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# 解有理式 3 (英): 解有理式 3

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- 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.

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