8.E: Review Exercises and Sample Exam
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Review Exercises
(Assume all variables represent nonnegative numbers.)
Simplify.
- √36
- √425
- √−16
- −√9
- 3√125
- 33√−8
- 3√164
- −53√−27
- √40
- −3√50
- √9881
- √1121
- 53√192
- 23√−54
- Answer
-
1. 6
3. Not a real number
5. 5
7. 14
9. 2√10
11. 7√29
13. 203√3
Simplify.
- √49x2
- √25a2b2
- √75x3y2
- √200m4n3
- √18x325y2
- √108x349y4
- 3√216x3
- 3√−125x6y3
- 3√27a7b5c3
- 3√120x9y4
- Answer
-
1. 7x
3. 5xy√3x
5. 3x√2x5y
7. 6x
9. 3a2bc3√ab2
Use the distance formula to calculate the distance between the given two points.
- (5,−8) and (2,−10)
- (−7,−1) and (−6,1)
- (−10,−1) and (0,−5)
- (5,−1) and (−2,−2)
- Answer
-
1. √13
3. 2√29
Simplify.
- 8√3+3√3
- 12√10−2√10
- 14√3+5√2−5√3−6√2
- 22√ab−5√ab+7√ab−2√ab
- 7√x−(3√x+2√y)
- (8y√x−7x√y)−(5x√y−12y√x)
- √45+√12−√20−√75
- √24−√32+√54−2√32
- 2√3x2+√45x−x√27+√20x
- √56a2b+√8a2b2−√224a2b−a√18b2
- 5y√4x2y−(x√16y3−2√9x2y3)
- (2b√9a2c−3a√16b2c)−(√64a2b2c−9b√a2c)
- 3√216x−3√125xy−3√8x
- 3√128x3−2x3√54+33√2x3
- 3√8x3y−2x3√8y+3√27x3y+x3√y
- 3√27a3b−33√8ab3+a3√64b−b3√a
- Answer
-
1. 11√3
3. 9√3−√2
5. 4√x−2√y
7.√5−3√3
9. −√3x+5√5√x
11. 12xy√y
13. 43√x−53√xy
15. 2x3√y
Multiply.
- √3⋅√6
- (3√5)2
- √2(√3−√6)
- (√2−√6)2
- (1−√5)(1+√5)
- (2√3+√5)(3√2−2√5)
- 3√2a2⋅3√4a
- 3√25a2b⋅3√5a2b2
- Answer
-
1. 3√2
3. √6−2√3
5. −4
7. 2a
Divide.
- √72√4
- 10√48√64
- √98x4y2√36x2
- 3√81x6y73√8y3
- Answer
-
1. 3√2
3. 7xy√26
Rationalize the denominator.
- 2√7
- √6√3
- √142x
- √1215
- 3√12x2
- 3√5a2b5ab2
- 1√3−√2
- √2−√6√2+√6
- Answer
-
1. 2√77
3. √7√xx
5. 2233√x2x
7. √3+√2
Express in radical form.
- 71/2
- 32/3
- x4/5
- y−3/4
- Answer
-
1. √7
3. 5√x4
Write as a radical and then simplify.
- 41/2
- 501/2
- 42/3
- 811/3
- (14)3/2
- (1216)−1/3
- Answer
-
1. 2
3. 23√2
5. 18
Perform the operations and simplify. Leave answers in exponential form.
- 31/2⋅33/2
- 21/2⋅21/3
- 432412
- 934914
- (36x4y2)12
- (8x6y9)1/3
- (a43a12)25
- (16x43y2)12
- Answer
-
1. 9
3. 4
5. 6x2y
7. a1/3
Solve.
- √x=5
- √2x−1=3
- √x−8+2=5
- √3x−5−1=11
- √5x−3=√2x+15
- √8x−15=x
- √x+41=x−1
- √7−3x=x−3
- 2(x+1)=√2(x+1)
- √x(x+6)=4
- 3√x(3x+10)=2
- 3√2x2−x+4=5
- 3√3(x+4)(x+1)=3√5x+37
- 3√3x2−9x+24=3√(x+2)2
- y1/2−3=0
- y1/3+3=0
- (x−5)1/2−2=0
- (2x−1)1/3−5=0
- Answer
-
1. 25
3. 17
5. 6
7. 8
9. −12,−1
11. 23,−4
13. −5,53
15. 9
17. 9
Sample Exam
In Exercises 12-16, assume all variables represent nonnegative numbers.
Simplify.
-
- √100
- √−100
- −√100
- Answer
-
1. a. 10 b. Not a real number c. -10
Simplify.
- g
- 3√27
- 3√−27
- −3√27
- √12825
- 3√192125
- 5√12x2y3z
- 23√50x2y3z5
- Answer
-
2. 8√25
4. 10xy√3yz
Perform the operations.
- 5√24−√108+√96−3√27
- 3√8x2y−(x√200y−√18x2y)
- 2√ab⋅(3√2a−√b)
- (√x−√2y)2
- Answer
-
1. 14√6−15√3
3. 6a√2b−2b√a
Rationalize the denominator.
- 10√2x
- 3√14xy2
- 1√x+5
- √2−√3√2+√3
- Answer
-
1. 5√2xx
3. √x−5x−25
Perform the operations and simplify. Leave answers in exponential form.
- 223⋅216
- 10451013
- (121a4b2)12
- (9y13x6)12y16
- Answer
-
1. 25/6
3. 11a2b
Solve.
- √x−7=0
- √3x+5=1
- √2x−1+2=x
- 3√1−10x=x−4
- √(2x+1)(3x+2)=√3(2x+1)
- 3√x(2x−15)=3
- The period, T, of a pendulum in seconds is given the formula T=2π√L/32, where L represents the length in feet. Calculate the length of a pendulum if the period is 11/2 seconds. Round off to the nearest tenth.
- Answer
-
1. 49
3. 5
5. −12,13
7. 1.8 feet