 × 2 = 200
(ii) (1.02)2 – (0.98)2 = (1.02 + 0.98) (1.02 – 0.98) = 2.00 × 0.04 = 0.08
(iii) 1532 – 1472 = (153 + 147) (153 – 147) = 300 × 6 = 1800
(iv) 12.12 – 7.92 = (12.1 + 7.9) (12.1 – 7.9) = 20.0 × 4.2 = 84

Question - 8 : -
Using (x + a) (x + b) = x2 + (a + b)x + ab, find
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8

Answer - 8 : -

(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 = 26.5 + 0.02 = 26.52
(iii) 103 × 98 = (100 + 3) (100 – 2) = (100)2 +(3 – 2) (100) + 3 × (-2) = 10000 + 100 – 6 = 10100 – 6 = 10094
(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2) = (10)2 –(0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 = 95 + 0.06 = 95.06

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