>> In this video we solve this
rational equation which leads
to solving a quadratic equation.
All right we have a rational
equation 3 over x plus 4
over x plus 2 equals 2.
So when you have an irrational
equation there are a couple
of ways to solve it.
One is to multiply both sides
by the least common denominator
to eliminate all the fractions.
I'm going to show you
another way to do this
which you might find easier.
What we could so is rewrite
the problem so all of the terms
on each side have the
same denominator--
the least common
denominator so watch what
that would look like-- 33 over x
and we'll leave a little space;
you've got 4 over x plus
2, leave a little space
and I'm going to write 2.
So here's the original
problem and I look at this
and I think well what would
the least common denominator--
remember 2 would be
written as 2 over 1.
So the least common
denominator is well I have an x
and an x plus 2 so
I'm going to make all
of them have the same
denominator so I would have
to multiply this first term
by x plus 2 over x plus 2;
this term I'd have to
multiply by x over x
and on the right hand side I'd
have to multiply the bottom
by x plus x plus 2 which means
I'd have one for the top also.
Okay now here's the key here.
This is an equation right?
So if you have an equation
where both sides really
have the same denominator,
then the numerator
should be equal.
Another way of thinking
about this is
since they all have the
least common denominator,
if you multiply both sides by
the least common denominator,
it would cancel with every
single one of its denominators.
So this becomes a
simpler equation--
3 times x plus 2 plus 4x
equals 2x times x plus 2.
This only is true if you
have an equation remember?
If there is no equals 2
you just can't make the
denominators disappear.
It's really because if 2
numerators are equal here
we are.
Let's say I have 2
over x equals y over x,
the denominators are
the same does it make
since that y would
have to equal 2?
That's really what
I'm saying here.
Okay. Now you would
get the same equation
if you would have taken
this original problem
and multiplied both sides
by x times x plus 2.
It does take a little
bit more space
but you will get
exactly the same thing.
All right?
Just wanted to show you
another way of looking
at the problem--
rational equations.
Okay so not let's solve this.
We've got 3x plus 6 plus 4x
equals 4x squared (I'm sorry--
2x squared) plus 4x.
By the way if I would have
made a mistake there later
when I checked my answer I would
have found out that it was wrong
so remember if you do make
a mistake that's why it's
important that you
always check your answers.
All right so what
do we have here?
Do you notice that you have
a plus 4x on both sides?
So basically when I subtract 4x
from both sides that's not
going to be there anymore.
What I've got here is
a quadratic equation--
3x plus 6 equals 2x squared.
So to solve a quadratic
equation you set to equal to 0
so I'm going to subtract 3x
from both sides and subtract 6.
And now I need to state the
values of a, b, and c; a is 2,
b is negative 3, c is negative
6 and now we're just going
to solve this quadratic
equation.
Okay. So I'm going to do
b squared minus 4ac first
because it just makes less
of a mess when I put it
into the quadratic formula.
b is squared while b is negative
3 so that will be 9 minus 4 ac;
and what's a times c, 2 times
negative 6 that is a negative 12
so I have 9 plus 48 or 57.
Okay remember b squared minus
4ac or 57 is what's going
to go underneath the square
root in the quadratic formula.
So the quadratic
formula is negative b
which is well b is negative
3 so to offset that is 3 plus
or minus the square
root of 57 all over 2a--
a is 2 so 2a is 4 and
so I have two solutions.
3 plus square to 57 over 4 and
3 minus square to 57 over 4.
Actually this would be
really hard to check
so I'm not going
to do it on here.
It's probably easier
to do the problem over.
So those are your solutions
and that's all you have to do.
Now I'm going to do that
problem again multiplying
by the least common
denominator for those of you
who may have gotten
a little confused
when I use this new method.
All right so notice the method
I use this is an equation,
I just made all the
denominators the same,
so I made them all have
a common denominator
and then I made the
numerators equal.
All right we're going to do
the same thing the traditional
method multiplying both sides
by the least common denominator
which it would be
the x times x plus 2.
All right so here's
the same problem.
The traditional method
is multiply both sides
by the least common
denominator which is l, c,
d is x times x plus 2 so I'm
going to multiply each term
by x times x plus 2; so I
could take the 3x, 3 over x
and multiply it right?
And then the next term--
it's easier if you write
that x plus 2 in parenthesis,
times x plus x plus
2 and then the last--
the number on the right hand
side also has to be multiplied
by the least common
denominator and then we'll see
if anything cancels
which of course anything
that was a fraction should
cancel if you multiply
by the least common denominator.
So the x's cancel here,
the x plus 2's cancel here,
and there is no fraction on
the right so nothing cancels.
So I get 3 times x plus 2,
4 times x equals
2x times x plus 2.
So I get exactly the same
thing as I got up here.
There it is 3 times x plus 2
plus 4x equals 2x times x plus 2
and here it is again.
3 times x plus 2 plus 4x
equals 2x times x plus 2.
So from this point
everything is the same.
You would still use the
distributive properties set the
equation equal to 0, state
the values of a, b, and c,
plug it in the quadratic
formula and solve
and so this was just verifying
that our answer here
was correct.
So that's the answer
to this problem.
So we started off with
a rational equation
and we eliminated the fractions
or two ways I showed you
to eliminate the fractions
and get this equation right
here without fractions.
And then we ended up with
a quadratic equation,
use the quadratic formula to
solve because it didn't factor.
If it factored I would try
factoring first usually,
doesn't factor so the quadratic
formula is the natural way
to go here.
So this is a rational
equation that leads you
to a quadratic equation then
you could use the method
for solving a quadratic
equation.
