Ex 4 4 Fsc Part1

Exercise 4.4 (Solutions)

mathcity.org

Textbook of Algebra and Trigonometry for Class XI

Merging man and maths

Available online @ http://www.mathcity.org, Version: 1.0.0

{ Question # 1 (i) Let x be a cube root of 8 then 1

x  (8)3

 x3  8

 x3  8  0

 (x )3  (2)3  0

 (x  2)(x 2  2x  4)  0

 x 2  0  x 2

or

or

x 2  2x  4  0

Review: 1  3  2 1  3 2  2

2  (2)2  4(1)(4) x 2(1)

2  4  16 2  12  2 2  2  2 3 1  3    2     2 2  1  3   1  3   x  2   or x  2       2 2  x  2 or x  2 2 

Hence cube root of 8 are 2,2 and 2 2 .

(ii) Hint

Considering x as a cube root of 8 and Solving as above you will get the following values of x 2  2 3 2  2 3 x  2 , x  , x 2 2  1  3   1  3   x  2   , x  2       2 2  x  2 2 , x  2 Hence cube root of 8 are 2,  2 and 2 2 .

(iii)

Do yourself as (iv) below.

(iv) Let x be a cube root of 27 then 1

x  (27)3

 x 3  27

 x 3  27  0

 (x )3  (3)3  0

 (x  3)(x 2  3x  9)  0

 x 3  0  x  3

x 2  3x  9  0

or or

3  (3)2  4(1)(9) x 2(1)

3  9  36 3  27 3  3 3   2 2 2 3  3 3 3  3 3  x or x  2 2 

FSC-I - 4.4 - 2

 1  3   x  3    2   x  3 2 or

or

Hence cube root of 27 are 3,  3 and 3 2 .

 1  3  x  3     2 x  3

(v) Let x be a cube root of 64 then 1

x  (64)3

 x 3  64

 x 3  64  0

 (x )3  (4)3  0

 (x  4)(x 2  4x  16)  0

 x 4  0  x 4

x 2  4x  16  0

or

4  (4)2  4(1)(16) x 2(1)

or

4  16  64 4  48  2 2 4  4 3  Q 48  16  3 2 4  4 3 4  4 3  x or x  2 2  1  3   1  3   or x  4    x  4      2 2  x  4 or x  4 2 

Hence cube root of 64 are 4, 4 and 4 2 .

{ Question # 2 8 8 (i) 1    2   1    w 2  2w 2   0  2w 2 

8

Q 1    2  0

 (2)8 (w 2 )8  256 16

 256  15    256  3    5

 256 15    256 

(ii)

Q 3  1

 28  w 29  1   27     27   2  1

  3      3    2  1 9

9

 1    1   2  1 9

9

   2  1  0 Answer

(iii)

Answer

1     2 1     2   1     2  2 2 1     2  2   0  2 2 0  2   2 2 2   4  3  4(1) = 4

Answer

Q 3  1 Q 1    2  0 Q 1    2  0 Q 3  1

Made By: Atiq ur Rehman ([email protected]), http://www.mathcity.org

FSC-I - 4.4 - 3

(iv)

 1  3 7  1  3 7          2 2

  7  (2 )7   7   14

1  3 2 1  3 2  2

Q 

  6    12   2   3      3    2 3

4

 13    14   2     2  1 Answer

(v)

1 

Q 1    2  0

3   1  3  5

5

 1  3 5  1  3 5  2    2       2 2

1  3 2 1  3 2  2

Q 

 2  5  2   2 

5

 32  5  32  10  32  3   2  32  9  1  32(1)   2  32(1)  

 32(  2 )  32(1)  32

Q  9   3   (1)3  1 3

Q 1    2  0

{ Question # 3 (i) R.H.S= x  y x  y  x   2y 

= x  y  x x   2y   y x   2y    2 2 3 2 = x  y  x   xy  xy   y  = x  y  x 2   2    xy  (1)y 2    2 2 = x  y  x  1 xy  y  = x  y  x 2  xy  y 2  =x 3  y 3  L.H.S

(ii)

Q 3  1 Q 1    2  0     2  1

R.H.S= x  y  z  x  y   2z x   2y  z 

= x  y  z [x 2   2xy  xz  xy   3y 2   2yz  2xz   4yz   3z 2 ] = x  y  z [x 2  ( 2   )xy  (   2 )xz  ( 2   4 )yz  3y 2   3z 2 ] = x  y  z [x 2  (1)xy  (1)xz  ( 2  )yz (1)y 2  (1)z 2 ] Q  4   &    2  1 = x  y  z [x 2  y 2  z 2  xy  (1)yz  xz ] = x  y  z [x 2  y 2  z 2  xy  yz  xz ]  x 3  y 3  z 3  3xyz  L.H.S

(iii)

L.H.S= (1  )(1   2 )(1   4 )(1   8 )............2n factors  [(1   )(1   2 )][(1   4 )(1   8 )]............ n factors  [(1   )(1   2 )][(1   3   )(1   6   2 )]............ n factors  [(1   )(1   2 )][(1   3  )(1  ( 3 )2   2 )]............ n factors

FSC-I - 4.4 - 4

 [(1   )(1   2 )][(1  1  )(1  (1)2   2 )]............ n factors  [(1  )(1   2 )][(1   )(1   2 )]............ n factors

n n  (1  )(1   2 )  1     2   3    0  1n Q 1    2  0 , 3  1

 1n

 1  R.H.S

{ Question # 4 (i)

x 2  x  1  0 ...........(i )

Let

Since  is root of (i ) therefore

 2    1  0...........(ii )

To prove  2 is root of (i )

 2 

2

Consider



  2  1   4  2 2  1   2

  2  1   2   2  1    2  1    2

 2 

2

 0   2  1   

from (i )

  2  1  0 ...........(iii )

  2 is the root of the equation (i ) .

Now subtracting (ii ) from (iii )

 2 

2



4

 2 

1  0

2    1  0 





0

   3  1  0  3  1  0

as

0

 3  1

{ Question # 5

Let x be a cube root of 1 then 1

x  (1)3

 x 3  1

 x3 1  0

 (x )3  (1)3  0

 (x  1)(x 2  x  1)  0

 x 1  0  x  1

or

x2  x 1  0

or

(1)  (1)2  4(1)(1) x 2(1)

1 1 4 1  3  2 2 1  3 1  3  x or x  2 2 1  3i 1  3i  x or x  2 2 1  3i 1  3i Hence complex cube root of 1 are and . 2 2 

FSC-I - 4.4 - 5

{ Question # 6

Since 2 and 2 2 are roots of required equation, therefore x  2 x  2 2   0  x 2  2x  2 2x  4  3  0  x 2  2x (   2 )  4(1)  0

Q 3  1

 x 2  2x (1)  4  0

Q 1    2  0

 x 2  2x  4  0 is the required equation.

{ Question # 7 (i) Let x be a fourth root of 16 then x

1 (16)4

 x 4  16

 x 4  16  0 



x 2 

2

x 2  4x 2  4  0

 (4)2  0

 x2  4  0

or

x2  4  0

 x 2  4

or

x2  4

 x   4 or

 x  2i

or

x  4 x  2

Hence the four fourth root of 16 are 2,  2,2i ,  2i .

(ii)

Hint: 81  (9)2

Do yourself as above.

(iii) Let x be a fourth root of 625 then x

1 (625)4

 x 4  625

 x 4  625  0 



x 2  25x 2  25  0

x 2 

2

 (25)2  0

 x 2  25  0

or

x 2  25  0

 x 2  25

or

x 2  25

 x   25 or

 x  25i

or

x   25 x  25

Hence the four fourth root of 625 are 25,  25,25i ,  25i .

Question # 8 (i) 2x 4  32  0

 2(x 4  16)  0

 x 4  16  0

Now do you as in Question # 7 (i)

(ii)

3y 5  243y  0

 3y y 4  81  0

 3y  0

or

y 4  81  0

FSC-I - 4.4 - 6

y 2   (9)2  0 y 2  9y 2  9  0

 y0 

2

or

 y2  9  0

or

y2  9  0

 y 2  9

or

y2  9

 y   9

or

y 9

 y  3i



y  3

or



Hence S.Set  0, 3,  3i

(iii)

x3  x2  x  1  0  x 2 x  1  1(x  1)  0  (x  1)(x 2  1)  0

 x 1  0

or

x2  1  0

 x  1

or

x 2  1

Hence S.Set  1, i 

(iv)

 x  i

5x 5  5x  0  5x (x 4  1)  0

 5x  0

or

 x 0

or 



Hence S.Set  0, 1, i



x4 1  0

x 2   (1)2  0 x 2  1x 2  1  0 2

 x2  1  0

or

x2 1  0

 x 2  1

or

x2  1

 x  i

or

Error Analyst Waiting for Someone

‫ﮐﺘﻢ ﺷﺪ‬

x  1