Mathematics 10 (Science) - KPK - Gain - Solution - PART - 16 - 24 Chapter 2

24 Chapter 2 

Exercise 2.2 

2 w = - 1 + i 3, 2 w 2 = - 1 - 

i 3 Comparing with the quadratic equation Taking LHS ( - 1 + i 3 ) 3 + ( - 1 - 

i 3 ) 

3 we have 1 , 2, 4 

Using quadratic formula = ( 2 w ) 3 + 

( 2 

w 

2 

) 

3 Putting values of a,b & c 

= 2 3 w 3 + 

2 

3 ( w 

3 

) 2 = 8 ( 1 ) + 

8. ( 1 

) 

2 

= 8 + 8= 16 RHS Exercise 2.2 Q1i). Find cube root of 

1- Sol: Let x be a cube root of 1- 

, then x = 3 - 1 = ( - 

1 

) 

31 

Either Or 

( x ) 3 = ⎧ │ ⎨ │ ⎩ ( 1 ) 

3 1 3 x 31 - ⎫ │ ⎬ │ ⎭ ⇒ = - ⇒ x 3+ 1 = 

0 

Solution Set 

Q1iii). Find cube root of - 27 Sol: Let x be a cube root of - 27, then 

x = 3 - 27 = ( - 27 )1 3 ⇒ x 3 + 1 3 

= 

0 ⇒ ( x + 1 )( x 2 - x.1 + 1 2 

) 

= 

0 

⇒ x 3 = - 

27 327 0 ⇒ ( x + 1 )( x 2 

- x + 1 ) 

= 

0 

3 3 30 Either x x x+ = 

x 

+ = 

+ 1 = 

0 = - 

1 

Or 

x 2- x + 1 = 

0 ( x + 3 )( x 2 - x .3 + 3 2 ) = 

0 Either 3 0 Comparing with the quadratic equation 

3 ax 2 + bx + c = 0 we have a = 1 , b = - 1, c = 

1 Using quadratic formula x b b 2 2a 

x+ = 

Or x 2 - 3 x + 9 = 0 x = - Comparing with we have 

1, -3 9 Using 

= 

- ± - 4ac Putting values of a,b & c 

( ) ( ) ( )( ) 

x ( 1 ) ( 1 ) 2 4 ( 1 )( 1 ) 

2 ( 1 )Either Or 

Solution Set 

Q1ii). Find cube root of 8 Sol: Let x be a cube root of 8, then 

Either Or 

x 1 1 4 2 

( ) x 1 2 3 1 i 3 2 

i 1 = 

- - 3 ± - 3 2 - 4 1 9 2 1 

= 3 ± 2 9 - 36 = 

3 ± 2 

- 27 

= 3 ± 9 - 3 2 = 

3 ± 3 2 

- 3 = - 3 ⎛ │ │ ⎝ x= - - ± - - x= ± - = ± - = ± ∴ = - xx = - 

( - 1 2 

i x 

- 1 2 

- 3 ⎞ │ │ ⎠ 

x = - 

3 ) ( - 1 - i 3 ) 2 x = - 

ω2x = - 

( - 1 + 2 i 3 ) = { - 1, - ,ω - ω 2 }x = - ωx = 3 8 =( 8 ) 31 ( x ) 3 = ⎧ │ ⎨ │ ⎩ ( 8 

) 

3 1 ⎫ │ ⎬ │ ⎭ 3 ⇒ x 3= 

8 ⇒ x 3- 8 = 

0 ⇒ x 3 - 2 3 

= 

0 ⇒ ( x - 2 )( x 2 + x.2 + 2 2 

) 

= 

0 ⇒ ( x - 2 )( x 2 

+ 2x + 4 ) 

= 

0 

x - 2 = 

0 x = 

2 

x 2+ 2x + 4 = 

0 

2 ax + bx + c = 0 a = b = c = 

x = 

- b ± b 2 - 4ac 2a 

x = 

- ( 2 ) ± ( 2 ) 2 - 4 ( 1 )( 4 ) 2 ( 1 

) 

x = - 2 ± 2 4 - 16 = 

- 2 ± 2 

( ) - 12 x = - 2 ± 2 - 3 × 4 = - 2 ± 2 

2i 3 ∴ i = - 1 - 1 ± 

i 3 x = 

2 x =2 

2 ( - 1 + 2 

i 3 ) x = 

2ωx =2 ( - 1 - 2 

i 3 

) 

= { 2,2 ω ,2 ω 2 }x = 2ω2ax 2 + bx + c = 0 a = b = c = 

x = 

- b ± b 2 - 4ac 2a 

Either x = - 3 ⎛ │ │ ⎝ - 1 - 2 ω 12 + ω 58 + 

ω 95 ω 12 + ω 58 + ω 95 ( ) ( ) ( ) ( ) ( ) ( ) 

- 3 ⎞ │ │ ⎠ 

Or x = - 3 ⎛ │ │ ⎝ - 1 + 2 - 3 ⎞ │ │ ⎠ Solution x Set = - 3 = w { 2 - 3, - 3 w , - 

3 w 2 }x =- 3 w Q2i). Evaluate Sol: Since = ω 12 + ω 57 + 1 + 

ω 

93 + 

2 

= ω 3 × 4 + ω 3 × 19 . ω 1 + 

ω 3 × 

31 . 

ω 

2 

= ω 3 4 + ω 3 19 . ω + ω 3 31 . ω 2 ∴ ω 

3 

= 

1 

= 1 4 + 1 19 . ω + 

1 31 . 

ω 

2 

= 1 + 1. ω + 1. ω 2 ∴ 1 + ω + ω 

2 

= 

0 = 1+ ω + 

ω 

2 

= 

0