NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

NCERT Solutions Class 8 Mathematics 
Chapter – 9 (Algebraic Expressions and Identities) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 9 Algebraic Expressions and Identities Exercise 9.5 has been provided here to help the students in solving the questions from this exercise. 

Chapter 9: Algebraic Expressions and Identities

Exercise – 9.5 

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{1}{2})(3a – \frac{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) (\frac{x}{2} + \frac{3y}{4}) (\frac{x}{2} + \frac{3y}{4})
(x) (7a – 9b) (7a – 9b)

Solution –

(i) (x + 3)(x + 3)
= (x + 3)2
= (x)2 + 2(x)(3) + (3)2 [Using the algebraic identity (a + b)2 = a2 + 2ab + b2]
= x2 + 6x + 9

(ii) (2y + 5)(2y + 5)
= (2y + 5)2
= (2y)2 + 2(2y)(5) + (5)2 [Since, (a + b)2 = a2 + 2ab + b2]
= 4y2 + 20y + 25

(iii) (2a – 7)(2a – 7)
= (2a – 7)2
= (2a)2 – 2(2a)(7) + (7)2 [Since, (a – b)2 = a2 – 2ab + b2]
= 4a2 – 28a + 49

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

= (3a – \frac{1}{2})2
= (3a)2 – 2(3a)\frac{1}{2} + (\frac{1}{2})2 [Since, (a – b)2 = a2 – 2ab + b2]

= 9a2 – 3a + \frac{1}{4}

(v) (1.1m – 0.4)(1.1m + 0.4)
= (1.1m)2 – (0.4)2 [Since, (a + b)(a – b) = a2 – b2]
= 1.21m2 – 0.16

(vi) (a2 + b2)(-a2 + b2)
= (b2 + a2)(b2 – a2)
= (b2)2 – (a2)2 [Since, (a + b)(a – b) = a2 – b2]
= b4 – a4

(vii) (6x – 7)(6x + 7)
= (6x)2 – (7)2 [Since, (a + b)(a – b) = a2 – b2]
= 36x2 – 49

(viii) (-a + c)(-a + c)
= (-a + c)2
= (-a)2 + 2(-a)(c) + (c)2 [Since, (a + b)2 = a2 + 2ab + b2]
= a2 – 2ac + c2

(ix) \mathbf{\left ( \frac{x}{2} +\frac{3y}{4}\right )}\mathbf{\left ( \frac{x}{2} +\frac{3y}{4}\right )}

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

= \left ( \frac{x}{2} \right )^2 + 2 \left ( \frac{x}{2} \right ) \left ( \frac{3y}{4} \right ) + \left ( \frac{3y}{4} \right )^2 [Since, (a + b)2 = a2 + 2ab + b2]

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

(x) (7a – 9b)(7a – 9b)
= (7a – 9b)2
= (7a)2 – 2(7a)(9b) + (9b)2 [Since, (a – b)2 = a2 – 2ab + b2]
= 49a2 -126ab + 81b2
= 4y2 + 20y + 25

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)

Solution –

(i) (x + 3) (x + 7)
= x2 + (3+7)x + 21
= x2 + 10x + 21

(ii) (4x + 5) (4x + 1)
= 16x2 + 4x + 20x + 5
= 16x2 + 24x + 5

(iii) (4x – 5) (4x – 1)
= 16x2 – 4x – 20x + 5
= 16x2 – 24x + 5

(iv) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x – 5
= 16x2 +16x – 5

(v) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)2x + 15y2
= 4x2 + 16xy + 15y2

(vi) (2a2+ 9) (2a2+ 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45

(vii) (xyz – 4) (xyz – 2)
= x2y2z2 + (-4 -2)xyz + 8
= x2y2z2 – 6xyz + 8

3. Find the following squares by using the identities.
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2 – 5y)2
(iv) \left ( \frac{2m}{3} +\frac{3n}{2}\right )^2
(v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2

Solution – Using identities:
(a – b) 2 = a2 + b2 – 2ab
(a + b) 2 = a2 + b2 + 2ab

(i) (b – 7)2
= b2 – 14b + 49

(ii) (xy + 3z)2
= x2y2 + 6xyz + 9z2

(iii) (6x2 – 5y)2
= 36x4 – 60x2y + 25y2

(iv) \left ( \frac{2m}{3} +\frac{3n}{2}\right )^2
= \frac{4m^2}{9} +\frac{9n^2}{4} + 2mn

(v) (0.4p – 0.5q)2
= 0.16p2 – 0.4pq + 0.25q2

(vi) (2xy + 5y)2
= 4x2y2 + 20xy2 + 25y2

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

Solution –

(i) (a2– b2)2
= a4 + b4 – 2a2b2

(ii) (2x + 5)2  – (2x – 5)2
= 4x2 + 20x + 25 – (4x2 – 20x + 25)
= 4x2 + 20x + 25 – 4x2 + 20x – 25
= 40x

(iii) (7m – 8n)2 + (7m + 8n)2
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2

(iv) (4m + 5n)2 + (5m + 4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2

(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25q2
= 4p2 – 4q2

(vi) (ab + bc)2– 2ab²c
= a2b2 + 2ab2c + b2c2 – 2ab2c
= a2b2 + b2c2

(vii) (m2 – n2m)2 + 2m3n2
= m4 – 2m3n2 + m2n4 + 2m3n2
= m4 + m2n4

5. Show that.
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2+ 180pq = (9p + 5q)2

(iii) \left ( \frac{4}{3}m -\frac{3}{4}n\right )^2  + 2mn = \frac{16}{9} m2 + \frac{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

Solution –

(i) (3x + 7)2 – 84x = (3x – 7)2
L.H.S. = (3x + 7)2 – 84x
= 9x2 + 42x + 49 – 84x
= 9x2 – 42x + 49
R.H.S = (3x – 7)2
= (3x)2 – 2(3x)(7) + (7)2
= 9x2 – 42x + 49
L.H .S = R.H .S

(ii) (9p – 5q)2+ 180pq = (9p + 5q)2
L.H.S. = (9p – 5q)2+ 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
R.H.S. = (9p + 5q)2
= 81p2 + 90pq + 25q2
LHS = RHS

(iii) \left ( \frac{4}{3}m -\frac{3}{4}n\right )^2  + 2mn = \frac{16}{9} m2 + \frac{9}{16} n2

L.H.S. =\left ( \frac{4}{3}m -\frac{3}{4}n\right )^2+ 2mn

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

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

R.H.S = \frac{16}{9}m^2 + \frac{9}{16}n^2 

L.H.S. = R.H.S.

(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
L.H.S = (4pq + 3q)2 – (4pq – 3q)2
= (4pq)2 + 2(4pq)(3q) + (3q)2 – (4pq)2 + 2(4pq)(3q) – (3q)2
= 16p2q2 + 24pq2 + 9q2 – 16p2q2 + 24pq2 – 9q2
= 48pq2
L.H.S = R.H.S

(v) (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)
= (a2 – b2) + (b2 – c2) + (c2 – a2)
= a2 – b2 + b2 – c2 + c2 – a2
= 0
L.H.S = R.H.S

6. Using identities, evaluate.
(i) 71²
(ii) 99²
(iii) 1022
(iv) 998²
(v) 5.2²
(vi) 297 x 303
(vii) 78 x 82
(viii) 8.92
(ix) 10.5 x 9.5

Solution – The three basic algebraic identities, which we will be using to evaluate the expressions are as follows.
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a – b) = a2 – b2

(i) 712
= (70 + 1)2
= (70)2 + 2(70)(1) + (1)2 [Since, (a + b)2 = a2 + 2ab + b2]
= 4900 + 140 + 1
= 5041

(ii) 99²
= (100 -1)2
= 1002 – 200 + 12
= 10000 – 200 + 1
= 9801

(iii) 1022
= (100 + 2)2
= 1002 + 400 + 22
= 10000 + 400 + 4 = 10404

(iv) 9982
= (1000 – 2)2
= 10002 – 4000 + 22
= 1000000 – 4000 + 4
= 996004

(v) 5.22
= (5 + 0.2)2
= 52 + 2 + 0.22
= 25 + 2 + 0.04 = 27.04

(vi) 297 × 303
= (300 – 3)(300 + 3)
= 3002 – 32
= 90000 – 9
= 89991

(vii) 78 × 82
= (80 – 2)(80 + 2)
= 802 – 22
= 6400 – 4
= 6396

(viii) 8.92
= (9 – 0.1)2
= 92 – 1.8 + 0.12
= 81 – 1.8 + 0.01
= 79.21

(ix) 10.5 × 9.5
= (10 + 0.5)(10 – 0.5)
= 102 – 0.52
= 100 – 0.25
= 99.75

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

Solution –

(i) 512– 492
= (51 + 49)(51 – 49)
= 100 x 2
= 200

(ii) (1.02)2– (0.98)2
= (1.02 + 0.98)(1.02 – 0.98)
= 2 × 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 × 4.2
= 84

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

Solution –

(i) 103 × 104
= (100 + 3)(100 + 4)
= 1002 + (3 + 4)100 + 12
= 10000 + 700 + 12
= 10712

(ii) 5.1 × 5.2
= (5 + 0.1)(5 + 0.2)
= 52 + (0.1 + 0.2)5 + 0.1 x 0.2
= 25 + 1.5 + 0.02
= 26.52

(iii) 103 × 98
= (100 + 3)(100 – 2)
= 1002 + (3-2)100 – 6
= 10000 + 100 – 6
= 10094

(iv) 9.7 × 9.8
= (9 + 0.7 )(9 + 0.8)
= 92 + (0.7 + 0.8)9 + 0.56
= 81 + 13.5 + 0.56
= 95.06

 

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