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

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

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Exercise 9.1 Page No: 140

Q1. 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 – p (v) (x/2) + (y/2) – xy (vi) 0.3a – 0.6ab + 0.5b

Solution :

Sl. No. Expression Term Coefficient
i) 5xyz2 – 3zy Term: 5xyz2

Term: -3zy

5 -3
ii) 1 + x + x2 Term: 1
Term: x
Term: x2
1 1 1
iii) 4x2y2 – 4x2y2z2 + z2 Term: 4x2y2
Term: -4 x2y2z2
Term :  z2
4 -4 1
iv) 3 – pq + qr – p Term : 3 -pq qr -p 3 -1 1 -1
v) (x/2) + (y/2) – xy Term : x/2 Y/2 -xy ½ 1/2 -1
vi) 0.3a – 0.6ab + 0.5b Term : 0.3a -0.6ab 0.5b 0.3 -0.6 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 + x4 , 7 + y + 5x, 2y – 3y2 , 2y – 3y2 + 4y3 , 5x – 4y + 3xy, 4z – 15z2 , ab + bc + cd + da, pqr, p2q + pq2 , 2p + 2q

Solution:

Let us first define the classifications of these 3 polynomials:

Monomials contain only one term.

Binomials contain only two terms.

Trinomials contain only three terms.

x + y two terms Binomial
1000 one term Monomial
x + x2 + x3 + x4 four terms Polynomial, and it does not fit in the listed three categories
2y – 3y2 two terms Binomial
2y – 3y2 + 4y3 three terms Trinomial
5x – 4y + 3xy three terms Trinomial
4z – 15z2 two terms Binomial
ab + bc + cd + da four terms Polynomial, and it does not fit in the listed three categories
pqr one term Monomial
p2q + pq2 two terms Binomial
2p + 2q two terms Binomial
7 + y + 5x three terms Trinomial

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 + l2, 2lm + 2mn + 2nl

Solution:

i) (ab – bc) + (bc – ca) + (ca-ab)

= ab – bc + bc – ca + ca – ab

= ab – ab – bc + bc – ca + ca

= 0

ii) (a – b + ab) + (b – c + bc) + (c – a + ac)

= a – b + ab + b – c + bc + c – a + ac

= a – a +b – b +c – c + ab + bc + ca

= 0 + 0 + 0 + ab + bc + ca

= ab + bc + ca

iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

= (2p2q2 – 3pq + 4) + (5 + 7pq – 3p2q2)

= 2p2q2 – 3p2q2 – 3pq + 7pq + 4 + 5

= – p2q2 + 4pq + 9

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

= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl

= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl

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) (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)

= 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12

= 12a – 4a -9ab + 7ab +5b – 3b -3 -12

= 8a – 2ab + 2b – 15

b) (5xy – 2yz – 2zx + 10xyz) – (3xy + 5yz – 7zx)

= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx

=5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz

= 2xy – 7yz + 5zx + 10xyz

c) (18 – 3p – 11q + 5pq – 2pq2 + 5p2q) – (4p2q – 3pq + 5pq2 – 8p + 7q – 10)

= 18 – 3p – 11q + 5pq – 2pq2 + 5p2q – 4p2q + 3pq – 5pq2 + 8p – 7q + 10

=18+10 -3p+8p -11q – 7q + 5 pq+ 3pq- 2pq^2 – 5pq^2 + 5 p^2 q – 4p^2 q

= 28 + 5p – 18q + 8pq – 7pq2 + p2q

 

Exercise 9.2 Page No: 143

1. Find the product of the following pairs of monomials.

(i) 4, 7p

(ii) – 4p, 7p

(iii) – 4p, 7pq

(iv)  4p3, – 3p

(v) 4p, 0

Solution:

(i) 4 , 7 p =  4 × 7 × p = 28p

(ii) – 4p × 7p = (-4 × 7 ) × (p × p )= -28p2

(iii) – 4p × 7pq =(-4 × 7 ) (p × pq) =  -28p2q

(iv) 4p3 × – 3p = (4 × -3 ) (p3 × p ) =  -12p4

(v) 4p ×  0 = 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:

Area of rectangle = Length x breadth. So, it is multiplication of two monomials.

The results can be written in square units.

(i) p × q = pq

(ii)10m ×  5n = 50mn

(iii) 20x2 ×  5y2 =  100x2y2

(iv) 4x × 3x2 = 12x3

(v) 3mn ×  4np = 12mn2p

3. Complete the following table of products:

ncert solution for class 8 maths chapter 09 fig 1

Solution:

ncert solutions for class 8 maths chapter 09 fig 2

4. Obtain the volume of rectangular boxes with the following length, breadth and height, respectively.

(i) 5a, 3a2, 7a4

(ii) 2p, 4q, 8r

(iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

Solution:

Volume of rectangle = length x  breadth x  height. To evaluate volume of rectangular boxes, multiply all the monomials.

(i) 5a x 3a2 x 7a4 = (5 × 3 × 7) (a × a2 × a4 ) = 105a7

(ii) 2p x 4q x 8r = (2 × 4 × 8 ) (p × q × r ) = 64pqr

(iii) y × 2x2y × 2xy2 =(1 × 2 × 2 )( x × x2 × x × y × y × y2 ) =  4x4y4

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

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 = x2 y2 z2

(ii) a × – a2  × a3 = – a6

(iii) 2 × 4y × 8y2 × 16y3 = 1024 y6

(iv) a × 2b × 3c × 6abc = 36a2 b2 c2

(v) m × – mn × mnp = –m3 n2 p

 

Exercise 9.3 Page No: 146

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) = 4pq + 4pr

(ii)ab(a – b) = a2 b – a b2

(iii)(a + b) (7a2b2) = 7a3b2 + 7a2b3

(iv) (a2 – 9)(4a) = 4a3 – 36a

(v) (pq + qr + rp) × 0 = 0 ( Anything multiplied by zero is zero )

2. Complete the table.

ncert solutions for class 8 maths chapter 09 fig 3

Solution:

First expression Second expression Product
(i) a b + c + d a(b+c+d)

= a×b + a×c + a×d

= ab + ac + ad

(ii) x + y – 5 5xy 5 xy (x + y – 5)

= 5 xy x x + 5 xy x y – 5 xy x 5

= 5 x2y + 5 xy2 – 25xy

(iii) p 6p2 – 7p + 5 p (6 p 2-7 p +5)

= p× 6 p2 – p× 7 p + p×5

= 6 p3 – 7 p2 + 5 p

(iv) 4 p2 q2 P2 – q2 4p2 q2 * (p2 – q2 )

=4 p4 q2– 4p2 q4

(v) a + b + c abc abc(a + b + c)

= abc × a + abc × b + abc × c

= a2bc + ab2c + abc2

3. Find the product.

i) a2 x (2a22) x (4a26)

ii) (2/3 xy) ×(-9/10 x2y2)

(iii) (-10/3 pq3/) × (6/5 p3q)

(iv) (x) × (x2) × (x3) × (x4)

Solution:

i) a2 x (2a22) x (4a26)

= (2 × 4) ( a2 × a22 × a26 )

= 8 × a2 + 22 + 26 

= 8a50

ii) (2xy/3) ×(-9x2y2/10)

=(2/3 × -9/10 ) ( x × x2 × y × y2 )

= (-3/5 x3y3)

iii) (-10pq3/3) ×(6p3q/5)

= ( -10/3 × 6/5 ) (p × p3× q3 × q)

= (-4p4q4)

iv)  ( x) x (x2) x (x3) x (x4)

= x 1 + 2 + 3 + 4 

=  x10

4. (a) Simplify 3x (4x – 5) + 3 and find its 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) Putting x=3 in the equation we gets 12x2 – 15x + 3 =12(32) – 15 (3) +3

= 108 – 45 + 3

= 66

(ii) Putting x=1/2 in the equation we get

12x2 – 15x + 3 = 12 (1/2)2 – 15 (1/2) + 3

= 12 (1/4) – 15/2 +3

= 3 – 15/2 + 3

= 6- 15/2

= (12- 15 ) /2

= -3/2

b) a(a2 +a +1)+5

= a x a2 + a x a + a x 1 + 5 =a3+a2+a+ 5

(i) putting a=0 in the equation we get 03+02+0+5=5

(ii) putting a=1 in the equation we get 1+ 1+ 1+5 = 1 + 1 + 1+5 = 8

(iii) Putting a = -1 in the equation we get (-1)3+(-1)+ (-1)+5 = -1 + 1 – 1+5 = 4

5. (a) Add: p ( p – q), q ( q – r) and r ( r – p) 

(b) Add: 2x (z – x – y) and 2y (z – y – x) 

(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l ) 

(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c)  from 4c ( – a + b + c )

Solution:

a) p ( p – q) + q ( q – r) + r ( r – p)

= (p2 – pq) + (q2 – qr) + (r2 – pr)

= p2 + q2 + r2 – pq – qr – pr

b) 2x (z – x – y) + 2y (z – y – x)

= (2xz – 2x2 – 2xy) + (2yz – 2y2 – 2xy)

= 2xz – 4xy + 2yz – 2x2 – 2y2

c) 4l ( 10 n – 3 m + 2 l ) – 3l (l – 4 m + 5 n)

= (40ln – 12lm + 8l2) – (3l2 – 12lm + 15ln)

= 40ln – 12lm + 8l2 – 3l2 +12lm -15 ln

= 25 ln + 5l2

d) 4c ( – a + b + c ) – (3a (a + b + c ) – 2 b (a – b + c))

= (-4ac + 4bc + 4c2) – (3a2 + 3ab + 3ac – ( 2ab – 2b2 + 2bc ))

=-4ac + 4bc + 4c2 – (3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc)

= -4ac + 4bc + 4c2 – 3a2 – 3ab – 3ac +2ab – 2b2 + 2bc

= -7ac + 6bc + 4c2 – 3a2 – ab – 2b2

 

Exercise 9.4 Page No: 148

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 x 4x – 2x x 3 + 5 x 4x – 5 x 3

8x² – 6x + 20x -15

8x² + 14x -15

ii) ( y – 8)(3y – 4)

= y x 3y – 4y – 8 x 3y + 32

= 3y2 – 4y – 24y + 32

= 3y2 – 28y + 32

(iii) (2.5l – 0.5m)(2.5l + 0.5m)

= 2.5l x 2.5 l + 2.5l x 0.5m – 0.5m x 2.5l – 0.5m x 0.5m

= 6.25l2 + 1.25 lm – 1.25 lm – 0.25 m2

= 6.25l2 – 0.25 m2

iv) (a + 3b) (x + 5)

= ax + 5a + 3bx + 15b

v) (2pq + 3q2(3pq – 2q2)

= 2pq x 3pq – 2pq x 2q2 + 3q2 x 3pq – 3q2 x 2q2

= 6p2q2 – 4pq3 + 9pq3 – 6q4

= 6p2q2 + 5pq3 – 6q4

(vi) (3/4 a² + 3b² ) and 4( a² – 2/3 b² )

=(3/4 a² + 3b² ) x 4( a² – 2/3 b² )

=(3/4 a² + 3b² ) x (4a² – 8/3 b² )

=3/4 a² x (4a² – 8/3 b² ) + 3b² x (4a² – 8/3 b² )

=3/4 a² x 4a² -3/4 a² x 8/3 b² + 3b² x 4a² – 3b² x 8/3 b²

=3a4 – 2a² b² + 12 a²  b² – 8b4

= 3a4 + 10a²  b² – 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)

=15 + 5x – 6x – 2x2

= 15 – x -2 x 2

(ii) (x + 7y) (7x – y)

= x(7x-y) + 7y ( 7x-y)

=7x2 – xy + 49xy – 7y2

= 7x2 – 7y2 + 48xy

iii) (a2+ b) (a + b2)

= a2  (a + b2) + b(a + b2)

= a3 + a2b2 + ab + b3

= a3 + b3 + a2b2 + ab

iv) (p2– q2) (2p + q)

= p2 (2p + q) – q2 (2p + q)

=2p3 + p2q – 2pq2 – q3

= 2p3 – q3 + p2q – 2pq2

3. Simplify.

(i) (x2– 5) (x + 5) + 25

(ii) (a2+ 5) (b3+ 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

= x3 + 5x2 – 5x – 25 + 25

= x3 + 5x2 – 5x

ii) (a2+ 5) (b3+ 3) + 5

= a2b3 + 3a2 + 5b3 + 15 + 5

= a2b3 + 5b3 + 3a2 + 20

iii) (t + s2)(t2 – s)

= t (t2 – s) + s2(t2 – s)

= t3 – st + s2t2 – s3

= t3 – s3 – st + s2t2

iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

= (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

=(ac – ad + bc – bd) + (ac + ad – bc – bd) + (2ac + 2bd)

= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd

= 4ac

v) (x + y)(2x + y) + (x + 2y)(x – y)

= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2

= 3x2 + 4xy – y2

vi) (x + y)(x2– xy + y2)

= x3 – x2y + xy2 + x2y – xy2 + y3

= x3 + y3

vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y = 2.25x2 – 16y2

viii) (a + b + c)(a + b – c)

= a2 + ab – ac + ab + b2 – bc + ac + bc – c2

= a2 + b2 – c2 + 2ab

 

Exercise 9.5 Page No: 151

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) (1/2x + 3/4y) (1/2x + 3/4y)

(x) (7a – 9b) (7a – 9b)

Solution:

(i) (x + 3) (x + 3) = (x + 3)2

= x2 + 6x + 9

Using (a+b) 2 = a2 + b2 + 2ab

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

= 4y2 + 20y + 25

Using (a+b) 2 = a2 + b2 + 2ab

iii) (2a – 7) (2a – 7) = (2a – 7)2

= 4a2 – 28a + 49

Using (a-b)= a2 + b2 – 2ab

iv) (3a – 1/2)(3a – 1/2) = (3a – 1/2)2

=  9a2 -3a+(1/4)

Using (a-b) 2  = a2 + b2 – 2ab

v)   (1.1m – 0.4) (1.1m + 0.4)

= 1.21m2 – 0.16

Using (a – b)(a + b) = a2 – b2

vi) (a2+ b2) (– a2+ b2)

= (b2 + a2 ) (b2 – a2)

= -a4 + b4

Using (a – b)(a + b) = a2 – b2

vii) (6x – 7) (6x + 7)

=36x2 – 49

Using (a – b)(a + b) = a2 – b2

viii) (– a + c) (– a + c) = (– a + c)2

= c2 + a2 – 2ac

Using (a-b) 2 = a2 + b2 – 2ab

ncert solution for class 8 maths chapter 09 fig 7

= (x2/4) + (9y2/16) + (3xy/4)

Using (a+b) 2 = a2 + b2 + 2ab

x) (7a – 9b) (7a – 9b) = (7a – 9b)2

= 49a2 – 126ab + 81b2

Using (a-b) 2 = a2 + b2 – 2ab

2. Use the identity (x + a) (x + b) = x+ (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)

= x2 + (3+7)x + 21

= x2 + 10x + 21

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

= 16x2 + 4x + 20x + 5

= 16x2 + 24x + 5

iii) (4x – 5) (4x – 1)

= 16x2 – 4x – 20x + 5

= 16x2 – 24x + 5

iv) (4x + 5) (4x – 1)

= 16x2 + (5-1)4x – 5

= 16x2 +16x – 5

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

= 4x2 + (5y + 3y)2x + 15y2

= 4x2 + 16xy + 15y2

vi) (2a2+ 9) (2a2+ 5)

= 4a4 + (9+5)2a2 + 45

= 4a4 + 28a2 + 45

vii) (xyz – 4) (xyz – 2)

= x2y2z2 + (-4 -2)xyz + 8

= x2y2z2 – 6xyz + 8

3. Find the following squares by using the identities.

(i) (b – 7)2

(ii) (xy + 3z)2

(iii) (6x2 – 5y)2

(iv) [(2m/3) + (3n/2)]2

(v) (0.4p – 0.5q)2

(vi) (2xy + 5y)2

Solution:

Using identities:

(a – b) 2 = a2 + b2 – 2ab (a + b) 2 = a2 + b2 + 2ab

(i) (b – 7)2 = b2 – 14b + 49

(ii) (xy + 3z)2 = x2y2 + 6xyz + 9z2

(iii) (6x2 – 5y)2 = 36x4 – 60x2y + 25y2

(iv) [(2m/3}) + (3n/2)]2 = (4m2/9) +(9n2/4) + 2mn

(v) (0.4p – 0.5q)2 = 0.16p2 – 0.4pq + 0.25q2

(vi) (2xy + 5y)2 = 4x2y2 + 20xy2 + 25y2

4. Simplify.

(i) (a2 – b2)2

(ii) (2x + 5) – (2x – 5)2

(iii) (7m – 8n)2 + (7m + 8n)2

(iv) (4m + 5n)+ (5m + 4n)2

(v) (2.5p – 1.5q)– (1.5p – 2.5q)2

(vi) (ab + bc)2– 2ab²c

(vii) (m– n2m)+ 2m3n2

Solution:

i) (a2– b2)2 = a4 + b4 – 2a2b2

ii) (2x + 5) – (2x – 5)2
= 4x2 + 20x + 25 – (4x2 – 20x + 25) = 4x2 + 20x + 25 – 4x2 + 20x – 25 = 40x

iii) (7m – 8n)2 + (7m + 8n)2
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2

iv) (4m + 5n)+ (5m + 4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2

v) (2.5p – 1.5q)– (1.5p – 2.5q)2
= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25q2
= 4p2 – 4q2

vi) (ab + bc)2– 2ab²c = a2b2 + 2ab2c + b2c2 – 2ab2c = a2b2 + b2c2

vii) (m– n2m)+ 2m3n2
= m4 – 2m3n2 + m2n4 + 2m3n2
= m4 + m2n4

5. Show that.

(i) (3x + 7)– 84x = (3x – 7)2

(ii) (9p – 5q)2+ 180pq = (9p + 5q)2

(iii) (4/3m – 3/4n)2 + 2mn = 16/9 m2 + 9/16 n2

(iv) (4pq + 3q)2– (4pq – 3q)= 48pq2

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

Solution:

i) LHS = (3x + 7)– 84x

= 9x2 + 42x + 49 – 84x
= 9x2 – 42x + 49
= RHS

LHS = RHS

ii)  LHS = (9p – 5q)2+ 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
RHS = (9p + 5q)2
= 81p2 + 90pq + 25q2
LHS = RHS

ncert solution for class 8 maths chapter 09 fig 8

LHS = RHS

iv)  LHS = (4pq + 3q)2– (4pq – 3q)2

= 16p2q2 + 24pq2 + 9q2 – 16p2q2 + 24pq2 – 9q2

= 48pq2

= RHS

LHS = RHS

v) LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

= a2 – b2 + b2 – c2 + c2 – a2

= 0

= RHS

6. Using identities, evaluate.

(i) 71²

(ii) 99²

(iii) 1022

(iv) 998²

(v) 5.2²

(vi) 297 x 303

(vii) 78 x 82

(viii) 8.92

(ix) 10.5 x 9.5

Solution:

i) 712

= (70+1)2

= 702 + 140 + 12

= 4900 + 140 +1

= 5041

ii) 99²

= (100 -1)2

= 1002 – 200 + 12

= 10000 – 200 + 1

= 9801

iii) 1022

= (100 + 2)2

= 1002 + 400 + 22

= 10000 + 400 + 4 = 10404

iv) 9982

= (1000 – 2)2

= 10002 – 4000 + 22

= 1000000 – 4000 + 4

= 996004

v) 5.22

= (5 + 0.2)2

= 52 + 2 + 0.22

= 25 + 2 + 0.04 = 27.04

vi) 297 x 303

= (300 – 3 )(300 + 3)

= 3002 – 32

= 90000 – 9

= 89991

vii) 78 x 82

= (80 – 2)(80 + 2)

= 802 – 22

= 6400 – 4

= 6396

viii) 8.92

= (9 – 0.1)2

= 92 – 1.8 + 0.12

= 81 – 1.8 + 0.01

= 79.21

ix) 10.5 x 9.5

= (10 + 0.5)(10 – 0.5)

= 102 – 0.52

= 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) = 100 x 2 = 200

ii) (1.02)2– (0.98)2

= (1.02 + 0.98)(1.02 – 0.98) = 2 x 0.04 = 0.08

iii) 1532 – 1472

= (153 + 147)(153 – 147) = 300 x 6 = 1800

iv) 12.12 – 7.92

= (12.1 + 7.9)(12.1 – 7.9) = 20 x 4.2= 84

8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find

(i) 103 x 104

(ii) 5.1 x 5.2

(iii) 103 x 98

(iv) 9.7 x 9.8

Solution:

i) 103 x 104

= (100 + 3)(100 + 4)

= 1002 + (3 + 4)100 + 12

= 10000 + 700 + 12

= 10712

ii) 5.1 x 5.2

= (5 + 0.1)(5 + 0.2)

= 52 + (0.1 + 0.2)5 + 0.1 x 0.2

= 25 + 1.5 + 0.02

= 26.52

iii) 103 x 98

= (100 + 3)(100 – 2)

= 1002 + (3-2)100 – 6

= 10000 + 100 – 6

= 10094

iv) 9.7 x 9.8

= (9 + 0.7 )(9 + 0.8)

= 92 + (0.7 + 0.8)9 + 0.56

= 81 + 13.5 + 0.56

= 95.06

Also Access 
NCERT Exemplar for class 8 Maths Chapter 9
CBSE Notes for class 8 Maths Chapter 9

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

The NCERT Solutions for Class 8 CBSE Maths Chapter 9, explains basic concepts like terms, factors, coefficients, like and unlike terms, addition and subtraction of algebraic expressions, and multiplication of two or more polynomials. Students will also learn about various algebraic expression identities and solve problems applying these identities. In NCERT Class 8, Chapter 9 – Algebraic Expressions and Identities carries a total weightage of 8 to 10 marks in the final examination. Students can utilise the NCERT Solutions for Class 8 Maths Chapter 9 while solving the exercise questions and preparing for their examination.

Get detailed solutions for all the questions listed under the below exercises:

Exercise 9.1 Solutions: 4 Questions (Short answers)
Exercise 9.2 Solutions: 5 Questions (Short answers)
Exercise 9.3 Solutions: 5 Questions (Short answers)
Exercise 9.4 Solutions : 3 Questions (Short answers)
Exercise 9.5 Solutions: 8 Questions (Short answers)

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

NCERT Solutions for Class 8 Maths Chapter 9 is mainly about the study of solving polynomial-related problems. The chapter builds a strong foundation for the students to deal with higher grade Maths problems.

The main topics covered in this chapter include:

Exercise Topic
9.1 What are Expressions?
9.2 Terms, Factors and Coefficients
9.3 Monomials, Binomials and Polynomials
9.4 Like and Unlike Terms
9.5 Addition and Subtraction of Algebraic Expressions
9.6 Multiplication of Algebraic Expressions: Introduction
9.7 Multiplying a Monomial by a Monomial
9.8 Multiplying a Monomial by a Polynomial
9.9 Multiplying a Polynomial by a Polynomial
9.10 What is an Identity?
9.11 Standard Identities
9.12 Applying Identities

Students can utilise the NCERT Solutions for Class 8 Maths Chapter 9 for any quick references to comprehend and solve complex problems.

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

  1. NCERT Solutions provides fully resolved step-by-step solutions to all textbook questions.
  2. Set of solutions contain a list of all important formulas on algebraic identities.
  3. These solutions are designed based on the latest syllabus.
  4. Solutions are prepared by subject experts.
  5. NCERT Solutions are also helpful for the preparation of competitive exams.

Disclaimer:

Dropped Topics – 9.1 Introduction, 9.2 Terms, Factors and Coefficients, 9.3 Monomials, Binomials and Polynomials, 9.4 Like and Unlike Terms, 9.10 What is an identity?, 9.11 Standard identities
and 9.12 Applying identities.

Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 9

Q1

How is NCERT Solutions for Class 8 Maths Chapter 9 helpful for board exams?

NCERT Solutions for Class 8 Maths Chapter 9 provides answers with detailed descriptions as per the term limit particularised by the Board for self-evaluation. Solving these solutions will provide excellent practice for the students, so that they can finish the paper on time. It is thus clear that the NCERT Solutions for Class 8 Maths Chapter 9 are essential to score high in the examinations. Students can get acquainted with writing exams and will be able to face the exams more confidently.
Q2

Is it necessary to practise all the exercises present in Chapter 9 of NCERT Solutions for Class 8 Maths?

Yes, it is crucial to practise all the exercises present in Chapter 9 of NCERT Solutions for Class 8 Maths. All exercises contain numerous questions to solve which may appear in their finals. This makes them more confident towards the syllabus.
Q3

Mention the topics that are covered in NCERT Solutions for Class 8 Maths Chapter 9.

The topics that are covered in NCERT Solutions for Class 8 Maths Chapter 9 are 9.1 – introduction to expressions, 9.2 – terms, factors and coefficients, 9.3 – monomials, binomials and polynomials, 9.4 – like and unlike terms, 9.5 – addition and subtraction of algebraic expressions, 9.6 – multiplication of algebraic expressions: introduction, 9.7 – multiplying a monomial by a monomial, 9.8 – multiplying a monomial by a polynomial, 9.9 – multiplying a polynomial by a polynomial, 9.10 – definition and meaning of identity, 9.11 – standard identities and 9.12 – applying identities.

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