Mathematics 10 (Science) - KPK - Gain - Solution - PART - 7 - 15 Chapter 1

15 Chapter 1 

Exercise 1.3 

Q1v). Solve x + 7 + x + 2 = 6x + 

13 ( x + 3x + 1 ) 2 = ( 5x + 

1 

) 

2 

Sol: Since x + 7 + x + 2 = 6x + 13 Taking square on both sides ( ) ( ) 

( x ) 2 + ( 3x + 1 ) 2 

+ 2 ( x )( 3x + 1 ) 

= 5x + 

1 

x + 7 + x + 2 2 = 6x + 

13 

2 

x + 3x + 1 + 2 x ( 3x + 1 ) = 5x + 

1 

( x + 7 ) 2 + ( x + 2 ) 2 

+ 2 ( x + 7 )( x + 2 ) 

= 6x + 

13 

x + 7 + x + 2 + 2 x ( x + 2 ) + 7 ( x + 2 ) = 6x + 13 

4x + 1 + 2 3x + x = 5x + 

1 

2 3x + x = 5x - 4x + 1 - 

1 

2 3x 2 + x = x 

+ + + + + = + 

+ + = - + - 

2 

2 

2x 9 2 x 2 

2x 7x 14 6x 13 

2 x 29x 14 6x 2x 13 9 

Taking ( 2 3x 2 

square + x ) 2 on both = 

( x 

) 2 sides 2 x 2 

+ 9x + 14 = 4x + 4 ÷ 

by2 

4 ( 3x 2 + x ) 

= 

x 

2 

x 2 

+ 9x + 14 = 2x + 

2 

12x 2 - x 2 

+ 4x = 

0 Taking square on both sides ( x 2 + 9x + 14 ) 2 = ( 2x + 

2 

) 2 11x 2 

+ 4x = 

0 x ( 11x + 4 ) 

=0 

x 2 + 9x + 14 = 4x 2 

+ 2.2x.2 + 

4 

4x 2 - x 2 

+ 8x - 9x + 4 - 14 = 

0 3x 2- x - 10 = 

0 

Either Or x = 0 11x + 4 = 

0 11x = - 43x 2 - 6x + 5x - 10 = 

0

x = 

- 11 4 3x ( x - 2 ) + 5 ( x - 2 ) 

= 

0 

Now it is necessary to verify the value of x in the ( 3x + 5 )( x - 2 ) 

= 

0 

given radical equation. x + 3x + 1 = 5x + 1 Either Or 3x 5 0 

If x = - 11 4 , then - 11 4 + 3 ⎛ │ ⎝ - 11 4 ⎞ │ ⎠ + = 

x - 2 = 

0 

+ 1 = 5 ⎛ │ ⎝ - 11 

4 ⎞ │ ⎠ + 1 3x = -5x = 

2 

x = 

- 3 

5 - 11 4 + - 12 11 + 11 = - 20 + 11 

11 Now it is necessary to verify value of x in given radical equation. x + 7 + x + 2 = 6x + 

13 - 11 4 + - 11 1 = - 9 11 False If x = - 3 5 ,then - 3 5 + 7 + - 3 5 + 2 = 6 ⎛ │ ⎝ - 3 

5 ⎞ │ ⎠ 

+ 13 Thus x = - 11 4 is an extraneous root 

If x = 0, then - 5 3 + 21 + - 5 3 

+ 6 = 2 ( - 5 ) + 

13 0 + 3 ( 0 ) + 1 = 5 ( 0 ) + 

1 

16 3 + 1 3 

= - 10 + 

13 0 + 0 + 1 = 0 + 

1 

0 + 1 = 

1 4 3 + 1 3 

= 3 5 3 

=3 false 1 = 

1 True Thus x = 0 is an Solution real root 

Set = { }0 Thus x = - 35 is an extraneous root, If x = 2, then 

Q1vii). Solve 6x + 40 - x + 21 = x + 5 Sol: Since 6x + 40 - x + 21 = x + 5 2 + 7 + 2 + 2 = 6 ( 2 ) + 

13 

9 + 4 = 12 + 

13 

3 + 2 = 

25 

Or 6x + 40 = x + 5 + x + 21 Taking square root on both sides ( ) ( ) 

5 = 

5 True Thus x = 2 is a Solution real root 

Set = 

{ }2 6x + 40 2 = x + 5 + x + 

21 

2 

( 6x + 40 ) 2 = ( x + 5 ) 2 + ( x + 21 ) 2 

+ 2 ( x + 5 )( x + 

21 

) 6x + 40 = x + 5 + x + 21 + 2 x ( x + 21 ) + 5 ( x + 

21 

) 

Q1vi). x + 3x + 1 = 5x + 

1 6x + 40 = 2x + 26 + 2 x 2+ 21x + 5x + 

105 Sol: Since x + 3x + 1 = 5x + 

1 6x - 2x + 40 - 26 = 2 x 2 

+ 26x + 

105 Taking square on both sides