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NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities

Orchid The International School presents comprehensive NCERT Solutions for Class 8 Maths Chapter 9 on Algebraic Expressions and Identities. Crafted by seasoned educators, these solutions offer a meticulous breakdown of each question within the NCERT textbook, aiding students in both homework assignments and exam preparation.

Exercise 9.1

Question 1 :

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

 

Answer :

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

 


Question 2 :

(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

 

Answer :

(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

 


Question 3 :

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

 

Answer :

 

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

 


Question 4 :

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

 

Answer :

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

 


Exercise 9.2

Question 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

 

Answer :

(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

 


Question 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)

 

Answer :

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

 


Question 3 :

Complete the following table of products:

ncert solution for class 8 maths chapter 09 fig 1

 

Answer :

 

ncert solutions for class 8 maths chapter 09 fig 2

 

 


Question 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

 

Answer :

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

 


Question 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

 

Answer :

(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

Question 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

 

Answer :

(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 )

 


Question 2 :

 Complete the table.

ncert solutions for class 8 maths chapter 09 fig 3

 

Answer :

 

 

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

 


Question 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)

 

Answer :

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

 


Question 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.

 

Answer :

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 13 + 12 + 1+5 = 1 + 1 + 1+5 = 8

(iii) Putting a = -1 in the equation we get (-1)3+(-1)2 + (-1)+5 = -1 + 1 – 1+5 = 4

 


Question 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 )

 

Answer :

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

Question 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)

 

Answer :

(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

 


Question 2 :

Find the product.

(i) (5 – 2x) (3 + x)

(ii) (x + 7y) (7x – y)

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

(iv) (p2 – q2) (2p + q)

 

Answer :

(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

 


Question 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)

 

Answer :

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

Question 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)

 

Answer :

(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) 2 = 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

 


Question 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)

 

Answer :

(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

 


Question 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

 

Answer :

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

 


Question 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

 

Answer :

i) (a2– b2)2 = a4 + b4 – 2a2b2

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

v) (2.5p – 1.5q)2 – (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) (m2 – n2m)2 + 2m3n2
= m4 – 2m3n2 + m2n4 + 2m3n2
= m4 + m2n4

 


Question 5 :

Show that.

(i) (3x + 7)2 – 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)2 = 48pq2

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

 

Answer :

i) LHS = (3x + 7)2 – 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

 


Question 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

 

Answer :

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

 


Question 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

 

Answer :

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

 


Question 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

 

Answer :

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

 


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