14 Chapter 1
Exercise 1.3
Sol: Since 4x + 5 = 2x - 5 Q1i). Solve 5x + 21 = x + 3 Sol: Since 5x + 21 = x + 3 Taking
square on both sides we get
Taking square on both sides we get ( 4x + 5 ) 2 = ( 2x -
5 ) 2 4x + 5 = ( 2x ) 2 - 2 ( 2x )
.5 +
5
2
( 5x + 21 ) 2 = ( x +
3
)
2
4x 2- 20x - 4x + 25 - 5 =
0
5x + 21 = x 2 + 2.x.3 +
3
2
4x 2
- 24x + 20 = 0 ÷
by4
x 2+ 6x - 5x + 9 - 21 =
0
x 2- 6x + 5 =
0 x 2+ x - 12 =
0
x 2- 5x - 1x + 5 =
0 x 2+ 4x - 3x - 12 =
0 x ( x + 4 ) - 3 ( x + 4 )
=
0 ( x - 3 )( x + 4 )
=
0 Either or x 3 0 x 3
x ( x - 5 ) - 1 ( x - 5 )
=
0 ( x - 1 )( x - 5 )
=0
Either or
- = =
+ = = - Now it is necessary to verify the value of x in the given radical equation. 5x + 21 = x + 3 If x = 3, then If x = - 4, then
5.3 21 3 3
x - 1 =
0
x - 5 =
0 x =
1
x =
5 Now it is necessary to verify the value of x in the given radical equation. 4x + 5 = 2x - 5 If x = 1, then If x = 5, then
2.1 1 1 2
15 21 6
2 1 1
36 6 6 6 True
x 4 0 x 4
+ = +
5 ( - 4 ) + 21 = - 4 +
3 + =
- 20 + 21 = -
1 =
1 = -
1 =
1 = -
1 False Thus x = 3 is a thus x = - 4 is Real root Solution an Set extraneous =
{ }3 root
- = -
2.5 - 1 = 5 -
2
- = -
10 - 1 =
3
1 = -
1
9 =
3 1 = -
1 False
3 =
3 True Thus x = 1 is a Thus x = 1 is an Real root Solution Set =
extraneous { }5 root
Q1ii). Solve 2x - 1 = x - 2 Sol: Since 2x - 1 = x - 2 Taking square on both sides we get ( ) ( )
Q1iv). Solve 29 - 4x = 2x + 3 Sol: Since 29 - 4x = 2x + 3 Taking square on both sides we get
2x - 1 2 = x -
2
2
( 29 - 4x ) 2 = ( 2x +
3
)
2
2x - 1 = x 2 - 2.x.2 +
2
2
29 - 4x = 4x 2 + 2.2x.3 +
3
2
x 2- 4x - 2x + 4 + 1 =
0
4x 2+ 12x + 4x + 9 - 29 =
0
x 2- 6x + 5 =
0
4x 2
+ 16x - 20 = 0 ÷ by4
+ x 2- 5x - 1x + 5 =
0 x ( x - 5 ) - 1 ( x - 5 )
=
0 ( x - 1 )( x - 5 )
=
0
x 2+ 4x - 5 =
0 x 2+ 5x - 1x - 5 =
0 x ( x + 5 ) - 1 ( x + 5 )
=
0 Either or x - 1 = 0
x - 5 =
0 x = 1
x =
5 Now it is necessary to verify the value of x in the
( x - 1 )( x + 5 )
=
0 Either or x - 1 = 0
x + 5 = 0
given radical equation. 2x - 1 = x - 2 If x = 1, then If x = 5, then
2.1 - 1 = 1 -
2
x =
1
x = -
5 Now it is necessary to verify the value of x in the given radical equation. 29 - 4x = 2x + 3 - = -
If x = 1, then If x = - 5, then 2 - 1 = -
1
- =
29 4.1 2.1 3 1 = -
1
=
29 4 2 3 1 = -
1 False
= Thus x = 1 is an thus s x = 5 is a extraneous root so Solution real Set root
=
{ }5 25 5
5 5 True
Q1iii). 4x + 5 = 2x - 5
2.5 1 5 2
10 1 3
9 3 3 3 True
- = +
- = +
= =
29 - 4 ( - 5 ) = 2 ( - 5 ) +
3
29 + 20 = - 10 +
3
49 = -
7 7 = -
7 False Thus x = 1 is Thus x = - 5 is an Real root Solution extraneous Set =
{ }1 root
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