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 a² – b² = (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