12 Chapter 1
Exercise 1.2
x 2+ 2x - 8 = 0 x 2+ 2x - 70 = 0
( y + 7 )( y +
15 ) + 16 = 0 y ( y + 15 ) + 7 ( y + 15 )
+ 16 =
0
y 2+ 15y + 7y + 105 + 16 = 0
y 2+ 22y + 121 = 0
( ) ( ) ( )( )
x =
- ( 2 ) ± ( 2 ) 2 - 4 ( 1 )( 70 ) 2 ( 1
)
x 2+ 4x - 2x - 8 =
0
x =
- 2 ± 4 + 280 2 x x + 4 - 2 x + 4 =
0
x - 2 x + 4 =
0
x =
- 2 ± 2
284 y 2 + 2.y.11 + 11 2 = 0
( + ) = either x - 2 = 0 y 11 2 0
Or x + 4 =
0
x =
- 2 ± 2
4 × 71 ⇒ y + 11 =
0 Putting back value of y = x 2 + 8x , then x = 2 x = -
4
x = - 2 ± 2 2 71 = - 2 2 ±
2 2 71 Solution set = { 2, - 4, - 1 x ±
= - 71 1 ±
} 71 Q1xiv). ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) + 1 =
0 x 2+ 8x + 11 = 0 Comparing with
ax 2 + bx + c = 0 soa = 1, b = 8, c = 11 Using
x = - b ± b 2 - 4ac 2a
putting
Solution: ( x + 1 )( x + Rearranging 4 )( x + 2 )( x with + 3 )
+ respect 1 =
0 to a+b =c+d { x ( x + 4 ) + 1 ( x + 4 ) } { x ( x + 3 ) + 2 ( x + 3 ) }
+ 1 =
0
x =
- ( 8 ) ± ( 8 ) 2 - 4 ( 1 )( 11 ) 2 ( 1 )x = - 8 ± 64 - 44 2 ( x 2 + 4x + 1x + 4 )( x 2
+ 3x + 2x + 6 )
+ 1 =
0 ( x 2 + 5x + 4 )( x 2
+ 5x + 6 )
+ 1 =
0
x = - 8 ± 2 20 x = - 8 ± 2 2 5 =
- 8 ± =
( 4 × 5 2 2 - 4 ± 5 ) Let y = x 2 + 5x then we have
2 ( y + 4 )( y + 6 )
+ 1 =
0 y ( y + 6 ) + 4 ( y + 6 )
+ 1 =
0
x = - 4 ±
5 Solution Set = { - 4 - 5, - 4 + 5 } y 2+ 6y + 4y + 24 + 1 =
0
Q2 Solve x 4 - 2 x 3 - 2 x 2 + 2 x + 1 = 0 y 2+ 10Y + 25 =
0 y 2 + 2.Y.5 + 5 2
=
0 ( Y + 5 )
2
=
0
Solution we have x 4 - 2 x 3 - 2 x 2 + 2 x + 1 = 0 Dividing each term by x
2x 2
- 2 x - 2 + 2 x + x 1 2 = 0 ⇒ Y + 5 =
0 Putting back value of y = x 2 + 5x then we have
x 2
+ x 1 2- 2 x + 2 x - 2 = 0 rearranging
x 2
+ 5x + 5 =
0 x =- ( 5 ) ± ( 5 ) 2
- 41 ( )( 5 ) 2 ( 1
)
⎛ │ ⎝ x 2
+ x 1 2⎞ │ ⎠ - 2 ⎛ │ ⎝ x - 1 x
⎞ │ ⎠ - 2 = 0 .......(1)
Let
x =
- 5 ± 25 - 20 2
x =
- 5 ± 2 y = x - x 1 y 2 = ⎛ │ ⎝ x - x 1 ⎞ │ ⎠
2
= + -
5 = + -
+ = + Solution set = ⎧ │ ⎨ │
- - - + ⎫ │ ⎬ │ ⎭
y 2 x 2 1 x
2.x. X 1y 2 x 2 1x
2
y 2 2 x
2 1x Putting into equation (1) we get
( )
22
2 5 2 5 , 5 2 5 Q1xv). ( x + 1 )( x + 3 )( x + 5 )( x + 7 ) + 16 = 0 Sol: Since ( x + 1 )( x + 3 )( x + 5 )( x + 7 ) + 16 = 0 ( )( )( )( ) { ( ) ( ) } { ( ) ( ) } ( )( ) ( )( )
y 2+ 2 - 2 y
- 2 =
0
y 2
- 2 y
=
0 y y- 2 =
0 Either y = 0 or y- 2 = 0 x + 1 x + 7 x + 3 x + 5 + 16 =
0
Putting back value of y x x + 7 + 1 x + 7 x x + 5 + 3 x + 5 + 16 =
0
x 2 + 7x + 1x + 7 x 2
+ 5x + 3x + 15 + 16 =
0
x - 1 x = 0 x - 1 x - 2 = 0 Multiplying each term by x x 2 + 8x + 7 x 2
+ 8x + 15 + 16 =
0
x 2 - 1 = 0 x 2 - 1 - 2 x = 0 Let y = x 2 + 8x Then we have
2 11 x=x 2 - 2 x
- 1 =
0 x
= ±
a = 1, b = - 2, c
= -
1
Chapter 1
= - - ± - - = ± + = ± = ± Solution set = { ± 1,1 ± 2 } Radical equations: An equation in which the variable occurs in one or more radical sign. e.g. x + a = x + b Radical equations Type 1: ax + b = cx + d Step1: Squaring both sides Example 13 solve 27 - 3 x = x - 3 Solution: we have 27 - 3 x = x - 3 Taking square on both sides ( 27 - 3 x ) 2 = ( x
-
3
)
2
27 - 3 x = x 2
- 6 x
+
9
0 = x 2
- 6 x + 3 x
+ 9 -
27 x 2- 3 x
- 18 =
0
x 2
- 6 x + 3 x
- 18 =
0 x ( x - 6 ) + 3 ( x
- 6 )
=
0 ( x + 3 )( x
- 6 )
=
0 Either x+ 3 =
0 x= - 3
or x- 6 =
0 x=
6 To verify the roots for given equation put x = - 3 put x =
6 27 - 3 ( - 3 ) = ( - 3 ) -
3
27 - 3 ( 6 ) = ( 6 ) -
3
27 + 9 = - 3 -
3
27 - 18 = 6 -
3
36 = -
6
9 = 36 = - 6 False 3 3= true Therefore x = - 3 is a extraneous root And x = 6 is a real root Solution set = { }6 Type 2: x + a + x + b = x + c Step 1, squaring both sides Step 2, separate the radical term Step 3, Again squaring both sides Example 14 Solve x + 2 + x + 7 = x + 23 Sol: Since x + 2 + x + 7 = x + 23 squaring ( x + 2 + x + 7 ) 2 = ( x
+
23
)
2
x + 2 + x + 7 + 2 x + 2 x + 7 = x
+
23
2 x + 9 + 2 x 2
+ 7 x + 2 x + 14 = x
+
23
2 x 2+ 9 x + 14 = x - 2 x
+ 23 -
9
2 x 2
+ 9 x + 14 = - x
+
14 2 2 ( x 2
+ 9 x + 14 ) 2 = ( 14
-
x
) 2 4 ( x 2 + 9 x + 14 ) = 14 2 - 2 ( 14
)( x )
+
x
2
xxx
13 Exercise 1.2
( 2 ) 4 4 ( 1 )( 1
)
4 x 2 - x 2
+ 36 x + 28 x
+ 56 - 196 =
0 2 ( 1
)
3 x 2+ 64 x
- 140 =
0 2 4 4 2 2 2
3 x 2
+ 70 x - 6 x
- 140 =
0 2 2 1 2
x ( 3 x + 70 ) - 2 ( 3 x
+ 70 )
=
0 ( x - 2 )( 3 x
+ 70 )
=
0 Either x- 2 = 0 or 3 x+ 70 =
0 x = 2 x = - 3 70To verify put x = 2 in given equation
2 + 2 + 2 + 7 = 2 +
23
4 + 9 =
25 2 + 3 =
5 5 5= true so x = 2 is a real root Now to verify x = - 3 70in given equation
- 3 70 + 2 + - 3 70 + 7 = - 3
70 + 23 - 70 3 + 6 + - 70 3 + 21 = - 70 + 69 3 - 3 64 + - 3 49 = 1 3
false so x = - 3 70is extraneous root. Therefore Solution set = { }2 Type 3: ax 2 + bx + c + ax 2 + bx + d = e Step 1: Rearranging
ax 2 + bx + c = e - ax 2 + bx + d Step 2, separate the radical term Step 3, Again squaring both sides
Exp 15 Solve x 2 + 3 x + 5 + x 2 + 3 x + 1 = 2 Sol: we have x 2 + 3 x + 5 + x 2 + 3 x + 1 = 2 x
2 + 3 x + 5 = 2 - x 2 + 3 x + 1 Taking ( x 2 + square 3 x + 5 ) 2 on = ( both 2 - sides x 2 + 3 x +
1 ) 2 x 2 + 3 x + 5 = 4 + x 2 + 3 x + 1 - 4 x 2
+ 3 x
+
1
4 x 2
+ 3 x
+ 1 =
0
x 2
+ 3 x
+ 1 =
0 x 2 + 3 x + 1 = 0 Again squaring
x=
- 3 ± 2
9 - 4
x
=
- 3 ± 2
5
Solution set = ⎧ │ ⎨ │ ⎩ - 3 ± 2
5
⎫ │ ⎬ │ ⎭
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