College Algebra Tutorial 4


College Algebra
Answer/Discussion to Practice Problems
Tutorial 4: Radical



WTAMU > Virtual Math Lab > College Algebra > Tutorial 4: Radical


 

 

check markAnswer/Discussion to 1a

prob 1a
 

The thought behind this is that we are looking for the square root of 81. This means that we are looking for a number that when we square it we get 81. 

What do you think it is?

Let's find out if you are right:

answer 1a

Since 9 squared is 81, 9 is the square root of 81.

Note that we are only interested in the principal root and since 81 is positive and there is not a sign in front of the radical, our answer is positive 9.  If there had been a negative in front of the radical our answer would have been -9.
 

(return to problem 1a)

 


 

check markAnswer/Discussion to 1b

prob 1b
 

The thought behind this is that we are looking for the cube root of -27. This means that we are looking for a number that when we cube it we get -27. 

What do you think it is?

Let's find out if you are right:

answer 1b

Since -3 cubed is -27, -3 is the cube root of -27.
 

(return to problem 1b)

 


 

check markAnswer/Discussion to 1c

prob 1c
 

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

answer 1c

Since the root number and the exponent inside are equal and are the even number 2, we need to put an absolute value around x for our answer. 

The reason for the absolute value is that we do not know if x is positive or negative.  So if we put x as our answer and it was negative, it would not be a true statement. 
 

(return to problem 1c)

 


 

check markAnswer/Discussion to 1d

prob 1d
 

Now we are looking for the square root of -4, which means we are looking for a number that when we square it we get -4.

What do you think it is?

Let's find out if you are right:

Since there is no such real number that when we square it we get -4, then the answer is not a real number.
 

(return to problem 1d)

 


 

check markAnswer/Discussion to 2a

prob 2a
 

answer 2a
*Use the prod. rule of radicals to rewrite

 
 

Note that both radicals have an index number of 2, so we were able to put their product together under one radical keeping the 2 as its index number. 

Since we cannot take the square root of 14 and 14 does not have any factors we can take the square root of, this is as simplified as it gets.
 

(return to problem 2a)

 


 

check markAnswer/Discussion to 2b

prob 2b
 

answer 2b
*Use the prod. rule of radicals to rewrite

 
 

In this example, we are using the product rule of radicals in reverse to help us simplify the cube root of answer 2b2.  When you simplify a radical, you want to take out as much as possible.  The factor of answer 2b2 that we can take the square root of is answer 2b3 .  We can write answer 2b2as answer 2b4 and then use the product rule of radicals to separate the two numbers.  We can take the square root of answer 2b3, which is answer 2b5, but we will have to leave the rest of it under the square root.

 
(return to problem 2b)

 


 

check markAnswer/Discussion to 3a

prob 3a
 

answer 3a
*Use the  quotient rule of radicals to rewrite
 

*The cube root of -1 is -1 and the cube root of 8 is 2
 
 

(return to problem 3a)

 


 

check markAnswer/Discussion to 3b

prob 3b
 

answer 3b
*Use the  quotient rule of radicals to rewrite

*Simplify the fraction

*The square root of 9 a squared  is 3|a|
 

(return to problem 3b)

 


 

check markAnswer/Discussion to 4a

prob 4a
 

Step 1: Simplify the radicals.

 
Both radicals are as simplified as it gets.

 
Step 2: Combine like radicals.

 
Note how both radicals are the square root of x.  These two radicals are like radicals.

 
answer 4a
*Combine like radicals: 2 - 5 = -3

 
(return to problem 4a)

 


 

check markAnswer/Discussion to 4b

prob 4b
 

Step 1: Simplify the radicals.

 
The 20 in the first radical has a factor that we can take the square root of. 

Can you think of what that factor is?

Let's see what we get when we simplify the first radical:
 

answer 4b

*Rewrite 20 as (4)(5)
*Use Prod. Rule of Radicals

*Square root of 4 is 2
 
 

Step 2: Combine like radicals.

 
Note how both radicals are the square root of 5.  These two radicals are like radicals.

 
answer 4b2
*Combine like radicals: 14 + 8 = 22

 
(return to problem 4b)

 


 

check markAnswer/Discussion to 5a

prob 5a
 

Step 1: Multiply numerator and denominator by a radical that will get rid of the radical in the denominator.

 
Since we have a square root in the denominator,  we need to multiply by the square root of an expression that will give us a perfect square under the radical in the denominator.

Square roots are nice to work with in this type of problem because if the radicand is not a perfect square to begin with, we just have to multiply it by itself and then we have a perfect square.

So in this case we can accomplish this by multiplying top and bottom by the square root of 7:
 

answer 5a1

*Mult. num. and den. by sq. root of 7
 
 

*Den. now has a perfect square under sq. root
 
 

Step 2: Simplify the radicals.

AND
 

Step 3: Simplify the fraction if needed.

 
answer 5a2
*Sq. root of 49 is 7
 

 
 

Be careful when you reduce a fraction like this.  It is real tempting to cancel the 7 which is on the outside of the radical with the 7 which is inside the radical on the last fraction.  You cannot do that unless they are both inside the same radical or both outside the radical.

 
(return to problem 5a)

 


 

Check markAnswer/Discussion to 5b

prob 5b
 

Step 1: Find the conjugate of the denominator.

 
In general the conjugate of a + b is a - b and vice versa.

So what would the conjugate of our denominator be?

It looks like the conjugate is answer b1.
 

Step 2: Multiply the numerator and the denominator of the fraction by the conjugate found in Step 1.

 
answer 5b2

*Mult. num. and den. by conjugate of den.

*Use distributive prop. to multiply the numerators
*In general, product of conjugates is conjugate

 
 

Step 3: Simplify the radicals.

AND
 

Step 4: Simplify the fraction if needed.

 
answer 5b3

 

*Square root of 2 squared is 2 and the square root of 3 squared is 3
 
 

*Divide BOTH terms of num. by -1
 
 

(return to problem 5b)

 
 

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WTAMU > Virtual Math Lab > College Algebra > Tutorial 4: Radicals



Last revised on Dec. 6, 2009 by Kim Seward.
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