Mathematics 10 (Science) - KPK - Gain - Solution - PART - 8 - 16 Chapter 1

16 Chapter 1 

Exercise 1.3 

4x + 14 = 2 x 2 

+ 26x + 105 ÷ 

by2 

2x + 7 = x 2 

+ 26x + 

105 Taking ( x 2 

+ 26x square + 105 on ) both 2 sides = ( 2x + 7 

) 2 x 2 + 26x + 105 = 4x 2 

+ 2.2x.7 + 

49 

4x 2 - x 2 

+ 28x - 26x + 49 - 105 = 

0 3x 2+ 2x - 56 = 

0 3x 2+ 14x - 12x - 56 = 

0 x ( 3x + 14 ) - 4 ( 3x + 14 ) 

= 

0 ( 3x + 14 )( x - 4 ) 

= 

0 

3x + 14 = 

0 Either 

3x = -14 

x =

- 3 14 =

Now it is necessary to verify value of x in given radical equation. 6x + 40 - x + 21 = x + 5 If x = - 3 14 then 

⎛ │ ⎝ - ⎞ │ ⎠ 

Or x 4 0 

x 4 

6 14 3 + 40 - - 3 14 + 21 = - 3 14 + 5 - 84 3 + 120 - - 14 3 + 63 = - 14 + 15 

3 

36 3 - 49 3 = 

1 3 6 - 7 = 

1 3 3 3 - 1 3 = 

1 3 False Thus x = - 3 14 is an extraneous root so If x = 4, then 

6 ( 4 ) + 40 - 4 + 21 = 4 + 

5 

24 + 40 - 25 = 

9 

64 - 5 = 

3 8 - 5 = 

3 3 = 

3 True Thus x = 4 is a real root. 

Solution Set = { }4 Q1viii). 2x - 3 + 2x + 4 = 6x + 13 Sol: Since 2x - 3 + 2x + 4 = 6x + 13 Taking square root on both sides ( 2x - 3 + 2x + 4 ) 2 = ( 6x + 

13 

) 

2 

( 2x - 3 ) 2 + ( 2x + 4 ) 2 

+ 2 ( 2x - 3 )( 2x + 4 ) 

= 6x + 

13 2x - 3 + 2x + 4 + 2 2x ( 2x + 4 ) - 3 ( 2x + 4 ) 

= 6x + 

13 

4x + 1 + 2 4x 2 

+ 8x - 6x - 12 = 6x + 

13 

2 4x 2 

+ 2x - 12 = 6x - 4x + 13 - 

1 

2 4x 2 

+ 2x - 12 = 2x + 12 ÷ 

by2 

4x 2 

+ 2x - 12 = x + 

6 

Taking square on both sides ( ) ( ) 

( ) ( ) ( )( ) 

4x 2 

+ 2x - 12 2 = x + 

6 2 4x 2 + 2x - 12 = x 2 + 2.x.6 + 

36 

4x 2 - x 2 

+ 2x - 12x - 12 - 36 = 

0 3x 2- 10x - 48 = 

0 3x 2- 18x + 8x - 48 = 

0 3x x - 6 + 8 x - 6 = 

0 

3x + 8 x - 6 = 

0 

Either Or 3x + 8 = 

0 3x = -8x = - 3 8 x - 6 = 

0 x = 

6 

Now it is necessary to verify value of x in given radical equation. 2x - 3 + 2x + 4 = 6x + 13 If x = - 3 8 , then 

2 ⎛ │ ⎝ - 3 8 ⎞ │ ⎠ - 3 + 2 ⎛ │ ⎝ - 3 8 ⎞ │ ⎠ + 4 = 6 ⎛ │ ⎝ - 3 

8 ⎞ │ ⎠ + 13 - 16 3 - 9 + - 16 3 + 12 = - + - + - = 

- + - + - = 

- 48 3 13 25 4 48 39 3 3 3 5 1 3 2 3 1 9 3 False Thus x = - 3 8 is an extraneous root If x = 6, then 

2 ( 6 ) - 3 + 2 ( 6 ) + 4 = 6 ( 6 ) + 13 

12 - 3 + 12 + 4 = 36 + 

13 

9 + 16 = 

49 3 + 4 = 

7 7 =7 True Thus x = 6 is a Solution real root Set so 

= { }6 Q1ix). x 2 + 2x + 4 + x 2 + 2x + 9 = 5 Sol: Since x 2 + 2x + 4 + x 2 + 2x + 9 = 5 Or x 2 + 2x

+ 4 = 5 - x 2 + 2x + 9 Taking ( x 2 + 2x square + 4 ) 2 on = both ( 5 - sides x 2 

+ 2x + 

9 

) 2 x 2 + 2x + 4 = 25 + ( x 2 + 2x + 9 ) 

2 - 2.5. x 2 

+ 2x + 

9 

x 2 + 2x + 4 = 25 + x 2 + 2x + 9 - 10 x 2 

+ 2x + 

9 

4 = 34 - 10 x 2 

+ 2x + 

9 

10 x 2 

+ 2x + 9 = 30 ÷ 

by10 

x 2+ 2x + 9 = 

3 Again ( x 2 + taking 2x + 9 square ) 2 both sides = 

3 

2