>> Welcome to the Cypress College Math Review on Radicals.
Objective one, taking nth roots with numbers.
B is the nth root of a if b to the nth power is equal to a. Here's a notation that we use
for radicals b is the nth root of a, n is called the index (the little number up to the left
of the radical sign), a is the radicand (the thing inside the radical)
and the whole thing nth root of a is called the radical expression.
Here's an example, 2 is equal to the cube root or 3rd root of 8.
Two is the 3rd root or cube root of 8.
Three is the index of the radical.
Eight is the radicand -- the thing inside the radical.
And the cube root of 8 is the radical, the whole radical expression.
In order to work with radicals, it's helpful to know as many perfect square numbers as possible.
Try to memorize as many of these relationships as you can.
Now, if b to the nth is equal to a then b is equal to the nth root of a. Well,
let's put some numbers in here to make that a little more clear.
If 3 squared is equal to 9, which it is, 3 times 3 is nine, then 3 is a 2nd root of 9.
Well, how do we write 2nd root?
Second root is the same thing as square root.
In other words, when the index of the radical is a 2, we don't write it.
So, 3 is the square root of 9.
So, this is a 3.
So, what would you square to get 4?
Two. What would you square to get 1?
One. What would you square to get 16?
Four. What would you square to get 25?
Five. These are the square roots of those numbers.
Now, let's look at cube roots.
Once again, it's helpful to know as many perfect cube numbers as possible.
So, learn as many of these relationships as you can.
What would you cube to get 1?
Well, that's just 1.
So, the cube root of 1 is just 1.
What would you cube to get 8?
Well, that's 2, so the cube root of 8 is 2.
What would you cube to get 27?
Well, that would be 3, 3 cubed is 27.
So, the cube root of 27 is 3.
What would you cube to get 64?
Well, that's 4, so the cube root of 64 is 4.
What would you cube to get 125?
That's 5, so the cube root of 125 is 5.
Example, simplify the following radical.
The square root of 49.
Well, the square root of 49 is 7.
Why? Because 7 squared is equal to 49.
Maybe you didn't know that.
Maybe you didn't know that perfect square.
What could you do then?
You could do a factor tree.
So, you take the number 49 and you do a factor tree and it turns out that it's 7 times 7.
So, you write the square root of 49 as the square root of 7 times 7.
Now, to be a perfect square, a number must have two of the same factor.
We have two of the same factor.
When this happens, you can take those two factors together
and allow one to escape the square root.
So, the square root of 7 times 7 is just 7.
Now, note that you do not write the radical symbol anymore.
Now we want to take the 3rd root or cube root of 216.
Well, that's equal to 6.
Why? Because if you take 6 and cube it, you get 216.
We could rewrite 216 as 6 times 6 times 6.
To be a perfect cube a number must have three of the same factor.
When this happens, we can take those three factors together
and allow one to escape the cube root.
So, three factors together to escape a cube root so we get the number 6.
Principal and negative square roots.
If you take the number 5 and you square it.
You get 25.
If you take the number negative 5 and square that you also get 25.
Well that means there's actually two different square roots
of the number 25 or any positive number.
So, 5 is a square root of 25 and negative 5 is a square root of 25.
What does this symbol stand for then?
This symbol stands for just the principal square root of 25.
The non-negative square root.
So, the symbol stands for just the non-negative or principal square root of 25.
If you put a negative in front of that symbol, that gives you the other square root of 25.
The square root of a negative number is not a real number.
So, the square root of negative 9 is not real.
The square root of 0 is 0 because if you square 0 you get 0.
The square root of 1 is 1 because 1 squared is 1.
If the index is even such as, square root which is the same thing as a 2nd root,
4th root or 6th root, then the radicand must be non-negative for the root to be a real number.
For example, the 4th root of 16 is equal to 2, because 2 to the 4th is equal to 16.
But the 4th root of negative 16 is not a real number.
A lot of people would say oh negative 2 works.
But no, if you take negative 2 and raise it to the 4th power, you have an even number
of negatives, which is going to give you positive 16, not negative 16.
So, the 4th root of negative 16 is not a real number.
The 6th root of 64 is equal to 2 because, 2 to the 6th is equal to 64.
But the 6th root of negative 64 is not a real number.
If the index is odd, such as cube root, which is 3rd root, 5th root or 7th root,
then the radicand can be any real number.
For example, the cube root of 64 is equal to 4 because 4 cubed is equal to 64.
The cube root of negative 64 is negative because if you take negative 4
and cube it that's an odd number of negatives, so that does give you a negative 64.
The 5th root of 32 is equal to 2 because 2 to the 5th is equal to 32.
And the 5th root of negative 32 is equal to negative 2 because if you take negative 2
and raise it to an odd power you will get a negative number.
If we do not get a perfect power that matches the index of the radical,
we can still take out any double factors that exist
and whatever's leftover will be stuck under the radical.
So, we're working with 18.
Eighteen is 2 times 9 and 9 is 3 times 3.
So, we do have a double factor in the 3's so, we have 2 times 3 times 3.
So, the pair of 3's will come out and give us one 3 on the outside and the 2 is left inside.
Forty is 4 times 10.
Four is 2 times 2 and 10 is 2 times 5.
We're working with the cube root of 40.
So, we have the cube root of 2 times 2 times 2 times 5.
We need three like factors to be able to come out of a cube root.
So, the cube root of 2 cubed is 2 it comes outside and we're left with a 5 on the inside.
Be sure to write the index of the radical, the little 3.
So, this is 2 times the cube root of 5.
Well, 162 is 2 times 81.
Eighty-one is 9 times 9 and 9 is 3 times 3.
So, here we have the square root of 2 times 3 times 3 times 3 times 3.
So, we have pair of 3's, so one 3 will escape.
Then we have another pair of 3's
so another 3 will escape.
The 2 is left inside.
So, the answer is 9 times the square root of 2.
Pause the video and try these problems.
Objective two, taking nth roots with variables.
Variables follow the same rules that numbers do.
It takes two like factors to escape the square root, three like factors
to escape the cube root and so on and so forth.
So, let's take the square root of x squared.
Well, the square root of x squared is equal to x. Why?
Because if you square x you get x squared.
What's the square root of x to the 4th?
Well that's x squared, why?
Because when you square x squared you get x to the 4th.
The square root of x to the 6th is x cubed
because if you square x cubed you get x to the 6th.
So, what's happening here?
If you take the exponent and divide by the index of the radical
and for a square root the index of the radical is 2.
You get the new exponent.
So, 8 divided by 2 is 4.
So, the square root of x to the 8th is x to the 4th.
Ten divided by 2 is 5.
So, the square root of x to the 10th is x to the 5th.
Let's take cube roots of variables.
You can rewrite this as x times itself three times and since we're doing a cube root,
it takes three like factors to pull it outside.
So, that gives us just x to the 1st on the outside and nothing left inside.
You can also think about this in terms of dividing the exponents.
If you take the exponent that's inside divide by the index of the radical and we get 1.
And that's the exponent you're left with.
Let's take the cube root of x to the 6th.
So, 6 divided by 3 is 2.
Because not too many people want to keep writing out all these x's.
So, there's one x that would come out.
There's another x that would come out.
Multiply times each other, that gives us x to the 2nd power.
Nine divided by 3 is 3, so the cube root of x to the 9th is x to the 3rd.
Twelve divided by 3 is 4.
Fifteen divided by 3 is 5.
Simplify the following radicals, assume variables represent positive real numbers.
Since this is a square root the index of the radical is 2,
so we divide by 4 by 2 and we get 2.
On this problem, we divide 10 by 2 and we get 5.
Now, we're going to divide 11 by 2, we get 5 with a remainder of 1.
So, you will end up with 5 x's on the outside and one that's left on the inside.
The remainder tells you the power of the variable left inside.
The quotient tells you how many x's are on the outside.
The other way to do this problem is to write out all of those 11 x's.
Then, two like factors will bring us an x on the outside,
another pair of like factors gives us another x, and another, and another, and another.
Those are all on the outside of the radical.
Left inside is that one x there.
Once again, that gives us the answer we have above x to the 5th times the square root
of x. Cube root of x to the 17th.
So, we divide our exponent 17 by the index of the radical which is 3.
We get 5 with the remainder of 2.
So, we will end up with x to the 5th power on the outside
and x to the 2nd power on the inside.
In general, for any division problem the remainder is always smaller than the divisor.
In this case, 2 of course is smaller than 3.
For a radical to be in simplified form,
the exponents on the inside must be smaller than index of the radical.
Once again, 2 of course is smaller than 3.
For this problem, we divide 23 by the index of the radical which is 5
and we get 4 with the remainder of 3.
So, the answer would be x to the 4th times the 5th root of x to the 3rd.
This problem has numbers and variables.
We've done the prime factorization of the coefficient that was there and we've written it
as 3 to the 5th times 5 to the 6th.
So, we can work with this just like the 3 and the 5 were variables.
So, we divide 5 by 4 and get 1 with a remainder of 1.
We divide 6 by 4 and we get 1 with a remainder of 2.
We divide 11 by 4 and get 2 with a remainder of 3.
We divide 12 by 4 and get 3 with no remainder, so no y's are left on the inside.
So, our answer is 15 x squared, y cubed times the 4th root of - now what is this,
5 squared is 25 and 25 times 3 is 75, so 75 x cubed.
So, if you have your coefficient written in terms of a prime factorization,
you have it all factored, you can work with it just like you can variables.
We have 96 and we want to break it down into its prime factors.
So, that 3 times 32 and 32 is made up all 2's.
We've seen this number before.
So, we have the 5th root of 2 to the 5th power times 3 times a
to the 45th, times b to the 37th.
Now we divide the exponents by the index.
Five divided by 5 is 1 and there's no remainder, so there's no 2's left on the inside.
Three is just to the exponent of 1, which is already smaller
than the index, so the 3 stays on the inside.
Forty-five divided by 5 is 9, with no remainder, so there's no a's left on the inside.
Thirty-seven divided by 5 is 7, with a remainder of 2 and that's our answer.
Pause the video and try these problems.
Objective three, adding and subtracting radicals.
Like radicals have the same index and the same radicand.
We add or subtract radicals similar to the way that we combine like terms.
Add or subtract the coefficients then keep the radical.
Three x plus 7x is equal to 10x.
Three times the square root of 5 plus 7 times the square root of 5 is equal
to 10 times the square root of 5.
Three of these plus 7 of these is 10 of these.
If the radicals are not the same,
then we simplify them first before we combine like terms.
On this problem, the radicals are the same, they're both the square root of 7.
Four of these minus 9 of these is negative 5 of these.
On the next problem, the radicals are also the same, the cube root of 2y.
Nine of these plus 15 of these is 24, of these.
These radicals are not the same so first we have to simplify them.
The prime factorization of 50 is 5 square times 2, because 25 times 2.
The prime factorization of 18 is 3 square times 2, 9 times 2.
Now we divide the exponents, these are square roots that we're going to divide by 2.
Two divided by 2 is 1, with no remainder, so no 5's are left on the inside.
The exponent for 2 is a 1,
which is smaller than the index of the radical, which is 2.
Therefore, the square root of 2 stays on the inside, because 1 is smaller than 2.
We divide 2 by 2 and we get 3 to the 1st on the outside with no remainder.
The 2 is left on the inside.
So, we have 5 times the square root of 2 plus 15 times the square root 2.
Well 5 of these plus 15 of these is 20 of these.
For this problem, we're not going to do prime factorizations.
Here we recognize that 4 is a perfect square and it is a factor of 20.
That leaves us 5x on the inside.
Sixteen is a perfect square, leaves x on the inside.
Nine is a perfect square and that is a factor of 45 which leaves us
with 5 times x. The square root of 4 is 2.
The square root of 16 is 4.
The square root of 9 is 3.
So, we have 2 times the square root of 5x minus 24 times the square root
of x plus 3 times the square root of 5x.
Which of these are like terms?
Well the first and the last.
So, we have 5 times the square root of 5x the other term is not a
like term and it cannot be combined.
Here we recognize that the cube root of 8 is 2.
So, we bring that out.
You could also write that out as 2 cubed and then once again you end
up with 2 to the 1st on the outside.
Divide the exponent 5 by 3 and we get 1 with a remainder of 2.
The cube root of 27 is 3 because 27 is 3 cubed.
We divide 5 by 3 and we get 1 with a remainder of 2.
These are like terms.
Not only does the radical have to be the same but since this has a variable
in front, that has to be the same also.
This whole expression has to be the same.
We have two of these plus three of these which gives us five of these.
Pause the video and try these problems.
Objective four, rationalizing monomial denominators.
First let's look at the rules for a radical to be in simplified form.
So, we're taking the nth root.
First you can't have any perfect nth powers in the radicand.
So, if we had 2 to the 4th times x squared, we would have to rewrite that as 2
to the 1st times the cube root of 2 times x squared.
Notice that the exponents that are left on the inside
of the radical are smaller than the index of the radical.
That's making sure that we don't have any perfect cubes inside a cube root.
Next, you can't have any fractions in the radicand.
So, this is not simplified form.
You can't leave any radicals in the denominator.
This is also not simplified form.
We're going to work on these two processes next.
We'll use the product rule and quotient rule for radicals to do these problems.
Rationalizing the denominator of a fraction with one term in the denominator.
The process of eliminating a radical
in the denominator is called rationalizing the denominator.
If the denominator is a single term, we multiply both the numerator and the denominator
so that the denominator is a perfect square, or a perfect cube, or perfect 4th root,
depending upon the index of the radical.
Examples, simplify, since each of these expressions have a radical
in the denominator simplify includes rationalizing the denominator.
Assume variables represent positive real numbers.
On the first problem, we have the square root of 5 in the denominator.
If we multiply both the top and the bottom by the square root of 5,
we end up with by the product rule, the square root of 5 squared in the denominator.
Well, the square root of 5 squared is simply 5.
So, the 5 is no longer in the radical and we don't have a radical in the denominator.
We check to see if 2/5 reduces and it doesn't, so we're done.
For the next problem, we have the cube root of 1/5.
First of all, since we have a radical of a fraction, we see if the fraction can be reduced.
It can't, 1/5 cannot be reduced.
So, we split it using the quotient rule.
The cube root of 1 over the cube root of 5.
The cube root of 1 is just one.
Now we have a radical in the denominator and we can't leave that.
We have 5 to the 1st power under a cube root.
We will need two more 5's, 5 to the 2nd power so that we have the 3
that we need to get out of the cube root.
So, the numerator is the cube root of 5 squared and 5 squared is simply 25.
The denominator is the cube root of 5 cubed.
What's the cube root of 5 cubed?
That's just 5.
And that gives us our answer, no more radical in the denominator.
For the next problem, we're doing a 4th root.
So, we have x to the 1st power, how many more x's do we need so we have x to the 4th power?
Well, that would be x cubed.
So, we have 6 times the 4th root of x cubed in the numerator
and the denominator we have the 4th root of x to the 4th.
But the 4th root of x to the 4th is simply x. For the next problem, we have the 7th root
of x to the 5th in the denominator.
We need two more x's to make it so that we will have a 7th root
of x to the 7th in the denominator.
Now our denominator is simply x. No more radical and the problem's done.
You could combine this into one radical and you would do this
if anything canceled out and simplified.
But nothing does, so there's no reason to do it that way.
Let's start working on that denominator.
Well, 32 is 2 to the 5th power.
Now, what do we need to multiply by to make this radical in the denominator disappear?
Well, the 5th root of 2 to the 5th is just 2, so there aren't going to be any 2's left inside.
So, we don't need any more 2's.
The power for b is 17.
The new power needs to be a multiple of 5.
Five, 10, 15, 20.
Ah... 20, so I need 20 b's altogether.
So, I need three more.
We multiply the numerators times each other.
In the denominator we have the 5th root of 2 to the 5th times b to the 20th.
In the numerator, can that be simplified?
Always check.
Two is smaller than 5, 3 is smaller than 5, so the numerator cannot be simplified.
How about the denominator?
The 5th root of 2 to the 5th is just 2, so 2 to the 1st comes outside.
And 20 divided by 5 is 4.
No more radical.
On the next problem, you should combine them into one radical because that fraction reduces.
Divide the numerator and denominator each by 3 and we get 6/25.
Now we split it using the quotient rule.
And the denominator is simply 5 because the square root of 25 is 5.
No more radical in the denominator.
And the square root of 6 is simplified.
There's no perfect squares that can come out of that.
Pause the video and try these problems.
Objective five, rationalizing binomial denominators.
Recall from factoring the formula the quantity a plus b times the quantity a minus b is equal
to a squared minus b squared.
This formula plays an integral part in allowing us to rationalize
if our denominator is a binomial involving square roots.
Where a plus b and a minus b are called conjugates of each other.
First, let's multiply a couple conjugates and see what we get.
When you multiply 4 minus the square root of 3 times 4 plus the square root of 3,
you get 4 squared minus the square root of 3 squared, a squared minus b squared.
The outer and inner terms in F.O.I.L. cancel out.
So, we end up with 16 minus 3 which is 13.
No more radical.
Simplify. And since these expressions have a radical in the denominator,
simplify includes rationalizing the denominator.
We have two terms in the denominator and at least one
of them has a square root, so we multiply by the conjugate.
The conjugate of the square root of 5 plus square root
of 3 is the square root of 5 minus the square root of 3.
What you do to the top, you must do to the bottom.
Because this expression has to equal the number 1.
You can only multiply an expression by the number 1.
Since this 6 does not have a radical in it, I leave it factored out.
So, I have 6 times that quantity.
In the denominator, I have a squared minus b squared.
So, I'm going to have the square root of 5 squared minus the square root of 3 squared,
because the outer and the inner terms once again in F.O.I.L., cancel out.
So, we will end up with a squared minus b squared.
We leave the 6 out there just in case there will be any canceling that we can do.
So, the square root of 5 squared is just 5.
The square root of 3 squared is 3.
So, we end up with 2 in the denominator and that does cancel with our 6,
and since our denominator is now just 1 we go ahead and distribute,
and our final answer is 3 times the square root of 5 minus 3 times the square root of 3.
Here we also have two terms in the denominator
with at least one radical, so we multiply by the conjugate.
So, 5 times the square root of 3 plus 4 times the square root of 2 over itself.
Once again, you can only multiply by the number 1.
This numerator has a radical in it so, in this case we do go ahead and distribute.
So, we have 3 times the square root of 6
and that's multiplied times 5 times the square root of 3.
Plus 3 times the square root of 6
and that's multiplied times 4 times the square root of 2, I distributed.
In the denominator, we have a squared minus b squared.
So, 5 rad 3 quantity squared minus 4 rad 2 quantity squared.
In the numerator, we have on the outside 3 times 5 which gives us 15,
on the inside we have 6 times three which is 18.
For the next term in the numerator, we have 3 times 4 which is 12.
Inside the radical we have 6 times 2 which is also 12.
In the denominator 5 squared is 25.
The square root of 3 times the square root of 3 is 3.
Minus 4 squared is 16.
The square root of 2 times the square root of 2 is 2.
No more radical in the denominator.
Eighteen is 9 times 2.
Twelve is 4 times 3, because we have to get perfect squares out of those square roots.
So, we have 15 times 3, square root of 2 plus 12 times 2 square root of 3.
All over 43.
So, we have 45 square root of 2 plus 24 times the square root of 3.
All over 43.
Can we factor anything out?
Yes, we can factor out a 3.
What are we left with?
Well, 15 times the square root of 2 plus 8 times the square root of 3 and that's over 43.
Nothing cancels, so we're done.
We once again have two terms in the denominator with at least one square root,
so we multiply by the conjugate of the denominator, over itself.
In our numerator, we are doing F.O.I.L. So,
the first terms times each other would be the square root
of x times the square root of x or squared.
The outer terms would be 10 times the square root
of x. The inner terms would be 1 times the square root of x
and the last terms would be 1 times 10.
In the denominator, we have a squared minus b squared.
So, we have the square root of x squared minus 10 squared.
The square root of x squared those cancel
and you just x. Ten rad x plus 1 rad x is 11 rad x. We added like terms.
There aren't any other like terms in the numerator.
In the denominator, we have x minus 100.
Nothing factors, so we can't cancel anything and we're done.
Pause the video and try these problems.