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Hello and welcome to Bay College's Intermediate Algebra
online lecture series.
Today we're going to talk about Section 6.4, which is
solving radical equations, equations
that contain radicals.
If we consider this example here, this is an equation, the
square root of x equals 4.
Some of you may be able to look that and say intuitively,
well, I know the square root of 16 is 4, but let's
actually see why.
What we can do to solve this, the property we can use, is a
property we've essentially been using for a long time.
What we do to one side of the equation, we
must do to the other.
The properties of equality.
So how do I get rid of the square root?
Well, I can essentially just square it.
But what I do to one side, I need to do to the other--
property of equality.
If I square a square root, I get what's under the radicand,
just the x.
And here, 4 squared is 16.
So we can check that.
Is the square root of 16 equal to 4?
Yes it is.
The square root of 16 is 4.
So this is our solution, because we checked our answer.
Essentially, if we have any equation, this side equal to
that side, if we raise both sides to the same power, it is
still a similar equation.
It will contain the solutions that we're looking for.
But it may also contain something
called extraneous solution.
It does not create an equivalent equation, but it
does create an equation that has the set of numbers that
contain the solution.
It may have additional ones.
That's why it's important to always check your work, no
what you're doing, in a math class.
Let's see an example here where we end up with an
equation that may have extraneous solutions.
We have x equals 8.
We're going to do a little bit of circular math here and come
back to this.
We can say, if x is 8, it is what it is, there's
nothing to do there.
We're going do some math to it and then undo it.
What happens if I square both sides?
I get x squared equals 64.
Now because I raised both sides to a power, it is not an
equivalent equation.
It's not the same as this anymore.
I did something different.
If I take this equation, and I go to solve it, I'm going to
take the square root of both sides.
To get rid something squared, you take the square root.
But when you introduce a square root, you have to
always remember, plus or minus.
There are two possible things that will give
me a positive 64.
If we take the square root of something squared, we just get
what's under the radicand, and we have plus or minus 8.
Now, notice we actually have two solutions, positive 8 and
negative 8.
Which one of these is the correct solution?
Just the positive one.
Positive 8 is positive 8, true statement.
Negative 8 is not the same.
This negative 8 is called an extraneous solution.
That's why working when with radicals and solving equations
that containing radicals, it's so important
to check your work.
So let's look at these steps here.
This is essentially what we want to do when solving
radical equations.
The first step is, isolate the radical to
one side of the equation.
Get it by itself.
Use your properties of equality, what you do to one
side, you do to the other, to get that radical by itself.
Then raise both sides to the index.
That means, this is a square root, so I squared both sides
of the equation.
So you look at the index.
Maybe it's a cube root, or a fifth power, or something of
that nature.
Then we go ahead and solve it as we would normally to get an
equation that contains the solutions and possible
extraneous solutions, and that's why we have to check
those, to make sure which ones work in the original equation.
Let's eliminate those extraneous solutions.
So let's look at some examples here.
Here I have the square root of 2x minus 3, minus 2 equals 1.
Step one, isolate that radical.
I'm going to add to both sides.
So I get 2x minus 3 equals 3.
Add 2 to both sides.
2 and 1 is 3.
Now the radical is isolated to one side of the equation.
I can now raise both sides to the index.
The index is 2, so I'm going to raise both sides to the
power of 2.
What I do to one side, I do to the other.
When you square a square root, you essentially get what's
under the radical, and 3 squared is 9.
Now this is a problem that I can go ahead and solve.
See, what do I get?
Add 3 to both sides.
And divide by 2.
x equals 6.
Check your answer.
Especially when it comes to radicals.
Let's see in the original equation if
that's a true statement.
2 times 6 is 12.
12 minus 3 is 9.
The square root of 9 is 3.
3 minus 2 is in fact 1.
So this does work.
I know it's the solution.
I've checked it.
I
Next example, the cube root-- a little bit different--
but our steps are still going to be the same.
I want to isolate the radical, so I'm going to
add 3 to both sides.
And by doing that, I get the cube root of x
minus 2 equals 3.
Now I'm going to raise both sides to the power of the
index, which is 3.
I'm going to cube both sides.
What I do to one side, I do to the other.
So cubing a cube root, again, just gives me
what's under the radical.
That's our tool to eliminate that radical.
And 3 cubed is 27.
And now I can solve it.
Add 2 to both sides.
x equals 29.
Let's check our work, make sure we got it.
29 minus 2 is 27.
The cube root of 27 is 3.
3 minus 3 is 0.
That's a true statement.
I checked my work.
I can't be wrong, which is a nice feeling.
Let's look at this one.
The square root of x minus 8 plus 2--
I'm going to subtract 2 from both sides.
And I'm going to stop right here.
Because what I know about square roots is
they're always positive.
Hopefully we recall that from the very first section when we
talked about radicals.
The square root of a radical has to be positive.
There's no value that times itself
is going to be negative.
Because a negative times a negative is a positive.
So there's no value that's going to give me a negative.
The answer to this is, there is no answer.
It's no solution.
Or we can just write our null set.
If we tried to continue on, and this is why it's so
important-- if we didn't recognize this and jumped to
no solution-- if I square both sides, let me just do it right
here, I get x minus 8 equals positive 4, because negative 2
squared is a positive 4.
And I could subtract 8 from both sides, and I'd get
negative 4.
Of, x is negative 4.
That's my answer .
No.
If we checked that, negative 4 minus 8 is negative 12.
Right?
Oh, I'm sorry I went the wrong way here.
If I add 8-- sign error.
We're all guilty of that one.
If I add 8, we get 12.
So if I put 12 in there, 12 minus 8 is 4.
The square root of 4 is 2.
A positive 2 plus 2 Is not equal to 0.
So this does not make a true statement.
So we can see, if we worked it out, sign error or not, we're
not going to get a true statement either way.
What if we have additional radicals?
What if we have more than one radical in the equation?
Don't panic.
Just take your time.
Essentially, follow those steps.
We want to isolate a radical.
Just pick one and isolate it.
And then go through the steps.
If there's still a radical that remains, start over.
Let's isolate that radical and work through it. u
So this problem is going to be a little bit longer and
tedious, but we're going to work through it.
I'm going to isolate this radical.
Essentially, add this value to both sides.
So I'm going to get the square root of x plus 7 equals 2 plus
the square root of x minus 1.
So I just added it to both sides.
Now I can square both sides.
Here my index is 2.
And hopefully we'll recall from the previous section how
to deal with something like this.
Treat it just like a binomial.
We're going to multiply it by itself and do a
little bit of FOILing.
Here, if you square the square root, you just get that value.
This is not equal to 4 plus x minus 1.
We have to actually FOIL it out.
It's this value times itself.
So if we do FOIL that out, we're going to get 4 plus 4
square root of x minus 1-- that's my inner and outer
terms combined--
plus the square root of x minus 1.
Oh, we're squaring that last term.
Sorry about that.
And we can see, there is some like terms we can deal with.
So we'll work through that.
We have an x on both sides.
They cancel out.
If I subtract x from both sides, it's gone.
And I have 4 minus 1 is 3.
I'm going to subtract 3 from both sides.
And if we see that, we get 4 equals 4 square
root of x minus 1.
Just by doing that combining on like terms, we can see,
hey, guess what?
I still have a radical.
But one nice thing, it's on one side all by itself.
Maybe I can do something with this 4.
I don't have to, but I can.
I can say, hey, these are both divisible by 4.
Let's divide that out and we get 1 equals the square root
of x minus 1.
If I square both sides here, 1 squared is 1.
This squared is x minus 1.
Add 1 to both sides.
And I get 2 equals x, or, if you prefer, x equals 2.
The order here doesn't matter.
Is this a true statement?
Does it work?
Is this the value I was looking for?
Well let's go back and check it.
But before we do, let's just review.
We isolated a radical, raised both sides by the index, then
we did what we had to do here.
We combined like terms to get the radical by itself because
it still contains a radical.
And then we started over.
Let's isolate the radical, let's raise it to the index
and solve like we normally work.
Now we're at the point where we can actually check it.
If I plug 2 in here, 2 plus 7 is 9.
The square root of 9 is 3.
I'll write it up there If you plug 2 into here, we get 2
minus 1 is 1.
The square root of 1 is 1.
So we have 3 minus 1 equals 2.
Well that's pretty true.
It doesn't get much more truer than that.
3 minus 1 is 2.
So that's true.
So I checked my answer and I know x is 2.
This one's a good problem because it reviews what we've
talked about in previous sections.
When we have fractional exponents, that's the same
thing as a radical.
So if we want, we could rewrite this.
This is just saying the cube root of x. x to the one third
power is the cube root of x--
minus 2 equals 4.
So that's good review from a previous section.
Make sure you remember, fractional exponents are
nothing more than radicals.
x to the first power, the numerator, the index of the
radical is in the denominator.
And now we can use our process before.
We're going to isolate the radical by
adding 2 to both sides.
Then we can cube on both sides to raise it to the
power of the index.
And 216, I believe.
Is that right?
I think so.
All right.
Well maybe I made a math error, maybe I didn't.
Let's check our answer.
Well if we take the cubed root, or 216 to the one third
power, do we get that?
Does this factor down to three factors of 6?
Well let's find out right here.
Does this factor down to three factors of 6?
Well 6 goes into 21 three times with a remainder of 3.
And 6 goes into 36 6 times.
So yeah, 36 and 3, and we know that this is two factors of 6,
so sure enough, it's 216.
So the cubed root of 216 is 6--
6 minus 2-- is 4.
So we checked it.
We know that this works.
Kind of a weird way to check it, but that's sometimes the
way I do my math.
That's how we work with equations
dealing with radicals.
Just to review, we isolate a radical, we raise both sides
to the index, we solve, and then we check our work.
If a radical remains after we solve--
or, excuse me, after we've raised both sides to the
index, re-isolate it.
Raise it to both powers, solve it, and then check the answer
in the original equation.
So that's how it's done for Section 6.4, equations
containing radicals.
Thank you.