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Multiplying and Dividing Radical Expressions - Sample Math Practice Problems

The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.

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Complexity=1, Mode=mul

Solve and simplify.

1.  
√5 • √45 =    
2.  
√6 • √30 =    

Complexity=2, Mode=mul

Solve and simplify.

1.  
   70p 8
 • √50p =    
2.  
√45j •
   27j 4
 =    

Complexity=1, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1.  
=
   
2.  
=
   

Complexity=2, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1.  
=
   
2.  
=
   

Complexity=3, Mode=mul

Solve and simplify.

1.  
   8b 5
 •
   6a 7b 8
 =    
2.  
   4tu 7
 •
   5t 2u 5
 =    

Complexity=3, Mode=div

Solve and simplify. Leave fractions as improper fractions.

1.  
=
   
2.  
    
 
=
   

Answers

Complexity=1, Mode=mul

Solve and simplify.

#Correct AnswerYour Answer
1
√5 • √45 =    
Solution
√5 • √45 = √5 • 45
=√5 • 5 • 9
=5 • √3 • 3
=5 • 3
=15
#Correct AnswerYour Answer
2
√6 • √30 =    
Solution
√6 • √30 = √6 • 30
=√6 • 6 • 5
=6√5

Complexity=2, Mode=mul

Solve and simplify.

#Correct AnswerYour Answer
1
   70p 8
 • √50p =    
Solution
   70p 8
 •
   50p
 =
   10 • 7 • p 8
 •
   10 • 5 • p
=
   5 • 7 • 10 • 10
•
    p 8 • p
=
10√5 • 7•
    p 9
=
10√35•p 4√ p
=10p 4√35p
#Correct AnswerYour Answer
2
√45j •
   27j 4
 =    
Solution
   45j
 •
   27j 4
 =
   9 • 5 • j
 •
   9 • 3 • j 4
=
   3 • 5 • 9 • 9
•
    j • j 4
=
9√3 • 5•
    j 5
=
9√15•j 2√ j
=9j 2√15j

Complexity=1, Mode=div

Solve and simplify. Leave fractions as improper fractions.

#Correct AnswerYour Answer
1
=
   
Solution
=
=
=
#Correct AnswerYour Answer
2
=
   
Solution
=
=

Complexity=2, Mode=div

Solve and simplify. Leave fractions as improper fractions.

#Correct AnswerYour Answer
1
=
   
Solution
The numerator and denominator are both divisible by 2
=
    
 
=
    
 
=
=
•
=
#Correct AnswerYour Answer
2
=
   
Solution
=
•
=

Complexity=3, Mode=mul

Solve and simplify.

#Correct AnswerYour Answer
1
   8b 5
 •
   6a 7b 8
 =    
Solution
   8b 5
 •
   6a 7b 8
 =
   2 • 4 • b 5
 •
   2 • 3 • a 7 • b 8
=
   2 • 2 • 3 • 4
•
    b 5 • b 8
•
    a 7
=
2√3 • 4•
    b 13
•a 3√ a
=
(2 • 2)√3•b 6√ b•a 3√ a
=
4√3•b 6√ b•a 3√ a
=4a 3b 6√3ab
#Correct AnswerYour Answer
2
   4tu 7
 •
   5t 2u 5
 =    
Solution
   4tu 7
 •
   5t 2u 5
 =
   4 • t • u 7
 •
   5 • t 2 • u 5
=
   4 • 5
•
    t • t 2
•
    u 7 • u 5
=
2√5•
    t 3
•
    u 12
=
2√5•t√ t•u 6
=2tu 6√5t

Complexity=3, Mode=div

Solve and simplify. Leave fractions as improper fractions.

#Correct AnswerYour Answer
1
=
   
Solution
=
=
•
=
#Correct AnswerYour Answer
2
    
 
=
   
Solution
The numerator and denominator are both divisible by q 2
    
 
=
    
 
=
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