Mathematics 10 (Science) - KPK - Gain - Solution - PART - 15 - 23 Chapter 2

23 Chapter 2 

Exercise 2.1 

⇒ ( 5 ) 2 - 4 ( 1 )( k ) 

≥ 

0 

⇒ 25 - 4k ≥ 

0 ⇒ 25 ≥ 

4k ⇒ 25 4 ≥ k Or k ≤ 

25 4 (b) if roots are imaginary then b 2- 4ac < 0 ⇒ ( 5 ) 2 - 4 ( 1 )( k ) 

< 

0 

⇒ 25 - 4k < 

0 ⇒25 < 4k ⇒ 25 4 < k Or k > 

25 4 

Solution Set = ⎧ │ │ ⎨ │ │ ⎩ Re Im al aginary k ≤ k >

25 425 4 ⎫ │ │ ⎬ │ │ ⎭ Cube root of unity Let x be a cube root of unity, then 

x= 3 

1 = 

( 1 

) 

13 

( x) 3 

= 

( 1 13 

) 3 ⇒ x 3= 

1 ⇒ x 3- 1 = 

0 ⇒ x 3 - 1 3 

= 

0 ⇒ ( x - 1 )( x 2 + x.1 + 1 2 

) 

= 

0 ⇒ ( x - 1 )( x 2 

+ x + 1 ) 

= 

0 

Either x x - 1 0 

= 1 

= 

Or x 2+ x + 1 = 0 Comparing with the quadratic equation ax 2 + bx + c = 0 we have a =1 , 1, 1 Using quadratic formula 

Putting values of a,b & c 

Solution Set 

Let then 

Solution Set 

Sum of the cube roots of unity 1). 

Sol: since 

Taking LHS and putting the values 1 + w + w 2 = 1 + - 1 + 2 - 3 + 

- 1 - 2 3 2 1 3 1 3 

2 

b = c = 

x = 

- b ± b 2 - 4ac 2a 

x = 

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

) 

x = 

- 1 ± 2 1 - 4 x = - 1 ± 2 - 3 = - 1 ± i 2 

3 ∴ i = - 1 = ⎧ │ ⎨ │ ⎩ 1, - 1 + 2 i 3 , - 1 - 2 

i 3 ⎫ │ ⎬ │ ⎭ 

ω = - 1 + i 2 = 

3 ⇒ ω 2 { 1, ,ω ω 2 }= 

- 1 - i 2 3 

1 + ω + ω 2 = 

0 ω = - 1 + i 2 3 ⇒ ω 2 = 

- 1 - i 2 3 

= - + - - - = 02 = 0 = RHS 

Product of the cube roots of unity 

Since Taking 1. ww . 2 = ( 1 ) ⎛ │ │ ⎝ - 1 + 2 - 3 ⎞⎛ ││ ││ ⎠⎝ - 1 - 2 ω = - 1 + i 2 3 ⇒ ω 2 = 

- 1 - i 2 3 ⎞ │ │ ⎠ ( ) ( ) 

( ) 

3 

= - 1 2 - 2 × 

2 2 ω = - 1 + i 2 - 3 = 

1 - 4 

- 3 

= 1 + 4 3 = 

4 4 w 3 = 1 Reciprocal of the cube roots of unity Since w 3 = 1 Taking 

w = w .w w 22 = w3w 2∴ w 3 = 1 w = w 1 2And w 2 = w 2.w w = w3w ∴ w 3 = 1 w 2 = w 1 Exp 7 Show x 3 + y 3 = ( x + that 

y )( x + wy )( x + w 2 y ) Sol: Taking RHS = ( x + y )( x + wy )( x + 

w 2 y ) = ( x + y )( x 2 + wxy + wxy 2 + 

wy 

3 2 

) = ( x + y ) ( x 2 + ( w + w 2 ) xy + 

y 

2 

) = ( x + y ) ( x 2 + ( - ) xy + 

y 

2 

) = ( x + y )( x 2 - xy + 

y 

2 

) 

= x 3 + 

y 

3 

Exp8.Evaluate w 15 , w 24 , w 90 , w 101 , w - 2 , w - 13 Sol: Given w 15 , w 24 , w 90 , w 101 , w - 2 , w - 13 ( ) ( ) ( ) ( ) ( ) ( ) 

3 ⇒ ω 2 = 

- 1 - i 2 3 

1. 

1 

w 15 = w 

3 

5 = 1 5 = 

1 

w 24 = w 

3 

8 = 1 8 = 

1 

w 90 = w 

3 

30 = 1 30 = 

1 

w 101 = w 99 . w 2 = ( w 3 )33 . w 2 = 1. 

w 2 = 

w 

2 

w - 2 = w - 2 .1 = w - 2 . 

w 3 = w - 2 + 

3 

= w w - 13 = w - 13 .1 = w - 13 .1 5 = w - 13 . w 3 × 5 = w - 13 + 15 = w 2 Exp9 Show that ( - 1 + i 3 ) 3 + ( - 1 - i 3 ) 

3 = 16 Sol: since