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Hi, class.
Today we're going to take a look at simplifying radicals.
Now in order to simplify radicals,
we're going to need some special numbers.
These are all the values
that are the answers to these problems over here
that I have on the right-hand side.
All of the red numbers here are special numbers.
These are numbers that we call 'perfect squares.'
Now, there's a reason why we call those 'perfect squares.'
You want to take a look at a square.
And a square has all sides equal.
So, if I have one side that is length x,
the other side is also length x.
And if I wanted to find the area of this square,
I would do length times width [ Area = L x W ]
which would be x times x or x squared. [ x · x = x² ]
So, relating to this square here,
the values that we have over here in red
are called perfect squares,
because those are the areas of all the squares
with nice whole number length sides up to 20.
So a square with sides both 1 would have an area of 1.
A square with both sides of 2 would have an area of 4.
A square with both sides 3 would have an area of 9 and so on,
up until we get to a square with sides 20
which would have an area of 400.
There are obviously squares with whole number length sides
that are larger than this,
but normally if you know these values up until 20,
you're doing pretty good.
Now, if you're on a test and you forget these values here,
you can go down the margin of your test
and just list all of them
and use your calculator to generate all of them
by just plugging in 1 squared,[ 1² ] 2 squared,[ 2² ] and so on
until you get to 20 squared. [ 20² ]
Then you're pretty much covered
for all the numbers you might need.
Now, let's take a look at how these relate to our topic today
of simplifying radicals.
We need to know what a radical is.
The radicals we're going to be looking at are square roots.
Square roots have this symbol here. [ √ ]
It's similar to a division symbol, but it's not.
You'll notice it has a little v out front here.
This is a 'square root' symbol. [ √ ] It's also called a 'radical.'
If we go to something where we're not looking for a square root,
but a cube root [ ∛ ] or a fourth root, [ ∜ ]
inside this v you'll see a little number.
But right now, we're just looking at square roots,
so there's no number in there.
There's an implied 2 there when there's no number.
When you go on to your next class,
we'll take a look at what happens
if you have something other than a square root.
Now, notice that this says square root of A. [ √A ]
The symbol here [ √ ] is called a radical.
The A is what we call a 'radicand.'
And that's what you're trying to find the square root of.
This particular symbol is asking you a specific question.
It's asking you, "What is the length of the side of the square
with area of this value underneath the symbol√"
So, for instance, if I had square root of 25, [ √25 ]
I'd be looking for the length of the side of the square
with area of 25.
Notice these numbers are handy,
because if I go down them right here, 25 is the area
and the side that we squared to get that was 5,
so my answer is 5. [ √25 = 5 ]
Now, there is often on your calculator
a button that you'll find that looks like this.
That is your square root button, and you can push in 25
and that button and equals
and it will give you an answer of 5.
Some calculators have you plug it in this way,
some calculators want you to push the button,
and then put in your 25 in order to give your 5.
So check your calculator
and see which way it likes you to plug those values in.
Now, if you don't have a value that is over here on this list,
then you will get a decimal approximation
if you use your calculator.
And we're actually only use the decimal approximation
if it's asked for.
If it's not asked for
or you're not doing a story problem where it makes sense,
then you're going to want to use an exact value
which is going to require us to know how to simplify things.
So, if you're not given something like square root of 256
where you're able to look on this list and say,
oh, there's 256 so my answer is 16, [ √256 = 16 ]
then you're going to need a few tools.
Now let's take a look at our first rule for square roots
so we can have some tools
that will help us with simplifying these.
Our first rule that we're going to take a look at
is the rule for multiplying.
When you are multiplying two square roots,
like the square root of a times square root of b, [ √a √b ]
what you want to do is just multiply the values underneath it
and leave the square root symbol over the top of them.
[ √a·√b = √ab ] So, for instance,
if I had square root of 3 times square root of 5,
that would be square root of 15. [ √3·√5 = √15 ]
Now, what we're going to do is
we're going to actually use this rule in reverse
when we simplify radicals.
So, for instance, let's say that you were told
What's the square root of 72? And you had to figure it out.
Well, the first thing you want to do
is go over here to your list and try to find 72.
And it's not on there.
But it's somewhere between 8 and 9.
And if you used your calculator
you'd get a decimal approximation that's in between 8 and 9.
But we don't want a decimal approximation.
We want an exact value.
So we need to simplify square root of 72.
To simplify square root of 72,
we need to make sure that there isn't any value on this list
that divides evenly into 72.
So, we don't want just any value.
For instance, many of you might be saying,
Well, 9 goes into 72. I can use 9.
We want the largest one on that list.
So what we're going to do is we're going to take that 72
and in order to find the largest value on here,
we're going to start by dividing it by very small numbers
and systematically go up
until we get one of these numbers on this list.
So, for instance,
2 is the smallest thing that will go into 72 evenly.
So 72 divided by 2 is 36. [ 72/2 = 36 ]
We check this list to see if 36 is on the list, and it is.
It's right here.
So that means we're going to use that 36
to split this 72 up into two things that can be multiplied
so we can start at this spot in our rule and move backwards.
So, this square root of 72 [ √72 ]
is going to turn into square root of 36 times 2 [ √72 = √36 · 2 ]
because we divided 72 by 2 to get 36. [ √72 = √36 · 2 ]
I like to always put my perfect square first.
That's just a preference.
But it does help in keeping things organized.
Now what we're going to do is
since we now have square root of 36 times 2,
we're going to go backwards with our rule
and separate those 2 square roots up. [ √ab = √a·√b ]
So we have square root of 36 times square root of 2. [ √36 · √2 ]
Now, we cannot simplify square root of 2. [ √2 ]
It's going to be a decimal.
But we can simplify square root of 36. [ √36 ]
So we're going to do that.
And square root of 36 is 6. [ √36 = 6 ]
So, simplified, square root of 72 [ √72 ]
is 6 square root of 2. [ √72 = 6√2 ]
Now, keep in mind, once you have done the square root,
it's just like an operation like multiply or divide.
Once you've done it, the symbol goes away.
So, since we did the square root of 36, we just write down 6.
There's no root left over the top of it.
That's a big mistake that a lot of people tend to make
and you want to be careful not to do that.
Let's do one more of these,
and then we'll check what we do if you have variables involved.
So how about if we have [ √150 ] square root of 150.
Well, let's look on our list and see if 150 is there.
And it's not.
It's going to be in between 12 and 13.
So, we first divide 150 by 2 and you get 75. [ 150/2 = 75 ]
And 75's not on there.
So then we divide it by 3, which is 50. [ 150/3 = 50 ]
And 50's not on there.
49 is, which is close, but it's not 50.
So then we divide 150 by 5. [ 150/5 ]
Notice that I skipped 4?
The reason why I skipped 4
is because 4 is a very small value on our list,
and we want the largest.
So, if 4 comes up later, that's okay,
but we don't want to start 4.
So I start by dividing by 2, then 3, then I go to 5.
And 5 goes into 150, 30 times. [ 150/5 = 30 ]
And that 30 is not on here either.
So I keep going
and I divide 150 by 6 and I get 25 [ 150/6 = 25 ]
and bingo. I've got a value on my list.
So that means 150 is 6 times 25 [ 150 = 6 · 25 ]
So I'm going to break it up
into square root of 25 times square root of 6, [ √25·6 ]
which we will separate into 2 roots. [ √25·√6 ]
And then I can do the square root of 25. It is 5. [ √25 = 5 ]
So I have 5, square root of 6. [ = 5√6 ]
And that is your answer.
Now, let's take a look at what happens if you have a variable.
If I have square root of x squared, [ √x² ]
I need to look for what times itself is x squared.
And that would be x. [ x² = x · x ]
If I have square root of x to the fourth, [ √x⁴ ]
I need to look at what times itself is x to the 4th.
And that's x to the second power. [ √x⁴ = x² ]
Notice that there's an invisible one up here [ x² = x¹ · x¹ ]
for our exponent√ [ √x² = x¹ ]
And that's going to come in handy
with the little trick we're going to come up with.
And square root of x to the 6th is x to the third. [ √x⁶ = x³ ]
Square root of x to the 8th is x to the 4th. [ √x⁸ = x⁴ ]
Notice that each time,
our exponent that is underneath the radical
is being divided by 2 to get what is out here [ the exponent ].
And there's a reason why.
Is square root is really like a one half power. [ √ = x¹′² ]
And half of the value under here [ 1/2 · A ] is what's over here.
Now, that works really nice if we have even exponents.
If we have an odd exponent, we need to see what happens.
Well, if you have an odd exponent,
you can do something similar to what we did with our numbers.
Let's say you have square root of x to the 7th. [ √x⁷ ]
We can separate x to the 7th
into x to the 6th times x. [ √x⁷ = √x⁶ · x ]
Which will make this x to the 6, [ √x⁷ = √x⁶ · x¹ ]
the square root of that times the square root of x. [ = √x⁶ √x ]
And square root of x to the 6th is x cubed. [ √x⁶ = x³
So it's x cubed times x for an answer. [ = x³√x ]
Now, over here, we were dividing all of these by 2
to get what's here.
Over here, if you look at what we have outside,
we have an exponent of 3 and inside we have an exponent of 1,
because there's an invisible 1.
If we divide 7 by 2 [ 7 ÷ 2 ]
and think of it as old fashioned division from grade school,
7 divided by 2 would be 3 with a remainder 1. [ 7 ÷ 2 = 3 r1 ]
The 3 is what's outside.[ 3√x ] The 1 is what's inside.
And that will always work.
It's a nice little trick for these.
And when you get on to the next class
and look at rational exponents,
you'll be able to see why that actually works.
And it does work for all of the radicals.
So let's take a look at one more of these.
What if I have square root of x to the 25th√ [ √x²⁵ ]
If we use our little trick we did
and divide 25 by 2, [ 25 ÷ 2 ]
25 divided by 2 is 12 [ 25 ÷ 2 = 12 r1 ]
with a remainder of 1 inside. [ √x²⁵ = 12x√x ]
Now let's combine our numbers and our letters together.
What if we have
square root of 108x to the 6th, y to the 9th. [ √108x⁶y⁹ ]
When I have numbers and variables all together in a radical,
what I like to do is just take care of the number first
and do the variables at the very end.
The reason why I do that
is because the variables we have our nice little short cut for,
and the numbers actually take a little bit more thought.
So, let's start with our number.
And if we look at our 108 here
and divide it by 2, we get 54. [ 108/2 = 54 ]
And notice that 54 is not on our list over here.
So divide your 108 by 3 [ 108/3 = 36] and you get 36.
And 36 is on here.
So, let's break our 108 up
into square root of 36 times 3. [ √36 · 3 ]
And we're going to let the variables [ √36 · 3x⁶y⁹ ]
just continue on with us for the time being.
And then we're going to separate this up
into the square root of 36 times square root of 3,
x to the 6th, y to the 9th. [ = √36 √3x⁶y⁹ ]
And then we're going to do
square root of 36 and get 6. [ √36 = 6 ]
And I'm going to leave a little space
between that and my square root of 3.
The reason why
is because now I'm going to take care of my variables.
So, I'm going to look at my x and my x has an exponent of 6. [ x⁶ ]
6 divided by 2 is 3 [ 6/2 = 3 r 0 ] with no remainder.
So that means I'm going to have an x cubed out here [ 6x³√3y⁹ ]
and there won't be any x's inside my radical.
So now I look at the y to the 9th. [ y⁹ ]
9 divided by 2 is 4 with a remainder of 1, [ 9/2 = 4 r 1 ]
so there's y to the 4th out front, [ y⁴√ ]
and a y to the first inside. [ = 6x³y⁴√3y ]
And this would be our simplified form [ = 6x³y⁴√3y ]
of this particular problem.
Now, that takes care of simplifying using our multiplying rule.
We need to take a look at also what happens
if we're adding these.
So let's take a look at adding our radicals.
Let's say you have 2 square root of 3 [ 2√3 ]
plus 4 square root of 3. [ 2√3 + 4√3 ]
This is very similar to 2x plus 4x. [ 2x + 4x ]
When you did something like this,
you just added your values in front.
So 2 plus 4 was 6x. [ 2x + 4x = 6x ]
The same thing's going to happen here.
The key is your radicals have to match.
So if one is a square root of 3, [ √3 ]
the other one has to be a square root of 3 [ √3 ]
or you cannot combine them.
Since they are both [ √3 ] square root of 3's, [ 2√3 + 4√3 ]
we can combine them
to get a 6 square root of 3 [ 2√3 + 4√3 = 6√3 ]
However, if we had a 2 square root of 3 [2√3 ]
plus a 4 square root of 5, [ 2√3 + 4√5 ] this would be finished.
We would not be able to combine anything
because the radicals do not match. [ 2√3 + 4 √5 ]
Now, sometimes, your radicals do not match,
but they're kind of disguised.
And if we simplify, they might actually match.
For instance, if I have [ √50 ] square root of 50,
and I'm adding it to square root of 98. [ √50 + √98 ]
At first glance, these do not look like you can add them.
However, they're not simplified.
So, if we look at square root of 50, [ √50 ]
half of 50 is 25. [ 50/2 = 25 ]
So 50 divided by 2 is 25 [ 50/2 = 25 ]
and 25 is a perfect square.
So the square root of 50
can be simplified into 25 times 2 [ √25 · 2 ]
which will break up into square root of 25
times square root of 2. [ √25 √2 ]
And the square root of 25 is 5. [ √25 = 5 ]
So this is really 5 square root of 2 [ 5√2 ] in disguise.
Now let's look at our square root of 98.
98 divided by 2 is 49. [ 98/2 = 49 ]
And 49 is a perfect square.
So this can be split up into 49 times 2. [ √49 · 2 ]
Which will in turn turn into square root of 49
times the square root of 2. [ √49 √2 ]
And the square root of 49 is 7, [ √49 = 7 ]
so this 7 square root of 2. [ √98 = 7√2 ]
Notice that once we simplify,
these actually do match. [ 5√2, 7√2 ]
And so we can actually combine them,
adding the 5 and the 7, [ 5√2 + 7√2 ]
we get 12 square root of 2. [ 5√2 + 7√2 = 12 √2 ]
So be careful to make sure that you simplify
using what we just learned before you add,
because you might have things
that look like they can't be added, but yet they actually can.
Now there's one other simplifying thing
that we need to take care of.
And that one other simplifying thing
that needs to be taken care of is What if you have a fraction?
For instance, let's say we have
a square root of 4 over 9. [ √4/9 ]
When you have a fraction like this,
you're going to use your dividing rule.
Now, the dividing rule, sometimes we will use one direction,
and sometimes we will use it the other.
I'll show you what that means.
When you're dividing radicals,
what you're going to do is basically simplify it
into the square root of the fraction. [ √a/√b = √a/b ]
Now, this rule will go this direction [ √a/√b to √a/b ]
if something happens to be the case.
You go this direction if a over b can be reduced,
because that will make your problem much easier.
You use your rule going the other direction [ √a/b to √a/√b ]
if a over b cannot be reduced.
The reason why is because
you don't want to leave a fraction underneath a radical.
So, if we look at this problem right here,
I cannot reduce 4 over 9. [ √4/9 ]
Since I can't reduce it,
I need to use my rule going this direction [ √a/b to √a/√b ]
and split this up.
So I'm going to split it up
into square root of 4 over square root of 9. [ √4/9 = √4/√9 ]
And I can actually do square root of 4. It's 2. [ √4 = 2 ]
And I can do square root of 9, which is 3. [ √9 = 3 ]
And that ends up
simplifying this problem for me. [ √4√9 = 2/3 ]
Now, let's take a look at another one.
What if I have square root of,
say, 55 over square root of 11√ [ √55/√11 ]
Well, I can reduce 55 over 11.
So I actually want to use this rule
going this direction [ √a/√b to √a/b ]
and put the fraction back underneath the radical.
Because now I can do 55 divided by 11 [ √55 / √11 = √55/11 ]
and get square root of 5. [ √55/11 = √5 ]
Now, let's take a look at one that's a little tricky.
What if I have
square root of 5 over 3√ [ √5/3 ]
Now, at first glance, I cannot reduce 5 over 3.
So you would say, oh, let's just split this up
into square root of 5 over square root of 3. [ √5/3 = √5/√3 ]
Well, that's nice but I can't do anything with square root of 5
and I can't do anything with square root of 3.
And since I can't do anything with them, I need to also recognize that
we don't really like having square roots
in the denominator of a fraction.
So that square root of 3 is not okay to have there.
So, what we're going to do
is a process that we call 'rationalizing the denominator.'
So, what we're going to do
is we're going to make the denominator a rational number
by multiplying it by something
that would make the root on the bottom a perfect square.
And the smallest perfect square 3 goes into is 9.
So if I multiply the bottom by square root of 3,
that will give me my square root of 9
which will make it so I can rationalize that
and turn it into a whole number.
But if multiply the bottom by square root of 3,
I need to multiply the top by square root of 3.
So, in doing that,
I get square root of 15 over square root of 9.[ = √15/√9 ]
And now I can actually square root the 9.
The numerator stays as square root of 15
and our denominator now is 3. [ = √15/3 ]
And that is a radical
with a denominator that is a rational number.
And that's what your goal is going to be is to make that happen.
So let's take a look at another one.
Let's say that we have
square root of the fraction 12 over 4x. [ √12/4x ]
Now, first thing I notice is
I can actually reduce that fraction.
So let's go ahead and do that.
4 goes into 4 once [ 4/4 =1 ] and into 12, 3 times. [ 12/4 = 3 ]
So this turns into square root of 3 over x. [ [ √12/4x = √3/x ]
So we were able to reduce it;
however we're not left with something that's not a fraction underneath
that square root.
So we can't have a fraction under there,
so let's split this up.
It's going to be square root of 3 over square root of x. [ √3/√x ]
Now we have another problem.
We have a square root in the bottom, and that's not okay.
So we need to make this square root of x here
something that we can actually take the square root of.
So how about if we multiply the top and bottom
by square root of x. [ = √3√x /√x√x ]
What we end up with is
square root of 3x [ √3x/√x² ] over square root of x squared.
And I can actually do the square root of x squared.
It's x. [ √x² = x ]
So we have square root of 3x over x. [ = √3x/x ]
This process where you're multiplying the top and the bottom
by whatever the square root on the bottom is
is called ‘rationalizing the denominator.’
And for square roots it does normally end up
being whatever's in the denominator.
As long as that is simplified.
It's normally whatever the radical is in the denominator. �