17 Chapter 1
Exercise 1.3
x 2+ 2x + 9 = 9
Thus x = - 1 is a real root
x 2+ 2x = 9 -
9
If x = - 12 ,then x ( x + 2 )
=
0 Either x = 0 or x 2 0
x 2
2 ⎛ │ ⎝ - 2 1 ⎞ │ ⎠ 2 + 3 ⎛ │ ⎝ - 2 1 ⎞ │ ⎠ + =
+ 5 + 2 ⎛ │ ⎝ - 2 1 ⎞ │ ⎠ 2 + 3 ⎛ │ ⎝ - 2
1 ⎞ │ ⎠ + 1 = 2 = - Now it is necessary to verify value of x in given
2 1 - 3 2 + 5 + 2 1 - 3 2
+ 1 = 2 radical equation. x 2 + 2x + 4 + x 2 + 2x + 9 = 5 1 - If x = 0 , ( 0 ) 2 + 2 ( 0 ) + 4 + ( 0 ) 2 + 2 ( 0 ) + 9 =
5
2 3 + 5 + 1 - 2
3 + 1 = 2 4 + 9 =
5
- 2 2 + 5 + - 2
2 + 1 = 2 2 + 3 =
5
- 1 + 5 + - 1 + 1 =
2
5 =5 True
4 + 0 =
2 Thus x = 0 is a real root If x = - 2 ,then
( - 2 ) 2 + 2 ( - 2 ) + 4 + ( - 2 ) 2 + 2 ( - 2 ) + 9 =
5
4 - 4 + 4 + 4 - 4 + 9 =
5
2 =
2 True Thus x = - 2 1is a real root
Solution set = { - 1, - 2 1} 4 + 9 =
5
Q2 Find x if 2 x+ 5 satisfies
5 =
5 True
40 - 9 x - 2 7 - x = - x Thus x = - 2 is Solution a real root
set = { 0, - 2 } Solution: we have 40 - 9 x - 2 7 - x = -
x Given that 2 x+ 5 = 0 satisfies given equation
Q1x). 2x 2 + 3x + 5 + 2x 2 + 3x + 1 =
2 x = - 2 5putting 40 - 9 x - 2 7 - x = - x Solution: 2x 2 + 3x + 5 + 2x 2 + 3x + 1 = 2 OR 2x 2 + 3x + 5 = 2 - 2x 2 + 3x +
1 40 9 2 5 2 7 2 5 2 5 Taking ( square ) on ( both sides ) 40 45 2 2 7 5 2 ( )
- ⎛ │ ⎝ 5 2 80 2 - ⎞ │ ⎠ - - ⎛ │ ⎝ 45 2 14 2 - ⎞ │ ⎠ = - ⎛ │ ⎝ 5 5 2
- ⎞ │ ⎠
2x 2 + 3x + 5 2 = 2 - 2x 2
+ 3x +
1 2 2x 2 + 3x + 5 = 4 + 2x 2 + 3x + 1 2 - 2.2. 2x 2
+ 3x +
1
2x 2 + 3x + 5 = 4 + 2x 2 + 3x + 1 - 4 2x 2
+ 3x +
1
+ - + =
+ - + =
125 2 - 2 19 2 = 5 2 which is false Thus 2 x+ 5 do not satisfies the given equation 0 = - 2 2x 2
+ 3x + 1 ÷ by -
2
Or 40 - 9 x - - x = 2 7 - x 2x 2
+ 3x + 1 =
0 Taking square on both sides Again taking square both sides ( 2x 2 + 3x + 1 )
2 =
0
2
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( )
40 - 9 x - - x 2 = 2 7
-
x
2
40 - 9 x 2 + - x 2
- 2 40 - 9 x . - x = 4 7
-
x 2x 2+ 3x + 1 =
0 2x 2+ 2x + x + 1 =
0
40 - 9 x + - x - 2 - 40 x + 9 x 2
= 28 -
4
x
2x x + 1 + 1 x + 1 =
0
40 - 28 - 9 x - x + 4 x = 2 9 x 2
-
40
x
2x + 1 x + 1 =
0
12 - 6 x = 2 9 x 2 - 40 x divided by 2 Either or
x 1 0 x 1
2x 1 0 2x 1x 2
1 6 - 3 x = 9 x 2 - 40 x taking square + = = -
( ) ( )
( )( )
Now it is necessary to verify value of x in given radical equation. 2x 2 + 3x + 5 + 2x 2 + 3x + 1 = 2 If x = - 1 ,then
2 ( 1 ) 2 3 ( 1 ) 5 2 ( 1 ) 2 3 ( 1 )
1 2
2 3 5 2 3 1 2
4 0 2 2 0 2 True
+ = = -6 - 3 x 2 = 9 x 2
-
40
x 2 = - 36 + 9 x 2 - 2 6 3 x = 9 x 2
-
40
x 36 - 36 x = -
40
x 40 x - 36 x
= -
36 4 x= -
36
- + - + + - + - + =
x
= - 9Put in given equation - + + - + =
40 - 9 ( - 9 ) - - ( - 9 ) = 2 7 - ( - 9 ) + + =
=
40 + 81 - 9 = 2 7 +
9
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