Two more examples of solving rational equations

⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.

Two more examples of solving rational equations

相關課程

0 / 750

Let's solve some more rational equations.
So let's say I have x plus 1/x is equal to 2.
Once again, we have an x in a denominator over here, it
looks kind of bizarre.
But you just have to remember, if we multiply every term in
this equation by this denominator, right here, by
this x, right there, then essentially, you'll have x
times 1/x and the x will disappear out of the
denominator.
So let's do that.
So we have to multiply every term by that, you can't just
multiply one term.
So multiply this term times x, this term times x,
and the 2 times x.
We're just multiplying both sides of this equation by this
denominator.
So what do we get?
We get x times x is x squared plus x times 1/x is just 1.
x's cancel out.
And then that's going to be equal to 2x.
And this is starting to look like our
standard quadratic equations.
We just have to subtract 2x from both sides.
And if we do that-- so let me subtract 2x from there,
subtract 2x from there-- the left-hand side of our equation
becomes x squared minus 2x plus 1.
And the right-hand side is equal to 0.
And we know how to solve an equation like this.
We could either factor or use the quadratic formula.
This actually seems pretty straightforward to factor.
Negative 1 times negative 1 is 1.
Negative 1 plus negative 1 is negative 2.
So this is x minus 1 times x minus 1 is equal to 0.
And we know, if two numbers, when you multiply them, they
equal 0, that means one or both of those numbers have to
be equal to 0.
In this situation, the number is the same, right?
This is x minus 1 squared is equal to 0, right?
So we know that x minus 1 has to be equal to 0.
Add 1 to both sides of this equation and we get
x is equal to 1.
And you can verify it.
Verify that this works.
1 plus 1/1 does, indeed, equal 2.
So it definitely does work.
Let's do a more involved one now, now that we're warmed up.
Let's say I have 2 over x squared plus 4x plus 3 is
equal to 2 plus x minus 2 over x minus 3.
So just like we've done in the last problem and in all of the
other ones, we want to get all of these x's out of the
denominator.
And the best way to do it is to multiply every term in this
equation by-- you can view it as the least common multiple
of the denominator.
So let me show you what I'm talking about.
The first thing we might want to do is factor this, to see
if this guy and this guy have anything in common.
So let's see if we can factor it.
What two numbers when you multiply them equal 3 and when
you add them equal 4?
Well, 3 and 1, right?
So this right here, I can rewrite it as x plus 3 and--
and really this is a plus over there, too.
I wrote the original problem wrong.
That is a [? product. ?]
So this over here on the left is x plus 3 times x plus 1.
And so, what can we multiply both sides of this equation by
to get rid of the x plus 3 and an x plus 1 here, and an x
plus 3 over there?
Well, we multiply everything by x plus 3 and
then by x plus 1.
So this expression right here is actually divisible by this.
So the least common multiple of x plus 3 and x squared plus
4x plus 3 is actually x squared plus 4x plus 3,
because this is a multiple of that.
It's divisible by it.
So let's multiply both sides of the equation by this
expression, right here.
So we're going to get-- let me just rewrite it-- x plus 3
times x plus 1-- and then we have this term over here--
times 2 over x plus 3 times x plus 1 is going to be equal to
x plus 3 times x plus 1 times this 2 times that 2, over
there, plus x plus 3 times x plus 1 times this term-- times
x minus 2 over x plus 3.
Now, the whole reason why we did that is because that will
cancel with that.
The left-hand side of our equation will just be a 2.
And 2 will be equal to-- well, 2 times this-- we know what
this is equal to-- x plus 3 times x plus 1 is x squared
plus 4x plus 3.
We know that.
We actually went the other way.
We factored it.
So it's going to be 2 times x squared plus 4x plus 3, which
is 2 times x squared is-- I'm going to do it in that green
color so you know we're dealing with
this 2 right here.
So this term right here is going to be 2 times x squared.
So it's 2x squared plus 2 times 4x plus 8x plus 2
times 3 plus 6.
And now this term.
We have an x plus 3 in the numerator.
Or it would be, if I multiplied it times this
fraction or this rational expression.
That x plus 3 will cancel with that x plus 3.
So we're just left with an x plus 1 times an x minus 2.
An x plus 1 times x minus 2-- let me just focus in on this
expression right here.
I'll just box the whole thing.
x plus 1 times x minus 2 is, let's just say it's x times x,
which is x squared.
x times negative 2, which is negative 2x.
And then you have 1 times x, so plus x, and then you have 1
times negative 2, so you have minus 2.
And now we can simplify it.
So once again, the left-hand side of this
equation is still 2.
Now on the right-hand side, let's look at the x squared
terms. We have a 2x squared, and we have an x squared.
Add them together, you get 3x squared.
Then we have an 8x, and we have a minus 2x plus x.
So what's that going to be?
This is an 8x, and then when you add these together, you
just get a minus-- you just get a negative x.
So when you add everything together, you're going
to get a plus 7x.
Those are the x terms. And then finally, you have a-- no,
I already used the orange-- you have a 6 and you
have a minus 2.
So that's going to be plus 4.
And now let's subtract 2 from both sides of this equation.
Let's subtract 2 from both sides.
And we are left with 0 is equal to 3x squared plus 7x.
We are subtracting 2 from both sides, so 4 minus 2 is plus 2.
So we've stumbled, then we've turned this bizarre-looking
thing into kind of a standard, plain
vanilla quadratic equation.
We have a non-zero coefficient out here.
We might-- well, we just want to solve for it.
We actually don't even have to factor it.
So we might as well just use the quadratic formula.
That's the easiest way to solve, especially a quadratic
equation that has a non-one coefficient on
the x squared term.
So our solutions, using the quadratic formula. x is going
to be equal to negative b-- which is negative 7-- plus or
minus the square root of b squared-- that's 7 squared--
which is 49 minus 4 times a-- which is 3--
times c, which is 2.
All of that over 2 times a, over 2 times 3 is 6.
So what does this equal to?
4 times 3 is 12 times 2 is 24.
So that, right there, is 24.
What's 49 minus 24?
It's 25.
So I don't want to make that too messy, so
let me rewrite it.
So x is equal to negative 7 plus or minus the square root
of 49 minus 24 is 25 over 6, which is equal to negative 7
plus or minus 5-- the square root of 25-- over 6.
And let's see, if we were to add 5, that would be negative
7 plus 5 is negative 2 over 6, which is
equal to negative 1/3.
That is one solution.
And then if I do negative 7 minus 5, that is negative 12
over 6, which is equal to negative 2.
So those are our two solutions to this original rational
equation right up here.
Hopefully you enjoyed that.

留言:

載入中...

新增一則留言(尚未登入)

two more examples of solving rational equations 討論區:

載入中...