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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities| PDF Download

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NCERT Solutions for Class 8 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+ 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

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


Chapter 9 Class 8 NCERT Solutions are very helpful in rectifying all those mistakes that can happen in the exams. Algebraic expressions are formed from the variables and constants.

• Expressions containing one, two or three terms are called monomial, binomial and trinomial respectively.

• The terms having the same literal factors are called like terms, otherwise they are called unlike terms.

• An identity is an equality, which is true for all values of the variables in the equality. The standard identities are:
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a – b) = a2 – b2
(x + a)(x + b) = x2 + (a + b)x + ab

Algebraic Expressions and Identities Class 8 NCERT Solutions contains total 5 chapters which will make learning exciting and cover the syllabus in less time. Below we have provided exercisewise Chapter 9 NCERT Solutions.


Studyrankers subject matter experts have prepared these NCERT Solutions to help you in exams and score maximum marks. Through the help of these solutions you can easily cover all the topics inside the chapter.


NCERT Solutions for Class 8 Maths Chapters:


Chapter 1 Rational Numbers
Chapter 2 Linear Equations in 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 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


FAQ on Chapter 9 Algebraic Expressions and Identities

How many exercises in Chapter 9 Algebraic Expressions and Identities?



There are total 5 exercises in the Chapter 9 that will make you aware of all those concepts embedded in the questions. These NCERT Solutions are updated as per the latest guidelines by CBSE.


Identify the variable and constant in the expression 24 – x.



In the expression above, 24 is the constant and x is the variable since 24 is fixed and cannot change its value whereas x can be anything.

The length and breadth of a rectangle are 3x2 – 2 and 2x + 5 respectively. Find its area.



Here,
Length = 3x2 – 2
Breadth = 2x + 5
∴  Area = (Length) × (Breadth)
= (3x2 – 2) × (2x + 5)
= 3x2(2x + 5) + (–2)(2x + 5)
= (3x2 × 2x) + (5 × 3x2) + [(–2) × 2x + (–2) × 5]
= 6x3 + 15x2 + (–4x) + (–10) = 6x3 + 15x2 – 4x – 10
Thus, the required area of the rectangle is 6x3 + 15x2 – 4x – 10 sq. units.

Verify the identity (x + a)( x + b) = x2 + (a + b)x + ab for a = 2, b = 3 and x = 4,



We have
(x + a)(x + b) = x2 + (a + b)x + ab
Puting x = 2, b = 3 and x = 4, we have
LHS = (x + a)(x + b)
= (4 + 2)(4 + 3)
= 6 × 7 = 42
RHS = x2 + (a + b)x + ab
= (4)2 + (2 + 3)4 + (2 * 3)
= 16 + 5 × 4 + 6
= 16 + 20 + 6 = 42
i.e. LHS = RHS
Thus, the given identity is true for the given values.
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