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>> Hi, this is Julie Harland
and I'm YourMathGal.
Please visit my website
at yourmathgal.com
where you could search for any
of my videos organized
by topic.
This is part 7
of solving an equation
containing a radical.
And notice in this equation
which we're going
to be solving,
there are two radicals.
So, the technique will be
to isolate one radical
on one side of the equation
and square both sides
to get started.
So here's our problem,
and it doesn't matter
which when we isolate.
We could keep it the way it is
which means the right side is
already isolated and begin
by squaring both sides,
or you could subtract one
from both sides
and square both sides.
Let's go ahead
and just leave it as it is
to get started.
So, how about if I just square
the entire left side
and I square the entire
right side?
Now, you have
to be careful here.
Remember, when I'm squaring
the left side,
I have a binomials.
So I'm going to write
out all these steps.
It's really, the square root
of negative 3X plus 16 plus 1
times the square root
of negative 3X plus 16 plus 1.
And on the right side,
when I square a square root,
I just get what's underneath
the square root sign.
So the right side will just
end up being the easy side
negative 4X plus 25.
That's what I'll get
when I square the right side.
Okay, but on the left side,
if I do the FOIL method here,
I will get first of all
and squaring the first terms
together which those are the
same thing.
So I get again what's
underneath the square
root sign.
So I'll simply get this square
root, let me see.
When I square the first term
in each, I'll get exactly
what's underneath the square
root, which will be the
negative 3X plus 16.
Now, the outer term will be a
plus square root
of negative 3X plus 16
and the inner term will
be another.
Square root
of negative 3X plus 16
and the last term will be a 1.
Now, you could use the formula
for squaring a binomial
and you will get exactly the
same thing.
You would have the negative 3X
plus 16, that would be
squaring the first term,
plus 2 times the first term
times the second term
which is 2, square root
of negative 3X plus 16 plus 1
which is the same
as just adding these two liked
terms here at this square root
of negative 3X plus 16 plus
square root
of negative 3X plus 16.
That's where these two square
roots of negative 3X plus 16
comes from.
[ Pause ]
Right, so I squared one side
of the equation
and I could simplify this
left-hand side a little bit
over, that's up to you
if you want to do that step
by adding the 16 and 17.
[ Pause ]
All right,
but the important thing
to note is I still have a
square root, so now I have
to isolate this square root.
So, I want to add 3X minus 17
to both sides, right?
If I add 3X minus 17
to both sides, I'll be--
I would isolate the left-hand
side square root
which is the two square roots
of negative 3X plus 16.
And on the right side,
that gives me a negative X
and a plus 8.
And by the way,
you could write
that as 8 minus X,
that's up to you.
Now, I haven't completely
isolated the left-hand side.
I have a 2 in front of it,
that's okay.
You're basically wanted
to isolate the term
that has a square root.
And then we're going
to do the process all
over again.
We will square both sides,
okay?
Okay, so on the left-hand
side, notice you have
to do the 2 squared
and the square root
of negative 3X plus
16 squared.
So I'm going to write
out all the steps just
so you can see that.
I know it takes a
little longer.
But in other words,
I'm going to do the 2 squared
and also the negative 3X plus
16 squared.
On the right-hand side,
I'm going to write
that as negative X plus 8
times negative X plus 8.
You don't have to write
out all these steps,
but I'm doing it for those
of you who would
like to see every single
step shown.
So this gets 2 squared is 4,
and when I squared the square
root, I get what's underneath
the square root
which is negative 3X plus 16.
And now, I'm going
to do the FOIL method
over here but of course you
can't just use the formula
for squaring a binomial.
So negative X times negative X
is X squared,
negative X times 8 is a
negative 8X.
The inner terms will be a
negative 8X and have plus 64
of the last terms.
And I have
to do the distributed property
on the left,
so I have negative 12X
plus 64.
And on the right,
I'm just going
to add the liked terms
at these two middle terms.
[ Pause ]
Okay, we're getting close.
I've gotten rid
of the radical signs.
I have an X squared term.
That means I have a quadratic
and when you have a quadratic,
you set the whole equation
equal to 0.
So since the X squared is
on the right,
how about we just add a 12X
minus 64 to both sides?
So I'm going
to add this at 12X.
Add the 12X and minus 64
to both sides will give me 0
equals X squared minus 4X,
all right because the 64 minus
64 is 0.
Oops, I didn't mean
to go that far.
Okay, here we are.
We're almost done.
Now we just factor the
right-hand side
and set each factor equal
to 0.
So either X equals 0
or I could say 0 equals X,
or 0 equals X minus 4
and of course,
0 equals X means X equals 0.
And here, if I add 4
to both sides,
I get 4 equals X
which of course means X
equals 4.
So it looks
like I have two solutions,
0 and 4.
Now the only way
to make sure that's correct is
by plugging 0
into the original equation
and making sure both sides are
the same when you simplify,
and then plugging in 4.
I'm going to go ahead
and do this.
Now I'm going
to make one more video,
solving the exact same
problem, isolating the other
square root
because you should still get
the same answers.
So we'll just go ahead
and check now,
but if you would
like to see this problem
solved by isolating the other
square root
for instance going back
up here.
If I want to subtract 1
from both sides
and then square both sides,
in a certain way,
it's a little bit easier
that problem.
I did it that way as well
on my own first.
You can see how that works.
All right, but for now,
let's go ahead
and take this
original problem.
We're going
to check the solutions
of 0 and 4.
All right, so we're going
to check X equals 0
so I write the original
equation and I'm going to plug
in 0 for X here
which you could see I'd put
in the 0 in red
and now let's simplify the
left-hand side.
So underneath the square root,
negative 3 times 0 is 0.
So 0 plus 16
and to the square root plus 1,
and that then is
of course the square of 16
which is 4.
So I get 5
on the left-hand side.
And on the right-hand side,
negative 4 times 0 is 0
so I have 0 plus 25 underneath
the square root
and that's a square root of 25
which is 5.
So, X equals 0 is a solution.
All right, that checks, right?
So now let's go ahead
and check X equals 4,
right here.
Again, my first set is
to write the equation and plug
in 4 in place of the Xs.
So on the left-hand side,
that gives me negative 12 plus
16 which is the square root
of 4 plus 1,
and the square root of 4 is 2
so that's 2 plus 1
and 2 plus 1 is 3.
So left side simplifies to 3
and let's see, right side.
Put the square root
of negative 16 plus 25
which is a square root of 9
and the square root of 9 is 3,
so X equals 4 also checks,
right?
They both check.
So that means it went away
back up here, when I solved it
but that is indeed the
correct solution.
0 and 4 is the solution.
So back up here,
original problem,
the answer-- oops--
[ Pause ]
-- for solution,
better way of putting
that is 0 and 4.
And if you'd
like to see this problem,
then again,
where you subtract one
from both sides
and instead you're isolating
this square root
of negative 3X plus 16,
then look at my video
of solving an equation
containing a radical, part 8.
The arithmetic is slightly
easier, the
out was slightly easier
when I do happened
to isolate that,
not much difference though.
All right, try this again all
on your own.
[ Silence ]
Please visit my website
at yourmathgal.com
where you can view all
of my videos
which are organized by topic.