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

Class 8 Maths Chapter 9 Algebraic Expressions and Identities

NCERT Solutions For Class 8 Maths Chapter 9 Algebraic Expressions And Identities, (Maths) exam are Students are taught thru NCERT books in some of state board and CBSE Schools.  As the chapter involves an end, there is an exercise provided to assist students prepare for evaluation.  Students need to clear up those exercises very well because the questions withinside the very last asked from those. 

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

Class 8 Maths Chapter 9 Algebraic Expressions and Identities

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 Term : -pq Term : qr Term : -p	3 -1 1 -1
v)	(x/2) + (y/2) – xy	Term : x/2 Term : Y/2 Term : -xy	½ 1/2 -1
vi)	0.3a – 0.6ab + 0.5b	Term : 0.3a Term : -0.6ab Term : 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 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 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:

Solution:

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.

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 × a^{2 + 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

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

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) 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

= (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) = 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)

= 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)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 = 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 + 49n²
= 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

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 – (4pq – 3q)² = 48pq²

(iv) (4pq + 3q)²– (4pq – 3q)² = 48pq²

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

Solution:

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

LHS = RHS

iv)  LHS = (4pq + 3q)²– (4pq – 3q)²

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

              = 48pq²

 RHS=48pq²

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 + 0.4

            = 27.4

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

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