NCERT Solutions for Chapter 9 Algebraic Expressions and Identities Class 8 Mathematics

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+ 4y, 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 + 4ycontains three terms. Therefore it is trinomial.
(vii) Since 5x - 4y + 3xy contains three terms. Therefore it is trinomial.
(viii) Since 4z - 15zcontains 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) a- 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 × a2+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 = x1+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 + q- 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)[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)[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)[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)[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 - a[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)[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)[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)[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

Go to Chapters list NCERT Solutions of Class 8 Maths
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