Intermediate Algebra Tutorial 37


Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 37: Radicals


WTAMU > Virtual Math Lab > Intermediate Algebra > Tutorial 37: Radicals


 

checkAnswer/Discussion to 1a

problem 1a
 

The thought behind this is that we are looking for the square root of ad1a2. This means that we are looking for an expression that, when we square it, we get ad1a2

What do you think it is?

Let's find out if you are right:

ad1a1



Since ad1a3  squared is ad1a2,ad1a3is the square root of ad1a2 
 

(return to problem 1a)

 


 
 

checkAnswer/Discussion to 1b

problem 1b

The thought behind this is that we are looking for the cube root of ad1b2. This means that we are looking for an expression that, when we cube it, we get ad1b2.

What do you think it is?
 
 

Let's find out if you are right:

ad1b1

Since ad1b3  cubed is ad1b2ad1b3is the cube root of ad1b2
 
 

(return to problem 1b)

 


 

checkAnswer/Discussion to 1c

problem 1c

The thought behind this is that we are looking for the fourth root of ad1c2. This means that we are looking for an expression that, when we raise it to the fourth power,  we get ad1c2.

What do you think it is?
 
 

Let's find out if you are right:

ad1c1



Since ad1c3  raised to the fourth power is ad1c2 and there is a negative in front of the radical, ad1c4is the negative fourth root of ad1c2
 

(return to problem 1c)

 


 

checkAnswer/Discussion to 1d

problem 1d
The thought behind this is that we are looking for the fifth root of ad1d2  .  This means that we are looking for an expression that, when we raise it to the fifth power, we get ad1d2

What do you think it is?

Let's find out if you are right:

ad1d1



Since ad1d3 raised to the fifth power  is ad1d2ad1d3  is the fifth root of ad1d2.  
 

(return to problem 1d)

 


 

checkAnswer/Discussion to 2a

problem 2a
 

Since it didn't say that x or y are positive, we have to assume that they can be either positive or negative.  Since the root number and exponent are equal, then we can use the nth rootrule.

ad2a2

Since the root number and the exponent inside are equal and are the even number 2, then we need to put an absolute value around x - y for our answer.  The reason for the absolute value is that we do not know if x or y are positive or negative.  So if we put x - y as our answer and it was negative, it would not be a true statement.
 

(return to problem 2a)


 

checkAnswer/Discussion to 2b

problem 2b
 

Since the root number and exponent are equal then we can use the nth rootrule.

ad2b

This time our root number and exponent were both the odd number 3.  When an odd numbered root and exponent match, then the answer is the base whether it is negative or positive.
 

(return to problem 2b)

 

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WTAMU > Virtual Math Lab >Intermediate Algebra >Tutorial 37: Radicals


Last revised on July 19, 2011 by Kim Seward.
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