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# Class 9 SELINA Solutions Maths Chapter 4 - Expansion

## Expansion Exercise Ex. 4(A)

### Solution 1(a)

Correct option: (iii) 166

Substituting  and  in ,

### Solution 1(b)

Correct option: (i) 11

Substituting  and  in ,

### Solution 1(c)

Correct option: (iv) 7

### Solution 1(d)

Correct option: (ii) 58

### Solution 1(e)

Correct option: (iii) 84

### Solution 1(f)

Correct option: (i) 8

### Solution 1(g)

Correct option: (ii)

Now,

### Solution 1(h)

Correct option: (ii)

Now,

### Solution 1(i)

Correct option: (ii) 2

(i)

(ii)

### Solution 5

(i)Consider the given expression:

(ii)Consider the given expression:

(i)

(ii)

(i)

(ii)

(i)

(ii)

(i)

(ii)

(i)

(ii)

(i)

(ii)

(i)

(ii)

(iii)

### Solution 16

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 the cases.

### Solution 17

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.

## Expansion Exercise Ex. 4(B)

### Solution 1(a)

Correct option: (iv) 2

### Solution 1(b)

Correct option: (ii) 90

### Solution 1(c)

Correct option: (i) 0

Given,

Now,

And,

### Solution 1(d)

Correct option: (iii)

(i)

(ii)

(iii)

(iv)

(i)

(ii)

(i)

(ii)

### Solution 9

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

(i)

(ii)

(i)

(ii)

(iii)

### Solution 12

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

### Solution 13

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

### Solution 14

(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 + 25y- 30xy - 4z2

= 9x2 +25y2 - 4z2 - 30xy

### Solution 15

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 both sides,

(a + b)2 = 92

a2 + b2 + 2ab = 81

a2 + b2 + 40 = 81

So sum of squares is 81 - 40 = 41

Cubing both sides,

(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 16

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 both sides,

(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 17

xy = b ….(i)

4x2 + y2 = a ….(ii)

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

= 4x2 + y2 + 4xy

= a + 4b ….[From (i) and (ii)]

## Expansion Exercise Ex. 4(C)

### Solution 1(a)

Correct option: (ii)

### Solution 1(b)

Correct option: (ii)

### Solution 1(c)

Correct option: (ii) 1

### Solution 1(d)

Correct option: (iii)

## Expansion Exercise Ex. 4(D)

### Solution 1(a)

Correct option: (i) 4

Now,

### Solution 1(b)

Correct option: (iii) -10

Therefore, the coefficient of  is -10.

### Solution 1(c)

Correct option: (ii)

### Solution 1(d)

Correct option: (i) -24

### Solution 2

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

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

Now

(i)

(ii)

Given that

### Solution 6

(i)

2(x2 + 1} = 5x

Dividing by x, we have

(ii)

### Solution 7

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)= 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

Given 3x -

We need to find

### Solution 9

Given that

We need to find the value of

Consider the given equation:

### Solution 10

By cross multiplication,

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

Dividing both sides by x,

### Solution 11

By cross multiplication,

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

Dividing both sides by x,

### Solution 12

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 13

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 14

(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

Given

### Solution 16

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.

## Expansion Exercise Test Yourself

### 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 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 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 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