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Algebraic Expressions And Identities : Exercise 9.5 (Mathematics NCERT Class 8th)


Q.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 - {1 \over 2})(3a - {1 \over 2}) (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) (ix)\left( {{x \over 2} + {{3y} \over 4}} \right)\left( {{x \over 2} + {{3y} \over 4}} \right)

(x)(7a - 9b)(7a - 9b)

Sol. (i) (x + 3)(x + 3)

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

= {(x)^2} + 2 \times x \times 3 + {(3)^2} (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

= {x^2} + 6x + 9

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

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

= {(2y)^2} + 2 \times 2y \times 5 + {(5)^2} (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

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

(iii)(2a - 7)(2a - 7)

(2a - 7)(2a - 7)={(2a - 7)^2}

= {(2a)^2} - 2 \times 2a \times 7 + {(7)^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= 4{a^2} - 28a + 49

(iv)(3a - {1 \over 2})(3a - {1 \over 2})

(3a - {1 \over 2})(3a - {1 \over 2})={\left( {3a - {1 \over 2}} \right)^2}

= {(3a)^2} - 2 \times 3a \times {1 \over 2} + {\left( {{1 \over 2}} \right)^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= 9{a^2} - 3a + {1 \over 4}

(v) (1.1m - 0.4)(1.1m + 0.4)

(1.1m - 0.4)(1.1m + 0.4)= {(1.1m)^2} - {(0.4)^2}

= 1.21{m^2} - 0.16 (Using identity(a - b)(a + b) = {a^2} - {b^2})

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

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

= {b^4} - {a^4}

(vii) (6x - 7)(6x + 7)

(6x - 7)(6x + 7)= {(6x)^2} - {(7)^2} (Using identity(a - b)(a + b) = {a^2} - {b^2})

= 36{x^2} - 49

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

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

= {(c - a)^2}

= {(c)^2} - 2 \times c \times a + {(a)^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= {c^2} - 2ca + {a^2}

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

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

= {\left( {{x \over 2}} \right)^2} + 2 \times {x \over 2} \times {{3y} \over 4} + {\left( {{{3y} \over 4}} \right)^2} (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

= {{{x^2}} \over 4} + {3 \over 4}xy + \left( {{9 \over {16}}} \right){y^2}

(x)(7a - 9b)(7a - 9b)

(7a - 9b)(7a - 9b)= {(7a - 9b)^2}

= {(7a)^2} - 2 \times 7a \times 9b + {(9b)^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= 49{a^2} - 126ab + 81{b^2}

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

(vii) (xyz - 4)(xyz - 2)

Sol. (i) (x + 3)(x + 7)

= {(x)^2} + (3 + 7)x + 3 \times 7(Using identity(x + a)(x + b) = {x^2} + (a + b)x + ab)

= {x^2} + 10x + 21

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

= {(4x)^2} + (5 + 1)4x + 5 \times 1(Using identity(x + a)(x + b) = {x^2} + (a + b)x + ab)

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

(iii)(4x - 5)(4x - 1)

= {(4x)^2} + ( - 5 - 1)4x + ( - 5) \times ( - 1)(Using identity(x + a)(x + b) = {x^2} + (a + b)x + ab)

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

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

= {(4x)^2} + (5 - 1)4x + (5) \times ( - 1)(Using identity(x + a)(x + b) = {x^2} + (a + b)x + ab)

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

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

= {(2x)^2} + (5y + 3y)2x + 5y \times 3y(Using identity(x + a)(x + b) = {x^2} + (a + b)x + ab)

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

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

= {(2{a^2})^2} + (9 + 5)2{a^2} + 9 \times 5(Using identity(x + a)(x + b) = {x^2} + (a + b)x + ab)

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

(vii) (xyz - 4)(xyz - 2)

= {(xyz)^2} + ( - 4 - 2)xyz + ( - 4) \times ( - 2)(Using identity(x + a)(x + b) = {x^2} + (a + b)x + ab)

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

Q.3 Find the following squares by using the identities.

(i){(b - 7)^2} (ii){(xy + 3z)^2} (iii){(6{x^2} - 5y)^2}

(iv) {\left( {{2 \over 3}m + {3 \over 2}n} \right)^2} (v) {(0.4p - 0.5q)^2} (vi){(2xy + 5y)^2}

Sol. (i){(b - 7)^2}

= {(b)^2} - 2 \times b \times 7 + {(7)^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= {b^2} - 14b + 49

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

= {(xy)^2} - 2 \times xy \times 3z + {(3z)^2} (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

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

(iii){(6{x^2} - 5y)^2}

= {(6{x^2})^2} - 2 \times 6{x^2} \times 5y + {(5y)^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= 36{x^4} - 60{x^2}y + 25{y^2}

(iv) {\left( {{2 \over 3}m + {3 \over 2}n} \right)^2}

= {\left( {{2 \over 3}m} \right)^2} + 2 \times {2 \over 3}m \times {3 \over 2}n + {\left( {{3 \over 2}n} \right)^2} (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

= {4 \over 9}{m^2} + 2mn + {9 \over 4}{n^2}

(v) {(0.4p - 0.5q)^2}

= {(0.4p)^2} - 2 \times 0.4p \times 0.5q + {(0.5q)^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= 0.16{p^2} - 0.40pq + 0.25{q^2}

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

= {(2xy)^2} + 2 \times 2xy \times 5y + {(5y)^2} (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

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

Q.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} - 2a{b^2}c

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

Sol. (i) {({a^2} - {b^2})^2}

= {({a^2})^2} - 2 \times {a^2} \times {b^2} + {({b^2})^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= {a^4} - 2{a^2}{b^2} + {b^4}

(ii) {(2x + 5)^2} - {(2x - 5)^2}

= {(2x)^2} + 2 \times 2x \times 5 + {(5)^2} - [{(2x)^2} - 2 \times 2x \times 5 + {(5)^2}] (Using identity {(a + b)^2} = {a^2} + 2ab + {b^2}and {(a - b)^2} = {a^2} - 2ab + {b^2})

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

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

= 40x

(iii) {(7m - 8n)^2} + {(7m + 8n)^2}

= {(7m)^2} - 2 \times 7m \times 8n + {(8n)^2} + [{(7m)^2} + 2 \times 7m \times 8n + {(8n)^2}] (Using identity {(a + b)^2} = {a^2} + 2ab + {b^2}and {(a - b)^2} = {a^2} - 2ab + {b^2})

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

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

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

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

= {(4m)^2} + 2 \times 4m \times 5n + {(5n)^2} + [{(5m)^2} + 2 \times 5m \times 4n + {(4n)^2}] (Using identity {(a + b)^2} = {a^2} + 2ab + {b^2})

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

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

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

= {(2.5p)^2} - 2 \times 2.5p \times 1.5q + {(1.5q)^2} - [{(1.5p)^2} - 2 \times 1.5p \times 2.5q + {(2.5q)^2}] (Using identity {(a - b)^2} = {a^2} - 2ab + {b^2})

= 6.25{p^2} - 7.50pq + 2.25{q^2} - [2.25{p^2} - 7.50pq + 6.25{q^2}]

= 4{p^2} - 4{q^2}

(vi) {(ab + bc)^2} - 2a{b^2}c

= {(ab)^2} + 2 \times ab \times bc + {(bc)^2} - 2a{b^2}c (Using identity {(a + b)^2} = {a^2} + 2ab + {b^2})

= {a^2}{b^2} + 2 \times a{b^2}c + {b^2}{c^2} - 2a{b^2}c

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

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

= {({m^2})^2} - 2 \times {m^2} \times {n^2}m + {({n^2}m)^2} + 2{m^3}{n^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

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

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

Q.5 Show that.

(i) {(3x + 7)^2} - 84x = {(3x - 7)^2} (ii) {(9p - 5q)^2} + 180pq = {(9p + 5q)^2}

(iii) {\left( {{4 \over 3}m - {3 \over 4}n} \right)^2} + 2mn = {{16} \over 9}{m^2} + {9 \over {16}}{n^2} (iv) {(4pq + 3q)^2} - {(4pq - 3q)^2} = 48p{q^2}

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

Sol. (i) {(3x + 7)^2} - 84x = {(3x - 7)^2}

LHS = {(3x + 7)^2} - 84x

= {(3x)^2} + 2 \times 3x \times 7 + {(7)^2} - 84x (Using identity {(a + b)^2} = {a^2} + 2ab + {b^2})

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

= 9{x^2} - 42x + 49

RHS = {(3x - 7)^2}

= {(3x)^2} - 2 \times 3x \times 7 + {( - 7)^2} (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= 9{x^2} - 42x + 49

We can see that, LHS = RHS

Hence, {(3x + 7)^2} - 84x = {(3x - 7)^2}

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

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

= {(9p)^2} - 2 \times 9p \times 5q + {(5q)^2} + 180pq (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= 81{p^2} - 90pq + 25{q^2} + 180pq

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

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

= {(9p)^2} + 2 \times 9p \times 5q + {(5q)^2} (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

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

We can see that, LHS = RHS

Hence, {(9p - 5q)^2} + 180pq = {(9p + 5q)^2}

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

LHS = {\left( {{4 \over 3}m - {3 \over 4}n} \right)^2} + 2mn

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

= {{16} \over 9}{m^2} - 2mn + {9 \over {16}}{n^2} + 2mn

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

= RHS

Hence, {\left( {{4 \over 3}m - {3 \over 4}n} \right)^2} + 2mn = {{16} \over 9}{m^2} + {9 \over {16}}{n^2}

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

LHS = {(4pq + 3q)^2} - {(4pq - 3q)^2}

= {(4pq)^2} + 2 \times 4pq \times 3q + {(3q)^2} - [{(4pq)^2} - 2 \times 4pq \times 3q + {(3q)^2}] (Using identity {(a + b)^2} = {a^2} + 2ab + {b^2}and {(a - b)^2} = {a^2} - 2ab + {b^2})

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

= 16{p^2}{q^2} + 24p{q^2} + 9{q^2} - 16{p^2}{q^2} + 24p{q^2} - 9{q^2}

= 48p{q^2}

= RHS

Hence, {(4pq + 3q)^2} - {(4pq - 3q)^2} = 48p{q^2}

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

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}

= 0

= RHS

Hence, (a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) = 0

Q.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

Sol. (i) 712

712 = (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) 992

992 = (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) 1022

1022 = (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) 9982

9982 = (1000 - 2)2 = (1000)2 - 2 × 1000 × 2 + (2)2 (Using identity{(a - b)^2} = {a^2} - 2ab + {b^2})

= 100000 - 4000 + 4 = 996004

(v)5.22

5.22 = (5 + 0.2)2 = (5)2 + 2 × 5 × 0.2 + (0.2)2 (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

= 25 + 2 + 0.04 = 27.04

(vi)297×303

297×303 = (300 - 3) × (300 + 3)

= (300)2 – (3)2 (Using identity(a - b)(a + b) = {a^2} - {b^2})

= 90000 – 9 = 89991

(vii) 78×82

78×82 = (80 - 2) × (80 + 2)

= (80)2 – (2)2 (Using identity(a - b)(a + b) = {a^2} - {b^2})

= 6400 – 4 = 6396

(viii)8.92

8.92 = (8 + 0.9)2 = (8)2 + 2 × 8 × 0.9 + (0.9)2 (Using identity{(a + b)^2} = {a^2} + 2ab + {b^2})

= 64 + 14.4 + 0.81 = 79.21

(ix) 1.05 × 9.5

1.05 × 9.5= (1 + 0.05) × (1 – 0.05) × 10

= [(1)2 - (0.05)2 ] × 10 (Using identity(a + b)(a - b) = {a^2} - {b^2})

= 0.9975 × 10 = 9.975

Q.7 Using a2b2 = (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}

Sol. (i) {51^2} - {49^2}

= (51 + 49)(51 - 49) (Using identity{a^2} - {b^2} = (a + b)(a - b))

= 100 \times 2

= 200

(ii) {(1.02)^2} - {(0.98)^2}

= (1.02 + 0.98)(1.02 - 0.98) (Using identity{a^2} - {b^2} = (a + b)(a - b))

= 2.00 \times 0.04

= 0.08

(iii){153^2} - {147^2}

= (153 + 147)(153 - 147) (Using identity{a^2} - {b^2} = (a + b)(a - b))

= 300 \times 6

= 1800

(iv) {12.1^2} - {7.9^2}

= (12.1 + 7.9)(12.1 - 7.9) (Using identity{a^2} - {b^2} = (a + b)(a - b))

= 20.0 \times 4.2

= 84

Q.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

Sol. (i) 103 × 104

= (100 + 3) × (100 + 4)

= (100)2 + (3 + 4) ×100 + 3 × 4 (Using identity (x + a) (x + b) = x2 + (a + b) x + ab)

= 10000 + 7 × 100 + 12

= 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 (Using identity (x + a) (x + b) = x2 + (a + b) x + ab)

= 25 + 0.3 × 5 + 0.02

= 25 + 1.5 + 0.02

= 26.52

(iii) 103 × 98

= (100 + 3) × (100 - 2)

= (100)2 + (3 - 2) ×100 + 3 × (-2) (Using identity (x + a) (x + b) = x2 + (a + b) x + ab)

= 10000 + 100 - 6

= 10094

(iv) 9.7 × 9.8

= (10 - 0.3) × (10 - 0.2)

= (10)2 + ((-0.3) + (-0.2)) ×10 + (-0.3) × (-0.2) (Using identity (x + a) (x + b) = x2 + (a + b) x + ab)

= 100 - 5 + 0.06

= 95.06

 



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