Chapter 4 : Expansion - Selina Solutions for Class 9 Maths ICSE

Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.

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Chapter 4 - Expansion Excercise Ex. 4(A)

Question 1

Find the square of:

(i) 2a + b

(ii) 3a + 7b

(iii) 3a - 4b

(iv) begin mathsize 11px style fraction numerator 3 straight a over denominator 2 straight b end fraction minus fraction numerator 2 straight b over denominator 3 straight a end fraction end style

Solution 1

Question 2

Use identities to evaluate:

(i) (101)2

 

(ii) (502)2

 

(iii) (97)2

 

(iv) (998)2

Solution 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Question 3

Evalute:

(i)

(ii) 

Solution 3

(i)


(ii) 

Question 4

Evaluate:

(i)

 

 

(ii) (4a +3b)2 - (4a - 3b)2 + 48ab.

Solution 4

(i)Consider the given expression:

 

 

 

 

 

 

 

 

 

 

(ii)Consider the given expression:

 

 

 

 

Question 5

If a + b = 7 and ab = 10; find a - b.

Solution 5

 

 

 

 

 

 

 

 

Question 6

If a -b = 7 and ab = 18; find a + b.

Solution 6

 

 

 

 

 

 

Question 7

If x + y = and xy = ; find:

(i) x - y

 

(ii) x2- y2

 

 

 

Solution 7

(i)

 

 

(ii)

 

 

 

 

 

 

 

 

 

Question 8

If a - b = 0.9 and ab = 0.36; find:

(i) a + b

(ii) a2 - b2.

Solution 8

(i)

 

(ii) 

 

 

 

 

 

 

Question 9

If a - b = 4 and a + b = 6; find

(i) a2 + b2

(ii) ab

Solution 9

(i)

 

(ii)

 

 

 

 

 

 

Question 10

If a + = 6 ans  a ≠ 0 find :

 

(i)

 

(ii)

 

Solution 10

(i)

 

 

 

 

 

 

 

 

 

 

(ii)

 

 

 

 

 

 

 

Question 11

If a - = 8 and a ≠0, find :

 

(i)

 

(ii)

 

Solution 11

(i)

 

 

 

 

 

 

 

 

 

(ii)

 

 

 

 

 

 

 

Question 12

If a2 - 3a + 1 = 0, and a≠ 0; find:

(i)

(ii)

Solution 12

(i)

 

 

 

 

 

(ii)

 

 

 

 

 

 

 

 

Question 13

If a2 - 5a - 1 = 0 and a ≠ 0; find:

(i)

(ii)

(iii)

Solution 13

(i)

 

 

 

 

 

(ii)

 

 

 

 

 

 

 

 

 

 

 

 

(iii)

 

 

 

 

 

 

 

 

 

Question 14

If 3x + 4y = 16 and xy = 4; find the value of 9x2 + 16y2.

Solution 14

 

 

 

 

 

 

 

 

 

 

 

Question 15

The number x is 2 more than the number y. If the sum of the squares of x and y is 34, then find the product of x and y.

 

Solution 15

Given x is 2 more than y, so x = y + 2



Sum of squares of x and y is 34, so x2 + y2 = 34.


Replace x = y + 2 in the above equation and solve for y.


We get (y + 2)2 + y2 = 34


2y2 + 4y - 30 = 0


y2 + 2y - 15 = 0


(y + 5)(y - 3) = 0


So y = -5 or 3


For y = -5, x =-3


For y = 3, x = 5




Product of x and y is 15 in both cases.

 

Question 16

The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.

 

Solution 16

Let the two positive numbers be a and b.

 

 

Given difference between them is 5 and sum of squares is 73.

 

 

So a - b = 5, a2 + b2 = 73

 

 

Squaring on both sides gives

 

 

(a - b)2 = 52

 

 

a2 + b2 - 2ab = 25

 

 

but a2 + b2 = 73

 

 

so 2ab = 73 - 25 = 48

 

 

ab = 24

 

 

 

 

 

So, the product of numbers is 24.

 

Chapter 4 - Expansion Excercise Ex. 4(B)

Question 1

Find the cube of :

(i) 3a- 2b

 

(ii) 5a + 3b

 

(iii) begin mathsize 11px style 2 straight a space plus space fraction numerator 1 over denominator 2 straight a end fraction left parenthesis straight a not equal to 0 right parenthesis end style

(iv) begin mathsize 11px style 3 straight a minus 1 over straight a left parenthesis straight a not equal to 0 right parenthesis end style 

Solution 1

(i) 


(ii) 


(iii) 


(iv) 

Question 2

If a2 + = 47 and ≠ 0 find:

(i)

 

(ii)

Solution 2

(i)

 

 

 

 

 

 

(ii)

 

 

 

 

 

Question 3

If a2 + = 18; a ≠ 0 find:

(i)

 

(ii)

Solution 3

(i)

 

 

 

 

 

 

 

 

(ii)

 

 

 

 

Question 4

If a + = p and a ≠ 0 ; then show that:

 

Solution 4

 

 

 

 

Question 5

If a + 2b = 5; then show that:

a3 + 8b3 + 30ab = 125.

Solution 5

 

 

 

 

 

Question 6

If  and a ≠ 0 ; then show: a3 +

 

Solution 6

 

 

 

Question 7

If a + 2b + c = 0; then show that:

a3 + 8b3 + c3 = 6abc.

Solution 7

 

 

 

Question 8

Use property to evaluate:

 

 

(i) 133 + (-8)3 + (-5)3

 

 

(ii)73 + 33 + (-10)3

 

 

(iii) 93 - 53 - 43

 

 

(iv) 383 + (-26)3 + (-12)3

 

Solution 8

Property is if a + b + c = 0 then a3 + b3 + c3 = 3abc

 

 

 

 

 

(i) a = 13, b = -8 and c = -5

 

 

133 + (-8)3 + (-5)3 = 3(13)(-8)(-5) = 1560

 

 

 

 

 

(ii) a = 7, b = 3, c = -10

 

 

73 + 33 + (-10)3 = 3(7)(3)(-10) = -630

 

 

 

 

 

(iii)a = 9, b = -5, c = -4

 

 

93 - 53 - 43 = 93 + (-5)3 + (-4)3 = 3(9)(-5)(-4) = 540

 

 

 

 

 

(iv) a = 38, b = -26, c = -12

 

 

383 + (-26)3 + (-12)3 = 3(38)(-26)(-12) = 35568

 

 

 

 

Question 9

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Solution 9

(i)

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

 

(ii)

a minus 1 over a equals 3
open parentheses a minus 1 over a close parentheses cubed equals 27
a cubed plus 1 over a cubed minus 3 open parentheses a minus 1 over a close parentheses equals 27
a cubed plus 1 over a cubed equals 27 plus 9 equals 36

 

 

 

Question 10

If a not equal to 0 and a - = 4; find:

(i)

 

(ii)

 

(iii)

 

Solution 10

(i)


(ii)


(iii)

Question 11

Ifx not equal to 0 and  x + = 2; then show that:

Solution 11

 

Thus from equations (1), (2) and (3), we have

 

Question 12

If 2x - 3y = 10 and xy = 16; find the value of 8x3 - 27y3.

Solution 12

Given that 2x - 3y = 10, xy = 16

:. (2x - 3y)3 = (10)3

Þ 8x3 - 27y3 - 3 (2x) (3y) (2x - 3y) = 1000 Þ 8x3 - 27 y3 -18xy (2x - 3y) = 1000

Þ 8x3 - 27 y3 - 18 (16) (10) = 1000

Þ 8x3 - 27 y3 - 2880 = 1000

Þ 8x3 - 27 y3 = 1000 + 2880

Þ8x3 - 27 y3 =3880

Question 13

Expand :

(i)  (3x + 5y + 2z) (3x - 5y + 2z)

(ii)  (3x - 5y - 2z) (3x - 5y + 2z)

Solution 13

(i)

(3x + 5y + 2z) (3x - 5y + 2z)

 

= {(3x + 2z) + (5y)} {(3x + 2z) - (5y)}

= (3x + 2z)2 - (5y)2

{since (a + b) (a - b) = a2 - b2}

= 9x2 + 4z2 + 2 × 3x × 2z - 25y2

= 9x2 + 4z2 + 12xz - 25y2

= 9x2 + 4z- 25y2 + 12xz

 

(ii)

(3x - 5y - 2z) (3x - 5y + 2z)

 

= {(3x - 5y) - (2z)} {(3x - 5y) + (2z)}

 

= (3x - 5y)2 - (2z)2{since(a + b) (a - b) = a2 - b2}

= 9x2 + 25y2 - 2 × 3x × 5y - 4z2

 

= 9x2 + 25y2- 30xy - 4z2

= 9x2 +25y2 - 4z2 - 30xy

 

Question 14

The sum of two numbers is 9 and their product is 20. Find the sum of their

(i) Squares (ii) Cubes

 

Solution 14

Given sum of two numbers is 9 and their product is 20.


Let the numbers be a and b.


a + b = 9


ab = 20





Squaring on both sides gives


(a+b)2 = 92


a2 + b2 + 2ab = 81


a2 + b2 + 40 = 81




So sum of squares is 81 - 40 = 41




Cubing on both sides gives


(a + b)3 = 93


a3 + b3 + 3ab(a + b) = 729


a3 + b3 + 60(9) = 729


a3 + b3 = 729 - 540 = 189




So the sum of cubes is 189.

 

 

 

 

Question 15

Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:

 

 

(i) Sum of these numbers

 

 

(ii) Difference of their cubes

 

 

(iii) Sum of their cubes.

 

Solution 15

Given x - y = 5 and xy = 24 (x>y)

 

 

(x + y)2 = (x - y)2 + 4xy = 25 + 96 = 121

 

 

So, x + y = 11; sum of these numbers is 11.

 

 

 

 

 

Cubing on both sides gives

 

 

(x - y)3 = 53

 

 

x3 - y3 - 3xy(x - y) = 125

 

 

x3 - y3 - 72(5) = 125

 

 

x3 - y3= 125 + 360 = 485

 

 

So, difference of their cubes is 485.

 

 

 

 

 

Cubing both sides, we get

 

 

(x + y)3 = 113

 

 

x3 + y3 + 3xy(x + y) = 1331

 

 

x3 + y3 = 1331 - 72(11) = 1331 - 792 = 539

 

 

So, sum of their cubes is 539.

 

 

 

 

Question 16

If 4x2 + y2 = a and xy = b, find the value of 2x + y.

Solution 16

xy = ab ….(i)

4x2 + y2 = a ….(ii)

Now, (2x + y)2 = (2x)2 + 4xy + y2

= 4x2 + y2 + 4xy

= a + 4b ….[From (i) and (ii)]

Chapter 4 - Expansion Excercise Ex. 4(C)

Question 1

Expand:

(i) (x + 8) (x + 10)

(ii) (x + 8) (x - 10)

(iii) (x - 8) (x + 10)

(iv) (x - 8) (x - 10) 

Solution 1

Question 2

Solution 2

  

Question 3

Solution 3

 

Question 4

If a + b + c = 12 and a2 + b2 + c2 = 50; find ab + bc + ca.

Solution 4

 

 

 

 

 

 

 

Question 5

If a2 + b2 + c2 = 35 and ab + bc + ca = 23; find a + b + c.

Solution 5

 

 

 

 

 

 

 

Question 6

If a + b + c = p and ab + bc + ca = q; find a2 + b2 + c2.

Solution 6

 

 

 

 

 

Question 7

If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c.

Solution 7

Question 8

If x+ y - z = 4 and x2 + y2 + z2 = 30, then find the value of xy - yz - zx.

Solution 8

Chapter 4 - Expansion Excercise Ex. 4(D)

Question 1

If x + 2y + 3z = 0 and x3 + 4y3 + 9z3 = 18xyz; evaluate:

 

Solution 1

Given that x3 + 4y3 + 9z3 = 18xyzand x + 2y + 3z = 0

\ x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y

Now

 

 

 

 

 

 

 

 

Question 2

If a + = m and ≠ 0 ; find in terms of 'm'; the value of :

(i)

 

(ii)

 

Solution 2

(i)

(ii)

 

Question 3

In the expansion of (2x2 - 8) (x - 4)2; find the value of

(i) coefficient of x3

(ii) coefficient of x2

(iii) constant term.

Solution 3

 open parentheses 2 straight x squared minus 8 close parentheses open parentheses straight x minus 4 close parentheses squared
equals open parentheses 2 straight x squared minus 8 close parentheses open parentheses straight x squared minus 8 straight x plus 16 close parentheses
equals 4 straight x to the power of 4 minus 16 straight x cubed plus 32 straight x squared minus 8 straight x squared plus 64 straight x minus 128
equals 4 straight x to the power of 4 minus 16 straight x cubed plus 24 straight x squared plus 64 straight x minus 128
Hence comma
coefficient space of space straight x cubed equals negative 16
coefficient space of space straight x squared equals 24
constant space term equals negative 128

 

 

 

 

 

 

 

 

 

 

Question 4

If x > 0 and find: .

 

 

Solution 4

Given that

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Question 5

If 2(x2 + 1) = 5x, find :

(i) (ii)

 

Solution 5

(i)

 

2(x2 + 1} = 5x

 

 

Dividing by x, we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii)

 

 

 

 

 

 

 

 

 

 

 

 

 

Question 6

If a2 + b2 = 34 and ab = 12; find:

(i) 3(a + b)2 + 5(a - b)2

(ii) 7(a - b)2 - 2(a + b)2

 

 

 

Solution 6

a2 + b2 = 34, ab= 12

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

 

 

= 34 + 2 x 12 = 34 + 24 = 58  

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

 

 

= 34 - 2 x 12 = 34- 24 = 10

(i) 3(a + b)2 + 5(a - b)2

 

 

= 3 x 58 + 5 x 10 = 174 + 50

= 224

 

 

 

 

(ii) 7(a - b)2 - 2(a + b)2

= 7 x 10 - 2 x 58 = 70 - 116 = -46

 

 

 

Question 7

If 3x -  and x ≠ 0 find : .

 

 

Solution 7

Given 3x -

 

We need to find

 

 

 

 

 

 

 

 

 

Question 8

If x2 + = 7 and  x ≠ 0; find the value of:

.

 

 

Solution 8

Given that

 

We need to find the value of

 

 

Consider the given equation:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Question 9

If x = and x ≠ 5 find .

 

Solution 9

 

By cross multiplication,

=> x (x - 5) = 1 => x2 - 5x = 1 => x2 - 1 = 5x

Dividing both sides by x,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Question 10

If x =  and x ≠ 5 find .

 

 

 

Solution 10

By cross multiplication,

=> x (5 - x) = 1 => x2 - 5x =-1 => x2 + 1 = 5x

Dividing both sides by x,

 

 

 

 

 

 

 

 

Question 11

If 3a + 5b + 4c = 0, show that:

27a3 + 125b3 + 64c3 = 180 abc

Solution 11

Given that 3a + 5b + 4c = 0

 

 

3a + 5b = -4c

 

Cubing both sides,

(3a + 5b)3 = (-4c)3

=>(3a)3 + (5b)3 + 3 x 3a x 5b (3a + 5b) = -64c3

=>27a3 + 125b3 + 45ab x (-4c) = -64c3

=>27a3 + 125b3 - 180abc = -64c3

=>27a3 + 125b3 + 64c3 = 180abc

 

 

Hence proved.

Question 12

The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.

Solution 12

Let a, b be the two numbers

.'. a + b = 7 and a3 + b3 = 133

 

(a + b)3 = a3 + b3 + 3ab (a + b)

=> (7)3 = 133 + 3ab (7)

 

=> 343 = 133 + 21ab => 21ab = 343 - 133 = 210

 

=> 21ab = 210 => ab= 2I

Now a2 + b2 = (a + b)2 - 2ab

=72 - 2 x 10 = 49 - 20 = 29

 

 

 

Question 13

In each of the following, find the value of 'a':

(i) 4x2 + ax + 9 = (2x + 3)2

(ii) 4x2 + ax + 9 = (2x - 3)2

(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2

Solution 13

(i) 4x2 + ax + 9 = (2x + 3)2

Comparing coefficients of x terms, we get

ax = 12x

so, a = 12

 

(ii) 4x2 + ax + 9 = (2x - 3)2

Comparing coefficients of x terms, we get

ax = -12x

so, a = -12

(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2

Comparing coefficients of x terms, we get

(7a - 5)x = 30x

7a - 5 = 30

7a = 35

a = 5

 

 

Question 14

If

 

(i)   (ii)

Solution 14

Given

 

 

 

 

 

 

 

 

 

 

 

 

 

Question 15

The difference between two positive numbers is 4 and the difference between their cubes is 316.

Find:

(i) Their product

(ii) The sum of their squares

Solution 15

Given difference between two positive numbers is 4 and difference between their cubes is 316.


Let the positive numbers be a and b

a - b = 4

a3 - b3 = 316

Cubing both sides,

(a - b)3 = 64

a3 - b3 - 3ab(a - b) = 64


Given a3 - b3 = 316

So 316 - 64 = 3ab(4)

252 = 12ab

So ab = 21; product of numbers is 21


Squaring both sides, we get

(a - b)2 = 16

a2 + b2 - 2ab = 16

a2 + b2 = 16 + 42 = 58

Sum of their squares is 58.

 

Chapter 4 - Expansion Excercise Ex. 4(E)

Question 1

Simplify:

(i) (x + 6)(x + 4)(x - 2)

(ii) (x - 6)(x - 4)(x + 2)

(iii) (x - 6)(x - 4)(x - 2)

(iv) (x + 6)(x - 4)(x - 2) 

Solution 1

Using identity:

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

(i) (x + 6)(x + 4)(x - 2)

= x3 + (6 + 4 - 2)x2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)

= x3 + 8x2 + (24 - 8 - 12)x - 48

= x3 + 8x2 + 4x - 48

 

(ii) (x - 6)(x - 4)(x + 2)

= x3 + (-6 - 4 + 2)x2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2

= x3 - 8x2 + (24 - 8 - 12)x + 48

= x3 - 8x2 + 4x + 48

(iii) (x - 6)(x - 4)(x - 2)

= x3 + (-6 - 4 - 2)x2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)

= x3 - 12x2 + (24 + 8 + 12)x - 48

= x3 - 12x2 + 44x - 48

 

(iv) (x + 6)(x - 4)(x - 2)

= x3 + (6 - 4 - 2)x2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)

= x3 - 0x2 + (-24 + 8 - 12)x + 48

= x3 - 28x + 48 

Question 2

Solution 2

  

Question 3

Using suitable identity, evaluate

(i) (104)3

(ii) (97)3 

Solution 3

Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)

(i) (104)3 = (100 + 4)3

= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)

= 1000000 + 64 + 1200 × 104

= 1000000 + 64 + 124800

= 1124864

 

(ii) (97)3 = (100 - 3)3

= (100)3 - (3)3 - 3 × 100 × 3(100 - 3)

= 1000000 - 27 - 900 × 97

= 1000000 - 27 - 87300

= 912673

Question 4

Solution 4

  

Question 5

Solution 5

  

Question 6

If a - 2b + 3c = 0; state the value of a3 - 8b3 + 27c3.

Solution 6

a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3

Since a - 2b + 3c = 0, we have

a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3

= 3(a)( -2b)(3c)

= -18abc 

Question 7

If x + 5y = 10; find the value of x3 + 125y3 + 150xy - 1000.

Solution 7

x + 5y = 10

(x + 5y)3 = 103

x3 + (5y)3 + 3(x)(5y)(x + 5y) = 1000

x3 + (5y)3 + 3(x)(5y)(10) = 1000

= x3 + (5y)3 + 150xy = 1000

= x3 + (5y)3 + 150xy - 1000 = 0 

Question 8

Solution 8

  

Question 9

If a + b = 11 and a2 + b2 = 65; find a3 + b3.

Solution 9

  

Question 10

Prove that:

x2+ y2 + z2 - xy - yz - zx  is always positive.

Solution 10

x2 + y2 + z2 - xy - yz - zx

= 2(x2 + y2 + z2 - xy - yz - zx)

= 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx

= x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx

= (x2 + y2 - 2xy) + (z2 + x2 - 2zx) + (y2 + z2 - 2yz)

= (x - y)2 + (z - x)2 + (y - z)2

Since square of any number is positive, the given equation is always positive.

Question 11

Find:

(i) (a + b)(a + b)

(ii) (a + b)(a + b)(a + b)

(iii) (a - b)(a - b)(a - b) by using the result of part (ii)

Solution 11

(i) (a + b)(a + b) = (a + b)2

= a × a + a × b + b × a + b × b

= a2 + ab + ab + b2

= a2 + b2 + 2ab

 

(ii) (a + b)(a + b)(a + b)

= (a × a + a × b + b × a + b × b)(a + b)

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

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

= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b

= a3 + a2 b + ab2 + b3 + 2a2b + 2ab2

= a3 + b3 + 3a2b + 3ab2

 

(iii) (a - b)(a - b)(a - b)

In result (ii), replacing b by -b, we get

(a - b)(a - b)(a - b)

= a3 + (-b)3 + 3a2(-b) + 3a(-b)2

= a3 - b3 - 3a2b + 3ab2

 

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