# 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.