math 100 half chapter 8 and chapter 9 test review selected ...flc.losrios.edu/~eitel/all pdfs/m-100...

Math 100 Half Chapter 8 and Chapter 9 Test Review Selected Solutions Perform the indicated operation and simplify the results.

1.

�

32x

+ 52x − 4

! 3.

�

2x + 5x2 − 2x

− 4x − 2

!

�

factor the denominators3

2x + 5

2(x − 2)

The LCD is 2x(x − 2)

multiply 32x

by (x − 2)(x − 2)

multiply 52(x − 2)

by xx

(x − 2)(x − 2)

•3

2x +xx

•5

2(x − 2)

3x − 62x(x − 2)

+5x

2x(x − 2)

combine the numerators over the LCD3x − 6 + 5x2x(x − 2)

simplify the numerator8x − 6

2x(x − 2)

factor the numerator2(4x − 3)2x(x − 2)

reduce(4x − 3)x(x − 2)

!

�

factor the denominators2x + 5x(x − 2)

− 4(x − 2)

The LCD is x(x − 2)

multiply 2x + 5x(x − 2)

by 11

multiply − 4(x − 2)

by xx

11

•2x + 5x(x − 2)

− xx

•4

(x − 2)

2x + 5x(x − 2)

− 4xx(x − 2)

combine the numerators over the LCD2x + 5 − 4xx(x − 2)

simplify the numerator−2x + 5

2x(x − 2)

Math 100! Half Chapter 8 and Chapter 9 Test Review WKD! © 2016Eitel

5.

�

3x2 + 6x + 5

+ 2x2 + 3x + 2

!

�


(x + 5)(x + 1) + 2

(x + 1)(x + 2)

The LCD is (x +5)(x +1)(x + 2)

multiply 3(x + 5)(x + 1)

by (x + 2)(x + 2)

multiply 2(x + 1)(x + 2)

by (x + 5)(x + 5)

(x + 2)(x + 2)

•3

(x + 5)(x + 1) +

(x + 5)(x + 5)

•2

(x + 1)(x + 2)

3x + 6(x +5)(x +1)(x + 2)

+2x + 10

(x +5)(x +1)(x + 2)

combine the numerators over the LCD3x + 6 + 2x + 10

(x +5)(x +1)(x + 2)

simplify the numerator5x + 16

(x +5)(x +1)(x + 2)

!


6.

�

2x2 + 3x −18

− 3x − 3

watch the negative sign

�


(x + 6)(x − 3) − 3

(x − 3)

The LCD is (x +6) (x − 3)

multiply 2(x + 6)(x − 3)

by 11

multiply 3(x − 3)

by (x + 6)(x + 6)

11

•2

(x + 6)(x − 3) − (x + 6)

(x + 6)•

3(x − 3)

2(x + 6)(x − 3)

− (3x + 18)(x + 6)(x − 3)

combine the numerators over the LCD2 − (3x + 18)(x + 6)(x − 3)

distribute the negitive2 − (3x + 18)(x + 6)(x − 3)

2 − 3x −18(x + 6)(x − 3)

simplify the numerator−3x −16

(x + 6)(x − 3)


7.

�

x + 26x − 3

− x −18x − 4

watch the negative

�

factor the denominatorsx + 2

3(2x −1) − x −1

4(2x −1)

The LCD is 12 (2x −1)

multiply (x + 2)3(2x −1)

by 44

multiply (x −1)4(2x −1)

by 33

44

•(x + 2)

3(2x −1) − 3

3•

(x −1)4(2x −1)

4x + 812 (2x −1))

− (3x − 3)13 (2x −1)

combine the numerators over the LCD4x + 8 − (3x − 3)

12 (2x −1)

distribute the negitive4x + 8 − 3x + 3

12 (2x −1)

simplify the numeratorx + 11

8 (2x −1)


Solve for x.

10.

�

x − 33x

+ 12

= 32x

! 11.

�

4 − 8xx + 1

= 8x + 1

! !

�

the LCD is 6xmultiply each termby 6x and cancel

6x1

•(x − 3)

3x + 6x

1•

12

= 32x

•6x1

/ 6 2 / x 1

•(x − 3)

/ 3 / x + / 6 3x

1•

1/ 2

= / 6 3 x 1

•3

/ 2 x

2(x − 3) + 3x •1 = 3• 3

2x − 6 + 3x = 9

5x − 6 = 9

5x = 15

x = 3

�

the LCD is (x +1)multiply each termby (x +1) and cancel

(x +1)1

•41

− (x +1)1

•8xx + 1

= (x +1)1

•8x + 1

(x +1)1

•41

− (x +1)1

•8xx + 1

= (x +1)1

•8x + 1

4(x + 1) −1• 8x = 8

4x + 4 − 8x = 8

−4x + 4 = 8

−4x = 4

x = −1

when you plug −1 into theoriginal equation you geta zero in the denominator

No Solution (NS)


Chapter 9 Simplify.

13.

�

6 48 ! ! 15.

�

−5 40x 6y 3

! !

�

factor 48

6 16 • 3

16 = 4the comes out of the radical sign as a 4

6 • 4 • 3

= 24 • 3

!

�

−5 40 = −5 4 • 5 = −10 5

x6 = x2 • x2 • x2 = x3

y3 = y2 • y = y y

−5 40x6y3 = −10x3y y

!

!16.

�

24 − 2 54 + 3 18 ! 19.

�

15x3y⎛ ⎝

⎞ ⎠ 20x2y⎛ ⎝

⎞ ⎠

�

24 = 4 • 6 = 2 6

−2 54 = −2 9• 6 = −6 9• 6

+3 18 = +3 2• 3• 3 = +9 2

24 − 2 54 + 3 18

= 2 6 − 6 6 + 9 2

combineliketerms

= −4 6 + 9 2

!

�

15• 20• x3 • x2 • y • y

3• 5• 4 • 5• x5 • y2

2• 5• x2 • y 3• x

10x2y 3x

!


21.

�

3 5 10 + 2 15( ) ! ! 22.

�

3 5 + 2( ) 5 −1( )

�

distrubite the 3 5to both terms

3 5 • 10 + 3 5 • 2 15

simplify

3 5 • 2 • 5 + 3• 2 5 • 5 • 3

15 2 + 30 3

!

�

F O I L

3 5 • 5 + 3 5 • −1( ) + 2 • 5 + 2(−1)

simplify

3 25 − 3 5 + 2 5 − 2

15− 3 5 + 2 5 − 2

13− 5

24.

�

4 − 5( ) 4 + 5( ) ! !

�

F O I L

4 • 4 + 4 • 5 − 5 • 4 − 5 • 5

simplify

16 + 4 5 − 4 5 − 25

16 + 4 5 − 4 5 − 5

11


26.

�

46! ! `! 28.

�

145 − 7

!

�

multiply by 66

46

•66

simplify

4 636

4 66

2 63

! ! !

�

multiply by5 + 7( )5 + 7( )

145 − 7

•5 + 7( )5 + 7( )

simplify

14 5 + 7( )25 − 5 7 + 5 7 − 49

14 5 + 7( )25 − 7

14 5 + 7( )18

7 5 + 7( )9

31.

�

a = 3 5

�

c = 7 Find b.

�

a2 + b2 = c2

3 5( )2 + b2 = 72

3 5( ) 3 5( ) + b2 = 72

9 25( ) + b2 = 49

45 + b2 = 49

b2 = 4b = 2


Solve for x.

32.

�

5x − 6 −1= 2! 33.

�

2x + 40 + 6 = −2 ! !

Get the radical alone on one side of the equation

5x − 6 = 3

Is the constant negitive?No, then proceed.

Square both sides of the equation

5x − 6 = 9

Solve for x

5x = 15x = 3

!

Get the radical alone on one side of the equation

2x + 40 = 3− 8Is the constant negitive?Yes, then STOP.

No Real Number works

NRN

34. 3x − 6 = −2x +14 ! 35. 5x −19 = 2x + 4 !

Both radicals are alone onthier side of the equation


3x − 6 = −2x +14

Solve for x

5x − 6 = 145x = 20x = 4

!

Both radicals are alone onthier side of the equation


5x +19 = 2x + 4

Solve for x

3x +19 = 43x = −15x = −5

When you plug in x = – 5it does not workNRN or ∅


36. 2x2 + 3x −10 = x ! 37. 2x2 − x −1 = 2x − 2


2x2 + 3x −10 = x2

Solve for x

x2 + 3x −10 = 0(x + 5)(x − 2) = 0

x = −5 or x = 2

When you plug in x = 2it does workWhen you plug in x = – 5it does NOT work

Answer: x = 2

!


2x2 − x −1= 2x − 2

Solve for x

2x2 − 3x +1= 0(2x −1)(x −1) = 0

x = 12

or x = 1

When you plug in x = 1it does work

When you plug in x = 12

it does NOT work

Answer: x = 1

!


�

x =−(B)± B( )2 − 4(A)(C)

2A

Quadratic Equations. Solve for x. You must use the quadratic equation for each problem and show the work. Use NRN or

�

∅ if needed.

38.

�

3x2 + 2x − 3 = 0 ! 39.

�

4x2 −8x +1 = 0 !

�

A B C

3x2 + 2x − 3 = 0

B2 − 4 A( ) C( ) = 2( )2 − 4 3( ) −3( )= 4 + 36 = 40

x =−(2) ± 40

6

x =−2 ± 4 •10

6

x =−2 ± 2 10

6

x =2 −1± 10( )

6

x =−1± 10

3

!

�

A B C

4x2 − 8x + 1 = 0

B2 − 4 A( ) C( ) = −8( )2 − 4 4( ) 1( )= 64 −16 = 48

x =−(−8) ± 48

8

x =8 ± 3•16

8

x =8 ± 4 3

8

x =4 2 ± 3( )

8

x =2 ± 3

2


40.

�

2x2 + 3x + 4 = 0 !

�

A B C

2x2 + 3x + 4 = 0

B2 − 4 A( ) C( ) = 3( )2 − 4 2( ) 4( )= 9 − 32 = −23

x =−(2) ± − 23

4

Thereis no real number

for − 23so the answer is no real number

NRN


math 100 half chapter 8 and chapter 9 test review selected ...flc.losrios.edu/~eitel/all pdfs/m-100...

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