Answer/Discussion on Radicals

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

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Answer/Discussion to 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:

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)

Answer/Discussion to 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:

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

(return to problem 1b)

Answer/Discussion to 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

rule.

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)

Answer/Discussion to 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)

Answer/Discussion to 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)

Answer/Discussion to 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

. When you simplify a radical, you want to take out as much as possible. The factor of

that we can take the square root of is

. We can write

and then use the product rule of radicals to separate the two numbers. We can take the square root of

, which is

, but we will have to leave the rest of it under the square root.

(return to problem 2b)

Answer/Discussion to 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)

Answer/Discussion to 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)

Answer/Discussion to 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.

*Combine like radicals: 2 - 5 = -3

(return to problem 4a)

Answer/Discussion to 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:

*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.

*Combine like radicals: 14 + 8 = 22

(return to problem 4b)

Answer/Discussion to 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:

*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.

*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)

Answer/Discussion to 5b