Mathematics 10 (Science) - KPK - Gain - Solution - PART - 3 - 8 Chapter 1

8 Chapter 1 

Exercise 1.1 

2 y 2 - 5 y - 4 y 

+ 10 = 

0 

1. Put a x= y y ( 2 y - 5 ) - 2 ( 2 y 

- 5 ) 

= 

0 

2. Find y & the find x by using log . ( y - 2 )( 2 y 

- 5 ) 

= 

0 Either y- 2 = 0 or 2 y- 5 = 

0 Example 10 solve 4.2 2 x - 10.2 x + 4 = 

0 Let y = 2x ⇒ y 2 = 2 2x given equation becomes 

Putting back value of y 

4 y 2- 10 y 

+ 4 = 

0 x + 1 x - 2 = 0 or 2 ⎛ │ ⎝ x + 1 x 

⎞ │ ⎠ - 5 = 0 Multiply each term by x x 2+ 1 - 2 x 

= 

0 

x 2 

- 2 x 

+ 1 = 

0 ( x ) 2 - 2 ( x 

)( 1 ) + ( 1 ) 

2 

= 

0 

4 y 2 

- 8 y - 2 y 

+ 4 = 

0 4 y ( y - 2 ) - 2 ( y 

- 2 ) 

= 

0 

2 x 2+ 2 - 5 x 

= 

0 2 x 2- 5 x 

+ 2 = 

0 

( 4 y - 2 )( y 

- 2 ) 

= 

0 

( x 

- 1 ) 

2 

= 

0 

2 x 2 

- 4 x - x 

+ 2 = 

0 2 x ( x - 2 ) - 1 ( x 

- 2 ) 

= 

0 

4 y- 2 = 

0 

y 

= 2 4 = 

1 2 

y- 2 = 

0 y 

= 

2 

Putting back value of y 

⇒ x- 1 = 

0 

( 2 x - 1 )( x 

- 2 ) 

= 

0 

x = 

1 2 x- 1 = 

0 

x- 2 = 

0 x 

= 

12 

x= 

2 

2 x = 2 - 1 2 x = 2 1 ⇒ x = - 1 ⇒ x = 1 Solution set = { - 1,1 } Example 11. Solve 2 2 + x + 2 2 - x =

10 Solution set = ⎧ ⎨ ⎩ 1 2,1, 2 ⎫ ⎬ ⎭ 

Solution: we have 2 2 + x + 2 2 - x = 

10 2 2 2 x + 2 2 2 - x 

= 

10 

Exp 9ii). 8 ⎛ │ ⎝ x 2 

+ 1 x 2⎞ │ ⎠ - 42 ⎛ │ ⎝ x - 1 x 

⎞ │ ⎠ 

+ 29 = 0 4.2 x 

+ 2 4 x 

= 10 Sol: we have 8 ⎛ │ ⎝ x 2 

+ x 1 2⎞ │ ⎠ - 42 ⎛ │ ⎝ x - 1 x Let y = x - 1 x

⎞ │ ⎠ + 29 = 0 ⇒ y 2 = x 2 

+ x 1 2- 

2 Let y = 2x given equation becomes 

4 y + 4 y = 10 multiply each term by y 4 y 2 + 4 = 10 y y 2 + 2 = x 2 

+ 

x 1 24 y 2 - 10 y + 4 = 0 divided by 2 Thus ( 2 

given ) equation becomes 2( ) ( ) 

2 

2 y 2- 5 y 

+ 2 = 

0 

8 y + 2 - 42 y 

+ 29 = 

0 

2 y 2 

- 4 y - 1 y 

+ 2 = 

0 

8 y + 16 - 42 y 

+ 29 = 

0 8 y - 42 y 

+ 45 = 

0 

2 y y - 2 - 1 y 

- 2 = 

0 ( 2 y - 1 )( y - 2 ) = 0 8 y 2 - 30 y - 12 y + 45 = 

0 ( ) ( ) 

Either 2 1 0 

12 ( )( ) 

y- = - 2 y 4 y - 15 - 3 4 y 

- 15 = 

0 

y 

= 

or y2 = 

0 y 

= 

2 

2 y - 3 4 y 

- 15 = 

0 

Putting back value of y 

Either 2 y- 3 = 0 or 4 y- 15 = 0 Putting back value of y 2 ⎛ │ ⎝ x - 1 x 

2 x = 2 - 1 2 x = 2 1 ⇒ x = - 1 ⇒ x = 1 ⎞ │ ⎠ - 3 = 0 or 4 ⎛ │ ⎝ x - 1 x 

⎞ │ ⎠ 

- 15 = 0 Solution set = { - 1,1 } Type 5: The Equation of the form, Multiply each term by x 

( x + a )( x + b )( x + c )( x + d ) = k 2 x 2- 2 - 3 x 

= 

0 

4 x 2- 4 - 15 x 

= 

0 

1.Multiply those factors such that a + b = c + d 2 x 2- 3 x 

- 2 = 

0 

4 x 2- 15 x 

- 4 = 

0 

2 x 2 

- 4 x + 1 x 

- 2 = 

0 

4 x 2 

- 16 x + 1 x 

- 4 = 

0 

2.Put the common terms = y 3.Find the value of y from the new equation 4Put the values of y in supposition to get roots. 2 x ( x - 2 ) + 1 ( x 

- 2 ) 

= 

0 

4 x x - 4 + 1 x 

- 4 = 

0 

Example 12:Solve ( x + 1 )( x + 3 )( x - 2 )( x - 4 ) = 24 ( 2 x + 1 )( x 

- 2 ) 

= 

0 

4 x + 1 x 

- 4 = 

0 

Sol: given ( x + 1 )( x + 3 )( x - 2 )( x - 4 ) = 24 2 1 0 

Here 1 - 2 = 3 - 4 1( )( )( )( ) 2 

( )( ) ( )( ) ( ) ( ) ( )( ) x+ = 

x 

= 

- x- 2 = 

0 x= 

2 + Solution set 4 x x= 4 1 - 1 4 , 1= 2 1 0 , 2, xx4 - = 

4 4 

= 

0 

x + 1 x - 2 x + 3 x 

- 4 = 

24 

= ⎧ ⎨ ⎩ - - ⎫ ⎬ ⎭ 

x 2 - 2 x + 1 x - 2 x 2 

- 4 x + 3 x 

- 12 = 

24 

x 2 - x - 2 x 2 

- x 

- 12 = 

24 

Type 4: The Equation of the form, a 2 x + a x + b = 

0 Let y = x 2 - x so 

9 Exercise 1.2 ( )( ) 

22 

Chapter 1 y - 2 y 

- 12 = 

24 

2 y = 1 3 y = 5 

y - 12 y - 2 y 

+ 24 = 

24 

y = 2 1 y = 5 3 y - 14 y 

= 

0 

Putting back value of y = x 2 y ( y - 14 ) = 0 y = 0 y- 14 = 0 Putting back value of y 

x 2 =21 ⇒ x =± 1 2 ( ) 

x 2 5 3 x 5 3 Solution Set = ⎧ │ ⎨ │ ⎩ ± ± ⎫ │ ⎬ │ ⎭ 

= 

x 2 - x 

= 

0 

⇒ = ± 

x x 

- 1 = 

0 x = 0 x- 1 = 

0 x 

= 

1 

x 2 

- x 

- = 

x= 

± - 2 - - x 

= ± + x = 1 ± 2 57 Solution set 0,1, 1 2 14 0 

1 ( 1 ) 4.1 ( 14 

) 2 ( 1 

) 

1 1 56 2 

57 , 1 2 

57 1 , 2 

5 3 Q1iv). Solve x + 2 - x 1 + 2 = 2 3 Sol: Since x + 2 - x 1 + 2 = 3 2 = ⎧ │ ⎨ │ ⎩ + - ⎫ │ ⎬ │ ⎭ 

Exercise 1.2 

Let y = x + 2 so y - 31y = 2 Multiply each term by 2y we get 

- = 

- = Q1i). Solve x 4 - 5x 2 + 4 = 0 Sol: Since x 4 - 5x 2 + 4 = 0 Suppose that y = x 2 then y 2 = x 4 so 

( ) ( ) ( )( ) 

2y.y 2y. y 12y. 

3 2 2y 2 

2 3y 

2y 2- 3y - 2 = 

0 R.W 2y 2 

- 4y + 1y - 2 = 

0 ( ) ( ) ( )( ) 

- 4 - × + 4 × 1 

y 2- 5y + 4 = 

0 R.W y 2- 4y - y + 4 = 0 

2y y - 2 + 1 y - 2 = 

0 

y - 2 2y + 1 = 

0 

y y - 4 - 1 y - 4 = 

0 

y - 1 y - 4 = 

0 

Sign of b + × a.c 

Result - 4 + × × 4 - 

Either y - 2 = 0 Or 2y + 1 =0 

1 

Either y - 1 = 0 Or y - 4 =0 

y = 1 y = 4 Putting back value of y = 

x 2 x 21 x 1 

y = 2 2y = - 1 

y = - 2 1 Putting back value of y = x + 2 we get 

x 2= 

4 

Solution Set = { ± 1, ± 

2 } ⇒ x = ± 

2 

x 2 2 x 2 2 x 0 

x 2 2 1 x 21 2 x 2 5 Solution Set = ⎧ ⎨ ⎩ 0, - 2 5 ⎫ ⎬ ⎭Q1v). Solve x - 4 x = 2 Sol: Since x - 4 x = 2 Multiply each term by x x 2- 4 = 2x x 2- 2x - 4 = 0 Comparing with ax 2 + bx + c = 0 so a = 1, b = - 2,c = - 4 Using 

+ = + =- = - = 

= - - 

= - Q1ii). Solve x 4 - 7x 2 + 12 = 0 Sol: Since x 4 - 7x 2 + 12 = 0 Suppose that y = x 2 then y 2 = x 4 so 

( ) ( ) ( )( ) 

= ⇒ = ± 

y 2- 7y + 12 = 

0 R.W y 2- 4y - 3y + 12 = 

0 y y - 4 - 3 y - 4 = 

0 

+ 12 - × - 6 × 1 5 × 2 y - 3 y - 4 = 

0 

4 × 3 Either y - 3 = 0 Or y - 4 =0 

y = 3 y = 4 Putting back value of y = 

x 2 x 2= 

3 

x 2= 

4 ⇒ x = ± 

3 

⇒ x = ± 

2 Solution Set = { ± 3, ± 

2 } x = - b ± b 2 - 4ac 2a 

putting 

( ) ( ) ( )( ) ( ) Q1iii). Solve 6x 4 - 13x 2 + 5 = 0 Sol: Since 6x 4 - 13x 2 + 5 = 0 Suppose that y = x 2 then y 2 = x 4 so 

( ) ( ) ( )( ) 

x = 

- - 2 ± - 2 2 - 4 1 - 4 2 1 

x = 2 ± 2 4 + 16 = 2 ± 2 20 x = 2 ± 2 4 5 x = 2 2 ± 

2 2 5 x = 1 ±Solution 6y 2- 13y + 5 = 

0 

R.W + 30 

6y 2 

- 10y - 3y + 5 = 

0 2y 3y - 5 - 1 3y - 5 = 

0

5 set = { 1 + 5,1 - 5 } - × - 12 × 1 11 × 2 2y - 1 3y - 5 = 

0 

10 × 3 Either 2 y - 1 = 0 Or 3 y - 5 =0 

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