NCERT Solutions for 8th Class Maths: Chapter 9-Algebraic Expressions and Identities

NCERT Solutions for 8th Class Maths: Chapter 9-Algebraic Expressions and Identities

Page No: 140

Exercise 9.1

1. Identify the terms, their coefficients for each of the following expressions:
(i) 5xyz² - 3zy
(ii) 1+x+x²
(iii) 4x²y² - 4x²y²z² +z²
(iv) 3-pq + qr - rp
(v) x/2 + y/2 -xy
(vi) 0.3a - 0.6ab + 0.5b

Solution

(i) Terms: 5xyz² and -3zy
Coefficient in 5xyz² is 5 and in -3zy is -3.

(ii) Terms: 1, x and x²
Coefficient of x and of x² is 1.

(iii) Terms: 4x²y² ,- 4x²y²z² and z².
Coefficient in 4x²y² is 4, coefficient of - 4x²y²z² is -4 and coefficient of z² is 1.

(iv) Terms: 3, -pq, qr and -rp
Coefficient of -pq is -1, coefficient of qr is 1 and coefficient of -rp is -1.

(v) Terms: x/2, y/2 and -xy
Coefficient of x/2 is 1/2, coefficient of y/2 is 1/2 and coefficient of -xy is -1.

(vi) Terms: 0.3a, -0.6ab and 0.5b
Coefficient of 0.3a is 0.3, coefficient of -0.6ab is -0.6 and coefficient of 0.5b is 0.5.

2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories:
x+y, 1000, x+ x² + x³, 7 + y +5x, 2y- 3y² , 2y-3y²+ 4y³ , 5x - 4y + 3xy, 4z-15z², ab+bc+cd+da, pqr, p²q+pq², 2p + 2q

Solution

(i) Since x+ y contains two terms. Therefore it is binomial.
(ii) Since 1000 contains one terms. Therefore it is monomial.
(iii) Since x+x²+x³+x⁴ contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(iv) Since 7+y+5x contains three terms. Therefore it is trinomial.
(v) Since 2y-3y² contains two terms. Therefore it is binomial.
(vi) Since 2y - 3y² + 4y³ contains three terms. Therefore it is trinomial.
(vii) Since 5x - 4y + 3xy contains three terms. Therefore it is trinomial.
(viii) Since 4z - 15z² contains two terms. Therefore it is binomial.
(ix) Since ab + bc+ cd+ da contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(x) Since pqr contains one terms. Therefore it is monomial.
(xi) Since p²q + pq² contains two terms. Therefore it is binomial.
(xii) Since 2p + 2q contains two terms. Therefore it is binomial.

3. Add the following:
(i) ab-bc, bc-ca, ca-ab
(ii) a-b+ab, b-c+bc, c-a+ac
(iii) 2p²q² -3pq +4, 5 +7pq - 3p²q²
(iv) l² +m², m²+n², n²+1² + 2lm+2mn+2nl

Solution

(i) ab-bc, bc-ca, ca-ab 

(ii) a-b+ab, b-c+bc, c-a+ac 

Hence the sum if 0.
Hence the sum is ab+bc+ac.
(iii) 2p²q² -3pq +4, 5 +7pq - 3p²q² 

(iv) l² +m², m²+n², n²+1² + 2lm+2mn+2nl

Hence the sum is 2(l² + m² + n² + lm + mn + nl)
4. (a) Subtract 4a-7ab+3b+ 12 from 12a-9ab+5b-3.
(b) Subtract 3xy+5yz - 7zx from 5xy-2yz -2zx+ 10xyz.
(c) Subtract 4p²q - 3pq + 5pq² - 8p + 7q - 10 from 18 - 3p - 11q + 5pq - 2pq² + 5p²q

Solution

(a)

(b)

(c)

Pg No. 143

Exercise 9.2

1. Find the product of the following pairs of monomials:
(i) 4, 7p
(ii) -4p, 7p
(iii) -4p, 7pq
(iv) 4p³ , -3p
(iv) 4p, 0

Solution

(i) 4 × 7p = 4 × 7 × p = 28p
(ii) -4p × 7p = (-4 × 7) × (p × q) = -28 p²
(iii) -4p × 7pq = (-4× 7)(p×pq) = -28p²q
(iv) 4p³ × -3p = (4× -3)(p³ ×p)  = -12p⁴
(v) 4p ×0 = (4×0)(p) = 0

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively:
(p,q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)

Solution

(i) Area of rectangle = length × breadth
= p× q = pq sq. units

(ii) Area of rectangle  = length × breadth
= 10m ×5n = (10 × 5)(m×n) = 50mn sq. units

(iii) Area of rectangle = length × breadth
=20 x² × 5y² = (20×5)(x² × y²) = 100x²y² sq. units

(iv) Area of rectangle = length × breadth
= 4x × 3x² = (4 × 3)( x × x²) = 12x³ sq. units

(v) Area of rectangle = length × breadth
= 3mn ×4np = (3 × 4)(mn × np)  = 12mn²p sq. units

3. Complete the table of products:

Solution

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively:
(i) 5a, 3a²7a⁴
(ii) 2p, 4q, 8r
(iii) xy, 2x²y, 2xy²
(iv) a, 2b, 3c

Solution

(i) Volume of rectangular box = length × breadth × height
= 5a ×3a² × 7a⁴ = (5×3×7)(a× a² × a⁴) = 105a⁷ cubic units

(ii) Volume of rectangular box = length × breadth × height
= 2p × 4q × 8r = (2×4×8)(p×q×r) = 64pqr cubic units

(iii) Volume of rectangular box = length × breadth × height
= xy × 2x²y × 2xy²  = (1×2×2)(x×x²×x×y×y×y²)
 = 4x⁴y⁴ cubic units

(iv) Volume of rectangular box = length × breadth × height
= a× 2b ×3c = (1×2×3)(a×b×c) = 6abc cubic units

5. Obtain the product of:
(i) xy, yz, zx
(ii) a, -a², a³
(iii) 2, 4y, 8y², 16y³
(iv) a, 2b, 3c, 6abc
(v) m, -mn, mnp

Solution

(i) xy × yz× zx = x×x×y×y×z×z = x²y²z²

(ii) a×(-a²) ×a³ = (-1)(a×a²×a³)  = -a⁶

(iii) 2×4y×8y²×16y³
= (2×4×8×16)(y×y²×y³) = 1024y⁶

(iv) a× 2b ×3c × 6abc
= (1×2×3×6)(a×b×c×abc) = 36a²b²c²

(v) m× -mn×mnp = (-1)(m ×m×m×n×n×p)
= -m³n²p

Pg No. 146

Exercise 9.3

1. Carry out the multiplication of the expressions in each of the following pairs:
(i) 4p, q+r
(ii) ab, a-b
(iii) a+b, 7a²b²
(iv) a² - 9, 4a
(v) pq+qr+ rp, 0

Solution

(i) 4p × (q +r) = 4p × q + 4p × r
= 4pq + 4pr

(ii) ab ×(a-b) = ab × a -ab × b
= a²b - ab²

(iii) (a+b) × 7a²b² = a× 7a²b² + b × 7a²b²
= 7a³b² + 7a²b³

(iv) (a² - 9) × 4a = a² × 4a -4a ×9
= 4a³ - 36a

(v) (pq +qr +rp) × 0 = pq × 0 + qr×0 + rp ×0
= 0 + 0 + 0 = 0

2. Complete the table:

Solution

3. Find the product:
(i) (a²) × (2a²²) × (4a²⁶)
(ii) (2/3 xy) × (-9/10 x²y²)
(iii) (-10/3 pq³) × (6/5 p³q)
(iv) x×x²×x³×x⁴

Solution

(i) (a²) × (2a²²) × (4ax²⁶)
= (2 × 4) (a²× a²²× a²⁶)
= 8 × a^2+22+26 = 8a⁵⁰

(ii) (2/3 xy) × (-9/10 x²y²)
= (2/3 × -9/19)(x×x²×y×y²)
= -3/5x³y³

(iii) (-10/3 pq³)(6/5 p³q)
= (-10/3 × 6/5)(p×p³×q³×q)
= -4p⁴q⁴

(iv) x×x²×x³×x⁴ = x^1+2+3+4 = x¹⁰

4. (a) Simplify: 3x(4x -5) +3 and find values for
(i) x= 3
(ii) x= 1/2

(b) Simplify: a(a²+a+1)+5 and find its value for
(i) a= 0
(ii) a=1
(iii) a= -1

Solution

(a) 3x(4x-5)+3
= 3x × 4x-3x × 5+3
= 12x² -15x + 3

(i) For x=3, 12x² +15x+3
= 12(3)² - 15×3+3
= 12× 9 - 45 + 3 = 108 - 45+ 3 = 66

(ii) For x= 1/2, 12x² -15x+3
= 12(1/2)² - 15×1/2 + 3
= 12×1/4 - 15/2 +3
= 6- 15/2 = (12-15)/2 = -3/2

(b) a(a²+a+2)+5
= a × a² + a×a + a ×1 + 5
= a³ +a² +a + 5

(i) For a= 0, a³ + a² + a+ 5
= (0)³ + (0)² + (0) + 5
 = 0 + 0 + 0 + 5 = 5

(ii) For a= 1, a³ + a² + a+ 5
= (1)³ + (1)² + (1) + 5
 = 1 + 1 + 1 + 5 = 8
(iii) For a= -1, a³ + a² + a+ 5
= (-1)³ + (-1)² + (-1) + 5
= -1 + 1 -1 + 5 = -2 + 6 = 4

5. (a) Add: p(p-q), q(q-r) and r(r-p)
(b) Add: 2x(z-x-y) and 2y(z-y-zx)
(c) Subtract: 3l(l -4m+5n) from 4l(10n - 3m + 2l)
(d) Subtract: 3a(a+b+c) - 2b(a-b+c) from 4c(-a + b+ c)

Solution

(a) p(p-q) + q(q-r) + r(r-p)
= p² - pq + q² - qr + r² - rp
= p² + q² + r² -pq - qr - rp

(b) 2x( z- x- y) + 2y(z-y-x)
= 2xz - 2x² - 2xy + 2yz - 2y² - 2xy
= 2xz - 2xy - 2xy + 2yz - 2x² - 2y²
= -2x² - 2y² - 4xy + 2yz + 2zx

(c) 4l (10n - 3m + 2l) - 3l (1- 4m + 5n)
= 40ln - 12lm + 8l² - 3l² + 12lm - 15ln
= 8l² - 3l² - 12lm + 12lm + 40ln - 15ln
=5l² + 25ln

(d) 4c(-a + b + c) - [ 3a(a+b+c) - 2b(a-b + c)]
= -4ac + 4bc + 4c² - [ 3a² + 3ab + 3ac - 2ab + 2b² - 2bc ]
= - 4ac + 4bc + 4c² - [ 3a² + 2b² + 3ab - 2bc + 3ac - 2ab]
= - 4ac + 4ac + 4c² - [3a² + 2b² + ab + 3ac - 2bc]
= - 4ac + 4bc + 4c² - 3a² - 2b² - ab - 3ac + 2bc
= -3a² - 2b² + 4c² - ab + 4bc + 2bc - 4ac - 3ac
= -3a² - 2b² + 4c² - ab + 6bc - 7ac

Pg No. 148

Exercise 9.4

1. Multiply the binomials:
(i) (2x+5) and (4x-3)
(ii) (y - 8) and (3y - 4)
(iii) (2.5l - 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q² ) and 3pq - 2q²)
(vi) (3/4 a² + 3b²) and 4 ( a² - 2/3 b²)

Solution

(i) (2x + 5) × (4x -3)
= 2x(4x - 3) + 5(4x -3)
= 2x × 4x - 2x × 3 + 5 × 4x - 5 × 3
= 8x² -6x + 20x - 15
= 8x² + 14x - 15

(ii) (y-8) × (3y - 4) = y(3y - 4) - 8 (3y -4)
= y × 3y - y × 4 - 8 × 3y - 8 × -4
= 3y² - 4y - 24y + 32
= 3y² - 28y + 32

(iii) (2.5l - 0.5m) × (2.5l + 0.5m)
= 2.5l × (2.5l + 0.5m) - 0.5m × (2.5l + 0.5m)
= 2.5l x 2.5l + 2.5l x 0.5m - 0.5m x 2.5l - 0.5m x0.5m
 = 6.25l² + 1.25lm - 1.25lm - 0.25m²
 = 6.25l² - 0.25m²

(iv) (a+3b) × (x +5) = a(x+5) + 3b(x+5)
= a × x + a × 5 + 3b × x + 3b × 5
= ax + 5a + 3bx + 15b

(v) (2pq + 3q²)(3pq - 2q²)
= 2pq × (3pq - 2q²) + 3q² (3pq - 2q²)
= 2pq × 3pq - 2pq × 2q² + 3q² × 3pq - 3q² × 2q²
= 6p²q² - 4pq³ + 9pq³ - 6q⁴
= 6p²q² + 5pq³ - 6q⁴

(vi) (3/4 a² + 3b²) × 4(a² - 2/3 b²)
= (3/4 a² + 3b²) × (4a² - 8/3 b²)
= 3/4 a² × (4a² - 8/3 b²) + 3b² × (4a² - 8/3 b²)
= 3/4 a² × 4a² - 3/4 a² × 8/3 b² + 3b² × 4a² - 3b² × 8/3b²
= 3a⁴ - 2a²b² + 12a²b² - 8b⁴
= 3a⁴ + 10a²b² - 8b⁴

2. Find the product:
(i) (5 - 2x)(3 + x)
(ii) (x + 7y) (7x - y)
(iii) (a² + b)(a + b²)
(iv) (p² - q²) (2p + q)

Solution

(i)  (5 - 2x)(3 + x)
= 5 × (3 + x) - 2x(3 + x)
= 5 × 3 + 5 ×x - 2x × 3 - 2x × x
= 15 + 5x - 6x - 2x² = 15 - x - 2x²

(ii) (x + 7y) (7x - y)
= x(7x - y) + 7y×(7x - y)
= x × 7x - x × y + 7y × 7x - 7y × y
= 7x² - xy + 49xy - 7y²
= 7x² + 48xy - 7y²

(iii) (a² + b)(a + b²)
= a² ×(a + b²) + b × (a + b²)
= a² × a + a² × b² × a + b × b²
= a³ + a²b² + ab + b³

(iv) (p² - q²) (2p + q)
= p² × (2p + q) - q²(2p + q)
= p² × 2p + p² × q -q² × 2p - q² × q
= 2p³ + p²q - 2pq² - q³

3. Simplify:
(i) (x² - 5)(x + 5) + 25
(ii) (a² + 5)(b² + 3) + 5
(iii) (t + s²)(t² - s)
(iv) (a + b)(c - d)+ (a - b)(c + d) + 2(ac + bd)
(v)(x +y)(2x + y) + (x + 2y)( x- y)
(vi) (x + y)(x² - xy + y²)
(vii) (1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y
(viii) (a + b + c)(a + b+ c)

Solution

 (i) (x² - 5)(x + 5) + 25
= x² (x +5 ) -5(x + 5) + 25
= x² × x + x² × 5 - 5 × x - 5 × 5 + 25
= x³ + 5x² - 5x - 25 + 25
= x³ + 5x² - 5x

(ii) (a² + 5)(b² + 3) + 5
= a²(b³ + 3) + 5(b³ + 3) + 5
= a² × b³ + a² × 3 + 5 × b³ + 5 × 3 + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5b³ + 20

(iii) (t + s²) (t² -s) = t(t² -s) + s²(t² - s)
= t × t² - t ×s + s² × t² - s² × s
= t³ - st + s²t² - s³

(iv) (a + b)(c - d)+ (a - b)(c + d) + 2(ac + bd)
= a( c - d) + b( c - d) + a( c + d) - b( c + d) + 2ac + 2bd
= ac - ad + bc - bd + ac + ad - bs - bd + 2ac + 2bd
= ac + ac - ad + ad + bc - bc - bd - bd + 2ac + 2bd
= 2ac - 2bd + 2ac + 2bd
= 4ac

(v) (x +y)(2x + y) + (x + 2y)( x- y)
= x(2x + y) + y( 2x + y) + x(x -y) + 2y( x - y)
= 2x² + xy + 2xy + y² + x² - xy + 2xy - 2y²
= 2x² + x² + xy + 2xy - xy + 2xy + y² - 2y²
= 3x² + 4xy - y²

(vi) (x + y)(x² - xy + y2)
= x (x² - xy + y²) + y( x² - xy + y²)
= x³ - x²y + xy² + x²y - xy² + y³
= x³ - x²y + x²y + xy² - xy² + y³
= x³ + y³

(vii) (1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y
= 1.5x (1.5x + 4y + 3) - 4y(1.5x + 4y + 3) - 4.5x + 12y
= 2.25x²+ 6.0xy + 4.5x - 6.0xy - 16y² - 12y - 4.5x + 12y
= 2.25x² + 6.0xy - 6.0 xy + 4.5x - 4.5x - 16y² - 12y + 12y
= 2.25x² - 16y²

(viii) (a + b + c)(a + b+ c)
 = a ( a + b - c) + b ( a + b - c) + c ( a + b - c)
= a² + ab - ac + ab + b² - bs + ac + bc - c²
= a² + ab + ab - ac + ac - bc + b² - c²
= a² + b² - c² + 2ab

Pg No. 151

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 - 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) (x/2 + 3y/4) (x/2 + 3y/4)
(x) (7a - 9b)( 7a - 9b)

Solution

(i) (x + 3) (x + 3) = (x + 3)²
= x² + 2 × x × 3 + (3)² [Using 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 [Using identity ( a + b )2 = a2 + 2ab + b2]
 = 4y² + 20y + 25

(iii) (2a - 7) (2a - 7) = (2a - 7)²
 = (2a)² - 2 × 2a × 7 + (7)² [Using identity ( a - b )2 = a2 - 2ab + b2]
 = 4a² - 28 a + 49

(iv) (3a - 1/2)(3a - 1/2) = (3a - 1/2)²
 = (3a)² - 2 × 3a × 1/2 + (1/2)²
[Using identity ( a - b )2 = a2 - 2ab + b2]

 = 9a² - 3a + 1/4

(v) (1.1m - 0.4)(1.1m + 0.4)
= (1.1m)² - (0.4)² [Using identity (a - b)(a +b) = a² - b² ]
= 1.21m² - 0.16

(vi) ( a² + b²)( -a² + b²) = ( b² + a²)(b² - a²)
= (b²)² - (a²)² [Using identity (a - b)(a +b) = a2 - b2 ]
= b⁴ - a⁴

(vii) (6x - 7) (6x + 7)
= (6x)² - (7)² [Using identity (a - b)(a +b) = a2 - b2 ]
= 36x² - 49

(viii) (-a + c)(-a + c)
(c - a)(c - a) = (c - a)²
= (c)² - 2× c × a + (a)² [Using identity ( a - b )2 = a2 - 2ab + b2]
= c² - 2ca + a²

(ix) (x/2 + 3y/4)(x/2 + 3y/4) = ( x/2 + 3y/4)²
= (x/2)² + 2 × x/2 × 3y/4 + (3y/4)² [Using identity ( a + b )2 = a2 + 2ab + b2]
= x²/4 + 3/4xy + 9/16y²

(x) (7a - 9b)( 7a - 9b) = (7a - 9b)²
= (7a)² - 2 × 7a × 9b +(9b)² [Using identity ( a - b )2 = a2 - 2ab + b2]
= 49a² - 126ab + 81b²

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)
= (x)² + ( 3 + 7)x + 3 × 7 [Using identity ( x + a)(x + b) = x² + (a + b)x + ab ]
= x² + 10x + 21

(ii) (4x + 5)(4x + 1)
= (4x)² + ( 5 + 1)4x + 5 × 1 [Using identity ( x + a)(x + b) = x² + (a + b)x + ab]
 = 16x² + 6 × 4x + 5 = 16x² + 24x + 5

(iii) (4x - 5)( 4x - 1)
= (4x)² + (-5-1)4x + (-5) × (-1)  [Using identity ( x + a)(x + b) = x² + (a + b)x + ab]
= 16x + (-6) × 4x + 5 = 16x² - 24x + 5

(iv) (4x + 5)(4x - 1)
=(4x)2+{5+(-1)}(4x)+(5)(-1) [Using identity ( x + a)(x + b) = x² + (a + b)x + ab]
 = 16x² + (5 -1) × 4x - 5
 = 16x² + 4 × 4x - 5
 = 16x² + 16x - 5

(v) (2x + 5y)(2x + 3y)
= (2x)² + ( 5y + 3y) × 2x + 5y × 3y [Using identity ( x + a)(x + b) = x² + (a + b)x + ab]
 = 4x² + 8y × 2x + 15y²
 = 4x² + 16xy + 15y²

(vi) (2a² + 9)(2a² + 5)
= (2a²)² + (9 + 5) × 2a² + 9 × 5 [Using identity ( x + a)(x + b) = x² + (a + b)x + ab ]
 = 4a⁴ + 14 × 2a² + 45
 = 4a⁴ + 28a² + 45

(vii) (xyz - 4)(xyz - 2)
= (xyz)² + (-4 -2) × xyz + (-4)×(-2) [Using identity ( x + a)(x + b) = x² + (a + b)x + ab ]
 = x²y²z² - 6xyz + 8

3. Find the following squares by using identities:
(i) (b - 7)²
(ii) (xy + 3z)²
(iii) (6x² - 5y)²
(iv) (2/3 m + 3/2 n)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 = a² - 2ab + b² ]
= b² - 14b + 49

(ii) (xy + 3z)²
= (xy)² + 2 × xy × 3z + (3z)² [Using identity (a + b)2 = a2 + 2ab + b2 ]
= x²y² + 6xyz + 9z²

(iii) (6x² - 5y)²
= ( 6x²)² - 2 × 6x² × 5y + (5y)² [Using identity (a - b)2 = a2 - 2ab + b2 ]
 = 36x⁴ - 60x²y + 25 y²

(iv) (2/3 m + 3/2 n)²
= (2/3 m)² + 2 × 2/3m × 3/2n + (3/2n)² [Using identity (a + b)2 = a2 + 2ab + b2 ]
 = 4/9 m² + 2mn + 9/4 n²

(v) (0.4p - 0.5q)²
= (0.4p)² - 2 × 0.4p × 0.5q + (0.5q)² [Using identity (a - b)2 = a2 - 2ab + b2 ]
 = 0.16p² - 0.40pq + 0.25q²

(vi) (2xy + 5y)²
= (2xy)² + 2 × 2xy × 5y + (5y)² [Using identity (a + b)2 = a2 + 2ab + b2 ]
= 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
= (a²)² - 2 × a² × b² + (b²)² [Using identity (a -b)² = a² - 2ab + b² ]
= a⁴ - 2a²b² + b⁴

(ii) (2x + 5)² - (2x - 5)²
={(2x+5)+(2x-5)}{(2x+5)-(2x-5)} [Using identity (a2-b2)=(a+b)(a-b)]
={4x}{2x+5-2x+5}
=(4x)(10)
=40x

(iii) (7m - 8n)² + (7m + 8n)²
= (7m)² - 2 × 7m × 8n + (8n)²+ [(7m)² + 2 × 7m ×8n + (8n)²] [Using identities (a + b)² = a² + 2ab + b² and (a -b)² = a² - 2ab + b²]
= 49m² - 112mn + 64n² + [49m² + 112mn + 64n²]
= 49m² - 112mn + 64n² + 49m² + 112mn + 64n²
= 98m² + 128n²

(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 = a² + 2ab + b²]
= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²
= 16m² + 25m² + 40mn + 40mn + 25n² + 16n²
= 41m² + 80mn + 41n²

(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 = a² - 2ab + b²]
= 6.25p² - 7.50pq + 2.25q² - [ 2.25p² - 7.50pq + 6.25q²]
= 6.25p² - 7.50pq + 2.25q² - 2.25p² + 7.50 pq - 6.25 q²
= 4p² - 4q²

(vi) (ab + bc)² - 2ab²c
= (ab)² + 2 × ab × bc + (bc)² - 2ab²c [Using identity (a + b)² = a² + 2ab + b²]
 = a²b² + 2ab²c + b²c² - 2ab²c
 = a²b² + b²c²

(vii) (m² - n²m) + 2m³n²
= (m²)² - 2 × m² × n²m + (n²m)² + 2m³n² [Using identity (a - b)² = a² - 2ab + b²]
 = m⁴ - 2m³n² + n⁴m² + 2m³n²
 = m⁴ + n⁴m²

5. Show that:

(i) (3x + 7)² - 84x = (3x - 7)²
(ii) (9p - 5q)² + 180pq = (9p + 5q)²
(iii) (4/3 m - 3/4 n)² + 2mn = 16/9m² + 9/16n²
(iv) (4pq + 3q)² - (4pq - 3q)² = 48pq²
(v) (a -b)(a+b) + (b - c)(b+c) + (c - a)(c + a) = 0

Solution

 (i) L.H.S. = (3x + 7)² - 84x
= (3x)² + 2 × 3x × 7 + (7)² - 84x [Using identity (a + b)² = a² + 2ab + b²]
 = 9x² + 42x + 49 - 84x
 = 9x² - 42x + 49
 = (3x - 7)² [ (a -b)² = a² - 2ab + b²]
 = R.H.S.

(ii) L.H.S. = (9p - 5q)² + 180pq
= (9p)² - 2 × 9p × 5q + (5q)² + 180pq [Using identity (a -b)² = a² - 2ab + b²]
 = 81p² - 90pq + 25q² + 180pq
 = 81p² + 90pq + 25q²
 = (9p + 5q)² [(a + b)² = a² + 2ab + b²]

(iii) L.H.S. = (4/3 m - 3/4 n)² + 2mn
= (4/3 m)² - 2 × 4/3m × 3/4n + (3/4 n)² + 2mn [Using identity (a -b)² = a² - 2ab + b²]
 = 16/9 m² - 2mn + 9/16 n² + 2mn
 = 16/9 m² + 9/16 n²
 = R.H.S.

(iv) L.H.S. = (4pq + 3q)² - (4pq - 3q)²
= (4pq)² + 2 × 4pq × 3q + (3q)² - [ (4pq)² - 2 × 4pq × 3q + (3q)²] [Using identities (a + b)² = a² + 2ab + b² and (a -b)² = a² - 2ab + b²]
= 16p²q² + 24pq² + 9q² - [16p²q² - 24pq² + 9q²]
= 16 p²q² + 24pq² + 9q² - 16p²q² + 24pq² - 9q² = 48pq²
= R.H.S.

(v) L.H.S.= (a - b)(a + b) + ( b - c)( b + c ) + (c - a)(c + a)
= a² - b²+ b² - c² + c² - a² [Using identity (a - b)(a + b) = a² - b²]
= 0
= R.H.S.

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

Solution

(i) 71² = (70 + 1)²
= (70)² + 2 × 70 × 1 + (1)² [Using identity (a + b)² = a² + 2ab + b²]
= 4900 + 140 + 1
= 5041

(ii) 99² = (100 - 1 )²
= (100)² - 2 × 100 × 1 + (1)² [Using identity (a -b)² = a² - 2ab + b²]
= 10000 – 200 + 1
= 9801

(iii) 102² = ( 100 + 2)²
= (100)² + 2 × 100 × 2 + (2)² [Using identity (a + b)² = a² + 2ab + b²]
= 10000 + 400 + 4
= 10404

(iv) 998² = (1000 - 2)²
= ( 1000)² - 2 × 1000 × 2 + (2)² [Using identity (a -b)² = a² - 2ab + b²]
= 1000000 – 4000 + 4
= 996004

(v) 5.2² = ( 5 + 0.2)²
= (5x² + 2 × 5 × 0.2 + (0.2)² [Using identity (a + b)² = a² + 2ab + b²]
= 25 + 2.0 + 0.04 = 27.04

(vi) 297 × 303
= (300 - 3) × (300 + 3)
= (300)² - (3)² [Using identity (a - b)(a + b) = a2 - b2 ]
 = 90000 – 9 = 89991

(vii) 78 × 82 = ( 80 - 2) × ( 80 + 2)
= (80)² - ( 2)² [Using identity (a - b)(a + b) = a² - b²]
 = 6400 – 4
= 6396

(viii) 8.9² = (8 + 0.9)²
= (8)² + 2 × 8 × 0.9 + (0.9)² [Using identity (a + b)² = a² + 2ab + b²]
= 64 + 14.4 + 0.81
= 79.21

(ix) 1.05 × 9.5
= ( 10 + 0.5) × ( 10 - 0.5)
= (10)² - (0.5)² [Using identity (a - b)(a + b) = a² - b²]
= 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) [Using identity (a - b)(a + b) = a² - b²]
= 100 × 2 = 200

(ii) (1.02)² - (0.98)²
= ( 1.02 + 0.98)(1.02 - 0.98)
[Using identity (a - b)(a + b) = a² - b²]
= 2.00 × 0.04 = 0.08

(iii) 153² - 147²
= (153 + 147)(153 - 147) [Using identity (a - b)(a + b) = a² - b²]
= 300 × 6 = 1800

(iv) 12.1² - 7.9²
= (12.1 + 7.9)(12.1 - 7.9) [Using identity (a - b)(a + b) = a² - b²]
= 20.0 × 4.2
= 84.0
= 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)
= (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 + (3 - 2) × 100 - 6
= 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 + {100 + {- 0.3 - 0.2} × 10 + 0.06
= 100 - 0.5 × 10 + 0.06
= 100 - 5 + 0.06 - 95. 06

Chapterwise NCERT Solutions for 8th Class Maths

Chapter 1 – Rational Numbers
Chapter 2 – Linear Equations in One Variable
Chapter 3 – Understanding Quadrilaterals
Chapter 4 – Practical Geometry
Chapter 5 – Data Handling
Chapter 6 – Squares and Square Roots
Chapter 7 – Cubes and Cube Roots
Chapter 8 – Comparing Quantities
Chapter 9 – Algebraic Expressions and Identities
Chapter 10 – Visualizing Solid Shapes
Chapter 11 – Mensuration
Chapter 12 – Exponents and Powers
Chapter 13 – Direct and Inverse Proportions
Chapter 14 – Factorization
Chapter 15 – Introduction to Graphs
Chapter 16 – Playing with Numbers

About

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website to learn more.