# NCERT Class 8 Maths Chapter 9 – Algebraic Expressions and Identities solutions

## 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: 1Term: xTerm: x2 1 1 1 iii) 4x2y2 – 4x2y2z2 + z2 Term: 4x2y2Term: -4 x2y2z2Term :  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 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

(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 = xyz2

(ii) a × – a2  × a= – a6

(iii) 2 × 4y × 8y2 × 16y= 1024 y6

(iv) a × 2b × 3c × 6abc = 36abc2

(v) m × – mn × mnp = –mnp

## 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– 9, 4a

(v) pq + qr + rp, 0

Solution:

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

(ii)ab(a – b) = ab – 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 × 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(a+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) (p– 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)

= p(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)

= t– st + s2t– 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) = 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) (2a+ 9) (2a+ 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)= b2 – 14b + 49

(ii) (xy + 3z)= x2y2 + 6xyz + 9z2

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

(iv) [(2m/3}) + (3n/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) (a– b2)2

(ii) (2x + 5) – (2x – 5)2

(iii) (7m – 8n)+ (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)+ (7m + 8n)2
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 49n2
= 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 – (4pq – 3q)2 = 48pq2

(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

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 + 0.4 = 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 a– 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) 153– 1472

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

iv) 12.1– 7.92

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

8. Using (x + a) (x + b) = x+ (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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