Class 9 SELINA Solutions Maths Chapter 4: Expansion

Correct option: (iii) 166

Substituting  and  in ,

Correct option: (i) 11

Substituting  and  in ,

Correct option: (iv) 7

Correct option: (ii) 58

Correct option: (iii) 84

Correct option: (i) 8

Correct option: (ii)

Now,

Correct option: (ii)

Now,

Correct option: (ii) 2

(i)

(ii) 

(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)

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.

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.

Correct option: (iv) 2

Correct option: (ii) 90

Correct option: (i) 0

Given,

Now,

And,

Correct option: (iii)

(i) 

(ii) 

(iii) 

(iv) 

(i)

(ii)

(i)

(ii)

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)

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

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

 (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 + 4z² - 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

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.

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.

xy = b ….(i)

4x² + y² = a ….(ii)

Now, (2x + y)² = (2x)² + 4xy + y²

= 4x² + y² + 4xy

= a + 4b ….[From (i) and (ii)]

Correct option: (ii)

Correct option: (ii)

Correct option: (ii) 1

Correct option: (iii)  

Correct option: (i) 4

Now,

Correct option: (iii) -10

Therefore, the coefficient of  is -10.

Correct option: (ii)

Correct option: (i) -24

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

 (i)

 2(x² + 1} = 5x

 Dividing by x, we have

 (ii)

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

Given 3x -

 We need to find

Given that

 We need to find the value of

 Consider the given equation:

By cross multiplication,

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

Dividing both sides by x,

By cross multiplication,

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

Dividing both sides by x,

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.

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

(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

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.

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 

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

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 

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 

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.

(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