# Class 9 SELINA Solutions Maths Chapter 4 - Expansion

## 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 x^{2 }+ y^{2 }= 34.

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

We get (y + 2)^{2 }+ y^{2 }= 34

2y^{2 }+ 4y - 30 = 0

y^{2 }+ 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, a^{2 }+ b^{2 }= 73

Squaring on both sides gives

(a - b)^{2 }= 5^{2}

a^{2 }+ b^{2 }- 2ab = 25

But a^{2 }+ b^{2 }= 73

So 2ab = 73 - 25 = 48

ab = 24

So, the product of numbers is 24.

## 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 a^{3 }+ b^{3 }+ c^{3 }= 3abc

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

13^{3} + (-8)^{3} + (-5)^{3 }= 3(13)(-8)(-5) = 1560

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

7^{3} + 3^{3} + (-10)^{3 }= 3(7)(3)(-10) = -630

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

9^{3} - 5^{3} - 4^{3 }= 9^{3} + (-5)^{3} + (-4)^{3 }= 3(9)(-5)(-4) = 540

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

38^{3} + (-26)^{3} + (-12)^{3 }= 3(38)(-26)(-12) = 35568

### Solution 9

(i)

(ii)

### 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

### 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) = a^{2} - b^{2}}

= 9x^{2} + 4z^{2} + 2 × 3x × 2z - 25y^{2}

= 9x^{2} + 4z^{2} + 12xz - 25y^{2}

= 9x^{2} + 4z^{2 }- 25y^{2} + 12xz

(ii)

(3x - 5y - 2z) (3x - 5y + 2z)

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

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

= 9x^{2} + 25y^{2} - 2 × 3x × 5y - 4z^{2}

= 9x^{2} + 25y^{2}- 30xy - 4z^{2}

= 9x^{2} +25y^{2} - 4z^{2} - 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 }= 9^{2}

a^{2 }+ b^{2 }+ 2ab = 81

a^{2 }+ b^{2 }+ 40 = 81

So sum of squares is 81 - 40 = 41

Cubing on both sides gives

(a + b)^{3 }= 9^{3}

a^{3 }+ b^{3 }+ 3ab(a + b) = 729

a^{3 }+ b^{3 }+ 60(9) = 729

a^{3 }+ b^{3 }= 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 }= 5^{3}

x^{3 }- y^{3 }- 3xy(x - y) = 125

x^{3 }- y^{3 }- 72(5) = 125

x^{3 }- y^{3}= 125 + 360 = 485

So, difference of their cubes is 485.

Cubing both sides, we get

(x + y)^{3 }= 11^{3}

x^{3 }+ y^{3 }+ 3xy(x + y) = 1331

x^{3 }+ y^{3 }= 1331 - 72(11) = 1331 - 792 = 539

So, sum of their cubes is 539.

### Solution 16

xy = b ….(i)

4x^{2 }+ y^{2 }= a ….(ii)

Now, (2x + y)^{2} = (2x)^{2} + 4xy + y^{2}

= 4x^{2} + y^{2} + 4xy

= a + 4b ….[From (i) and (ii)]

## Expansion Exercise Ex. 4(C)

### Solution 1

### Solution 2

### Solution 3

### Solution 4

_{}

### Solution 5

_{}

### Solution 6

_{}

### Solution 7

### Solution 8

## Expansion Exercise Ex. 4(D)

### Solution 1

Given that x^{3} + 4y^{3} + 9z^{3} = 18xyz and x + 2y + 3z = 0

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

Now

_{}

### Solution 2

(i)

_{}

(ii)

_{}

### Solution 3

### Solution 4

Given that

_{}

### Solution 5

(i)

2(x^{2} + 1} = 5x

_{}

Dividing by x, we have

_{}

_{}

_{}

(ii)

_{}

_{}

### Solution 6

a^{2} + b^{2} = 34, ab= 12

(a + b)^{2} = a^{2} + b^{2} + 2ab

= 34 + 2 x 12 = 34 + 24 = 58

(a - b)^{2} = a^{2} + b^{2} - 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 => x^{2} - 5x = 1 => x^{2} - 1 = 5x

Dividing both sides by x,

_{}

_{}

_{}

### Solution 10

_{}

By cross multiplication,

=> x (5 - x) = 1 => x^{2} - 5x =-1 => x^{2} + 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) = -64c^{3}

=>27a^{3} + 125b^{3} + 45ab x (-4c) = -64c^{3}

=>27a^{3} + 125b^{3} - 180abc = -64c^{3}

=>27a^{3} + 125b^{3} + 64c^{3} = 180abc

Hence proved.

### Solution 12

Let a, b be the two numbers

.'. a + b = 7 and a^{3} + b^{3} = 133

(a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)

=> (7)^{3} = 133 + 3ab (7)

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

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

Now a^{2} + b^{2} = (a + b)^{2} - 2ab

=7^{2} - 2 x 10 = 49 - 20 = 29

### Solution 13

(i) 4x^{2} + ax + 9 = (2x + 3)^{2}

Comparing coefficients of x terms, we get

ax = 12x

so, a = 12

(ii) 4x^{2} + ax + 9 = (2x - 3)^{2}

Comparing coefficients of x terms, we get

ax = -12x

so, a = -12

(iii) 9x^{2} + (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

a^{3 }- b^{3 }= 316

Cubing both sides,

(a - b)^{3 }= 64

a^{3 }- b^{3 }- 3ab(a - b) = 64

Given a^{3 }- b^{3 }= 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

a^{2 }+ b^{2 }- 2ab = 16

a^{2 }+ b^{2 }= 16 + 42 = 58

Sum of their squares is 58.

## Expansion Exercise Ex. 4(E)

### Solution 1

Using identity:

(x + a)(x + b)(x + c) = x^{3} + (a + b + c)x^{2} + (ab + bc + ca)x + abc

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

= x^{3} + (6 + 4 - 2)x^{2} + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)

= x^{3} + 8x^{2} + (24 - 8 - 12)x - 48

= x^{3} + 8x^{2} + 4x - 48

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

= x^{3} + (-6 - 4 + 2)x^{2} + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2

= x^{3} - 8x^{2} + (24 - 8 - 12)x + 48

= x^{3} - 8x^{2} + 4x + 48

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

= x^{3} + (-6 - 4 - 2)x^{2} + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)

= x^{3} - 12x^{2} + (24 + 8 + 12)x - 48

= x^{3} - 12x^{2} + 44x - 48

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

= x^{3} + (6 - 4 - 2)x^{2} + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)

= x^{3} - 0x^{2} + (-24 + 8 - 12)x + 48

= x^{3} - 28x + 48

### Solution 2

### Solution 3

Using identity: (a ± b)^{3} = a^{3} ± b^{3} ± 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

a^{3 }- 8b^{3} + 27c^{3} = a^{3} + (-2b)^{3} + (3c)^{3}

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

a^{3 }- 8b^{3} + 27c^{3 }= a^{3} + (-2b)^{3} + (3c)^{3 }

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

= -18abc

### Solution 7

x + 5y = 10

⇒ (x + 5y)^{3} = 10^{3}

⇒ x^{3} + (5y)^{3} + 3(x)(5y)(x + 5y) = 1000

⇒ x^{3} + (5y)^{3} + 3(x)(5y)(10) = 1000

= x^{3} + (5y)^{3} + 150xy = 1000

= x^{3} + (5y)^{3} + 150xy - 1000 = 0

### Solution 8

### Solution 9

### Solution 10

x^{2 }+ y^{2 }+ z^{2 }- xy - yz - zx

= 2(x^{2 }+ y^{2 }+ z^{2 }- xy - yz - zx)

= 2x^{2 }+ 2y^{2 }+ 2z^{2 }- 2xy - 2yz - 2zx

= x^{2 }+ x^{2} + y^{2 }+ y^{2} + z^{2} + z^{2 }- 2xy - 2yz - 2zx

= (x^{2} + y^{2} - 2xy) + (z^{2} + x^{2} - 2zx) + (y^{2} + z^{2} - 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

= a^{2} + ab + ab + b^{2}

= a^{2} + b^{2} + 2ab

(ii) (a + b)(a + b)(a + b)

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

= (a^{2} + ab + ab + b^{2})(a + b)

= (a^{2} + b^{2} + 2ab)(a + b)

= a^{2} × a + a^{2} × b + b^{2} × a + b^{2} × b + 2ab × a + 2ab × b

= a^{3} + a^{2} b + ab^{2} + b^{3} + 2a^{2}b + 2ab^{2}

= a^{3} + b^{3} + 3a^{2}b + 3ab^{2}

^{ }

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

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

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

= a^{3} + (-b)^{3} + 3a^{2}(-b) + 3a(-b)^{2 }

= a^{3} - b^{3} - 3a^{2}b + 3ab^{2}