Class 8 | NCERT Solution Maths Chapter 9 | Algebraic Expressions and Identities

Answers to NCERT 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 :

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.

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) l 2  + m 2 , m 2  + n 2 , n 2  + l 2 , 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

= (2p 2 q 2  – 3pq + 4) + (5 + 7pq – 3p 2 q 2 )

= 2p 2 q 2  – 3p 2 q 2  – 3pq + 7pq + 4 + 5

= – p2q2 + 4pq + 9

iv) (l 2  + m 2 ) + (m 2  + n 2 ) + (n 2  + l 2 ) + (2lm + 2mn + 2nl)

= l 2  + l 2  + m 2  + m 2  + n 2  + n 2  + 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) 4p 3 , - 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 = 12mn 2 p  

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, 3a 2 , 7a 4

(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  ×  2x 2 y ×  2xy 2  = (1 × 2 × 2) (x × x 2 × x × y × y × y 2 ) =   4x 4 y 4

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

5. Obtain the product of

(i) xy,  yz, zx

(ii) a, - a 2  , a 3

(iii) 2, 4y, 8y2 , 16y3

(iv) a, 2b, 3c, 6abc

(v) m, – mn, mnp

Solution:

(i) xy × yz × zx = x2 y2 z2

(ii) a × - a 2   × a 3 = - a 6

(iii) 2 × 4y × 8y2 × 16y3 = 1024 y6

(iv) a × 2b × 3c × 6abc = 36a2 b2 c2

(v) m × – mn × mnp = –m 3 n 2 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) a 2 - 9, 4a

(v) pq + qr + rp, 0

Solution:

(i)4p(q + r) = 4pq + 4pr

(ii) ab (a - b) = a 2 b - ab 2

(iii)(a + b) (7a 2 b 2 ) = 7a 3 b 2  + 7a 2 b 3

(iv) (a 2  - 9) (4a) = 4a 3  - 36a

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

2. Complete the table.

Solution:

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 x 4) ( a 2 x a 22 x a 26 )

= 8 × a 2 + 22 + 26 

= 8a50

ii) (2xy/3) ×(-9x2y2/10)

= (2/3 × -9/10) (x × x 2 × y × y 2 )

= (-3/5 x 3 y 3 )

iii) (-10pq3/3) ×(6p3q/5)

= ( -10/3 × 6/5 ) (p × p3× q3 × q)

= (-4p 4 q 4 )

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 + 5l 2

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

= (-4ac + 4bc + 4c 2 ) – (3a 2  + 3ab + 3ac – ( 2ab – 2b 2 + 2bc))

=-4ac + 4bc + 4c 2 – (3a 2  + 3ab + 3ac – 2ab + 2b 2 – 2bc)

= -4ac + 4bc + 4c 2  – 3a 2  – 3ab – 3ac + 2ab – 2b 2 + 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)

= yx 3y - 4y - 8 x 3y + 32

= 3y 2  - 4y - 24y + 32

= 3y 2  – 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

= 6.25l 2 + 1.25 lm – 1.25 lm – 0.25 m 2

= 6.25l 2 – 0.25 m 2

iv) (a + 3b) (x + 5)

= ax + 5a + 3bx + 15b

c) (2pq + 3q 2 )  (3pq - 2q 2 )

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

= 6p 2 q 2  – 4pq 3  + 9pq 3  – 6q 4

= 6p 2 q 2  + 5pq 3  – 6q 4

(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)

= a 3  + a 2 b 2 + ab + b 3

= a 3  + b 3  + a 2 b 2 + ab

iv) (p2– q2) (2p + q)

= p2 (2p + q) – q2 (2p + q)

=2p 3  + p 2 q – 2pq 2  – q 3

= 2p 3  – q 3  + p 2 q – 2pq 2

3. Simplify.

(i) (x2– 5) (x + 5) + 25

(ii) (a 2 + 5) (b 3 + 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) (x 2 - xy + y 2 )

(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) (a 2 + 5) (b 3 + 3) + 5

= a 2 b 3  + 3a 2  + 5b 3  + 15 + 5

= a 2 b 3  + 5b 3  + 3a 2  + 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) (x 2 - xy + y 2 )

= x 3  – x 2 y + xy 2  + x 2 y – xy 2  + y 3

= x 3  + y 3

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

= a 2  + b 2  – c 2  + 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) (a 2 + b 2 ) (- a 2 + b 2 )

(vii) (6x - 7) (6x + 7)

(Viii) (- 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

= 4a 2  – 28a + 49

Using (a-b) 2

= a 2 + b 2 – 2ab

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

= 9a 2  -3a+(1/4)

Using (a-b) 2

= a 2 + b 2 – 2ab

v) (1.1m - 0.4) (1.1m + 0.4)

= 1.21m 2  - 0.16

Using (a – b)(a + b)

= a 2 – b 2

vi) (a 2 + b 2 ) (- a 2 + b 2 )

= (b 2  + a 2  ) (b 2  – a 2 )

= -a 4  + b 4

Using (a – b)(a + b) = a2 – b2

vii) (6x - 7) (6x + 7)

=36x2 – 49 Using (a – b)(a + b)

= a 2 – b 2

viii) (- a +) (- + c) = (- + c) 2

= c2 + a2 – 2ac Using (a-b) 2

= a 2 + b 2 – 2ab

= (x 2 / 4) + (9y 2 /16) + (3xy/4)

Using (a+b) 2

= a 2 + b 2 + 2ab

x) (7a – 9b) (7a – 9b) = (7a – 9b) 2

= 49a 2  – 126ab + 81b 2

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) (2a 2 + 9) (2a 2 + 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)

= 4x 2  + (5y + 3y) 2x + 15y 2

= 4x2 + 16xy + 15y2

vi) (2a 2 + 9) (2a 2 + 5)

= 4a 4  + (9+5)2a 2  + 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 = (4m 2 /9) + (9n 2 / 4) + 2mn

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

(vi) (2xy + 5y) 2 = 4x 2 y 2  + 20xy 2  + 25y 2

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) (m 2  - n 2 m) 2  + 2m 3 n 2

Solution:

i) (a 2 - b 2 ) 2 = a 4  + b 4 - 2a 2 b 2

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
= 49m 2  - 112mn + 64n 2  + 49m 2  + 112mn + 49n 2
= 98m 2  + 128n 2

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

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 = a 2 b 2  + 2ab 2 c + b 2 c 2  – 2ab 2 c = a 2 b 2  + b 2 c 2

vii) (m 2  - n 2 m) 2  + 2m 3 n 2
= m 4  - 2m 3 n 2  + m 2 n 4  + 2m 3 n 2
= m 4  + m 2 n 4

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

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

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) 5.2 2

= (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