Algebra and Trigonometry 5th Edition Blitzer Solutions Manual

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Algebra and Trigonometry 5th Edition Blitzer Solutions Manual F u l l D o w n l ohat dt p: s : / / a l i b a b a d o w n l o a d . c o m / p r o d u c t / a l g e b r a - a n d - t r i g o n o m e t r y - 5 t h - e d i t i o n - b l i t z e Chapter 1 Equations and Inequalities

Section 1.6

x+5 − x−3 = 2

4.

x +5 = 2+ x−3

Check Point Exercises

(

4 x 4 = 12 x 2

1.

4 x 2 ( x 2 − 3) = 0

or

x=± 0

x=± 3

x=0

x=± 3

{

}

(1)

x2 − 4 = 0 5.

x2 = 4

a.

)

= ( x − 3)

(

=

x −3

)

2

5 x 3/ 2 − 25 = 0 x 3/ 2 = 5

(x )

3/ 2 2 3

= ( 5)

23

x = 52/3 or

3

25

Check:

x +3 = x−3 x+3

x−3

5 x 3/ 2 = 25

x+3 +3 = x

(

2

x = ±2

3   The solution set is  −2, − , 2  . 2  

2

) (

x−3 +

1= x −3 4=x The check indicates that 4 is a solution. The solution set is {4} .

(2 x + 3)( x 2 − 4) = 0

3.

2

4 4 x −3 = 4 4 1= x−3

x 2 (2 x + 3) − 4(2 x + 3) = 10

2 x= − 3 3 x=− 2

2

) + 2 ( 2) ( x −3

4 = 4 x−3

2 x 3 + 3x 2 = 8 x + 12

2 x + 3 = 0 or

2

x+5 = 4+ 4 x−3 + x −3

x2 − 3 = 0 x2 = 3

The solution set is − 3, 0, 3 . 2.

) = (2 +

x + 5 = ( 2)

4 x 4 − 12 x 2 = 0

4 x2 = 0 x2 = 0

x+5

5 ( 52/3 )

2

3/ 2

− 25 = 0

5 ( 5 ) − 25 = 0

x + 3 = x2 − 6x + 9

25 − 25 = 0

0 = x − 7x + 6 0 = ( x − 6)( x − 1) x − 6 = 0 or x − 1 = 0

0=0

2

The solution set is {52/3 } or

x=6 x =1 1 does not check and must be rejected. The solution set is {6} .

{ 25} . 3

2 x 3 − 8 = −4 x 2/3 = 4

b.

( x2/3 )

3/2

= 43/2

or

( )

x = 22

3/2

x = 23

x = (−2)3

x = −8 x=8 The solution set is {–8, 8}.

144

Copyright © 2014 Pearson Education, Inc.

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)

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