SELINA Solutions for Class 9 Maths Chapter 4 - Expansion

Chapter 4 - Expansion Exercise Ex. 4(A)

Solution 1

Solution 2

 

 

 

Solution 3

(i)


(ii) 

Solution 4

(i)Consider the given expression:

 

(ii)Consider the given expression:

Solution 5

Solution 6

Solution 7

(i)

 

(ii)

Solution 8

(i)

 

(ii) 

Solution 9

(i)

 

(ii)

Solution 10

(i)

(ii)

Solution 11

(i)

 

(ii)

 

Solution 12

(i)

 

(ii)

Solution 13

(i)

(ii)

(iii)

Solution 14

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.

 

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 Exercise Ex. 4(B)

Solution 1

(i) 


(ii) 


(iii) 


(iv) 

Solution 2

(i)

(ii)

Solution 3

(i)

(ii)

Solution 4

Solution 5

Solution 6

 

 

 

Solution 7

 

 

 

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

 

 

 

 

Solution 9

(i)

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

 

(ii)

begin mathsize 11px style a minus 1 over a equals 3
open parentheses a minus 1 over a close parentheses cubed equals 27
a cubed minus 1 over a cubed minus 3 open parentheses a minus 1 over a close parentheses equals 27
a cubed minus 1 over a cubed equals 27 plus 9 equals 36 end style

Solution 10

(i)


(ii)


(iii)

Solution 11

 

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

 

Solution 12

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

left parenthesis 2 x space minus space 3 y right parenthesis cubed space equals space left parenthesis 10 right parenthesis to the power of 3 end exponent space space
rightwards double arrow 8 x cubed space minus space 27 y cubed space minus space 3 space left parenthesis 2 x right parenthesis space left parenthesis 3 y right parenthesis space left parenthesis 2 x space minus space 3 y right parenthesis space equals space 1000 space space space space
rightwards double arrow 8 x cubed space minus space 27 space y cubed space minus 18 x y space left parenthesis 2 x space minus space 3 y right parenthesis space equals space 1000 space space
rightwards double arrow 8 x cubed space minus space 27 space y cubed space minus space 18 space left parenthesis 16 right parenthesis space left parenthesis 10 right parenthesis space equals space 1000 space space
rightwards double arrow 8 x cubed space minus space 27 space y cubed space minus space 2880 space equals space 1000 space space
rightwards double arrow 8 x cubed space minus space 27 space y cubed space equals space 1000 space plus space 2880 space space
rightwards double arrow 8 x cubed space minus space 27 space y cubed space equals space 3880

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

 

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.

 

 

 

 

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.

 

 

 

 

Solution 16

xy = b ….(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 Exercise Ex. 4(C)

Solution 1

Solution 2

Solution 3

 

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Chapter 4 - Expansion Exercise Ex. 4(D)

Solution 1

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

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

Now

Solution 2

(i)

(ii)

 

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

Solution 4

Given that

Solution 5

(i)

 

2(x2 + 1} = 5x

 

 

Dividing by x, we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii)

 

 

 

 

 

 

 

 

 

 

 

 

 

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

 

 

 

Solution 7

Given 3x -

 

We need to find

 

 

 

 

 

 

 

 

 

Solution 8

Given that

 

We need to find the value of

 

 

Consider the given equation:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 9

 

By cross multiplication,

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

Dividing both sides by x,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 10

By cross multiplication,

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

Dividing both sides by x,

 

 

 

 

 

 

 

 

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.

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

 

 

 

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

 

 

Solution 14

Given

 

 

 

 

 

 

 

 

 

 

 

 

 

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 Exercise Ex. 4(E)

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 

Solution 2

  

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

Solution 4

  

Solution 5

  

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 

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 

Solution 8

  

Solution 9

  

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.

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