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# Equations with one rational expression (advanced)

CCSS Math: HSA.REI.A.2

## Video transcript

Solve the equation: x squared
minus x squared minus 4 over x minus 2 is equal to 4. And they tell us that x won't
or cannot be 2 because if it was 2, then this would
be undefined. It would be dividing by 0. So let's see what x is. Let's see the x that actually
satisfies this equation. So you might be tempted to try
to express this with the common denominator of x minus
2 and then add these two expressions. But the thing that jumps out
at me initially is that we have x squared minus 4 on the
numerator, which is a difference of squares. Or if you factor it out, it's
x plus 2 times x minus 2. So we should be able to factor
this x minus 2 out. So let's do that. So if we were to rewrite it,
this is equivalent to x squared minus-- instead of
writing x squared minus 4, we know that's a difference
of squares. That is x plus 2 times
x minus 2. All of that over x minus 2. And that is equal to 4. And this whole time we're
assuming that x won't be equal to 2. And because x does not equal to
2, x minus 2 divided by x minus 2 is going
to be defined. And it will be 1. So those two will cancel out. And so we're left with x squared
minus x plus 2 is equal to 4. We can distribute the
negative sign. And I'll just arbitrarily
switch colors here. We can distribute the negative
sign, so we get x squared minus x minus 2 is equal to 4. And what we want to do is put
this in the form ax squared plus bx plus c is equal to 0. That allows us to either
factor it or apply the quadratic equation or complete
the square, or any of the ways that we know how to
solve quadratics. So let's do that. Let's get a 0 on the
right-hand side. The best way to do that is to
subtract 4 from both sides of this equation. Subtract 4 and we are left
with x squared minus x. Negative 2 minus 4
is negative 6. And then 4 minus 4 is 0. That was the whole point. So we have x squared minus
x minus 6 is equal to 0. Let me write it up here. x squared minus x minus
6 is equal to 0. And this looks factorable. We just have to think of two
numbers that when we multiply them give us negative 6. So they're going to have
different signs. When I add them I'll
get negative 1. So it looks like negative
3 and positive 2 work. So if we do x minus
3 times x plus 2. It's a little bit of trial and
error, but 6 doesn't have that many factors to deal with and
3 and 2 are one apart. They have different signs, so
that's how you can think of how we get to that conclusion. Negative 3 times 2
is negative 6. Negative 3 plus 2
is negative 1. So that is equal to 0. So we have two possible
ways to get 0. Either x minus 3 is equal to 0
or x plus 2 is equal to 0. And then if we take x minus 3 is
equal to 0, if we add 3 to both sides of that equation, we
get x is equal to 3 or, if we subtract 2 from both sides of
this equation, we get x is equal to negative 2. So both of these
are solutions. And let's apply them into
this equation to make sure that they work. Because these are solutions to
essentially, the situation where we got rid of
the x minus 2. Maybe it had some type
of side effects. So let's just make sure that
both of these actually work in the original equation. So let's try x is equal
to 3 first. So you get 3 squared minus
3 squared over 4. Sorry, 3 squared minus
4 over 3 minus 2. So this is equal to 9 minus--
3 squared is 9 --minus 4, which is 5, over 1. So 9 minus 5, which
is equal to 4. Which is exactly what we
needed it to equal. And let's try it with
negative 2. So if I take negative 2 squared
and I have a minus negative 2 squared minus
4, all of that over negative 2 minus 2. So negative 2 squared is 4. Minus negative 2 squared,
which is 4. Minus 4. All of that over negative
2 minus negative 2 over negative 4. Well 4 minus 4 is 0, so this
whole thing is just going to become 0. So this whole thing is
going to equal 4. So both of these
solutions work. And that makes sense, because
when we actually canceled this out, we actually didn't
fundamentally change anything about the equation. Only if you had the situation
where the x would have been equal to 2. That's the only thing that
you're really changing. So that's why it makes
sense that both of these solutions work.