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# 增根 (英): 增根

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- In this video, we're going to start having some experience
- solving radical equations or equations that involve square
- roots or maybe even higher-power roots, but we're
- going to also try to understand an interesting
- phenomena that occurs when we do these equations.
- Let me show you what I'm talking about.
- Let's say I have the equation the square root of x is equal
- to 2 times x minus 6.
- Now, one of the things you're going to see whenever we do
- these radical equations is we want to isolate at least one
- of the radicals.
- There's only one of them in this equation.
- And when you isolate one of the radicals on one side of
- the equation, this one starts off like that, I have the
- square root of x isolated on the left-hand side, then we
- square both sides of the equation.
- So let us square both sides of the equation.
- So I'll just rewrite it again.
- We'll do this one slowly.
- I'm going to square that and that's going to be equal to 2x
- minus 6 squared.
- And squaring seems like a valid operation.
- If that is equal to that, then that squared should also be
- equal to that squared.
- So we just keep on going.
- So when you take the square root of x and you square it,
- that'll just be x.
- And we get x is equal to-- this squared is going to be 2x
- squared, which is 4x squared.
- It's 2x squared, the whole thing.
- 4x squared, and then you multiply these two, which is
- negative 12x.
- And then it's going to be twice that, so minus 24x.
- And then negative 6 squared is plus 36.
- If you found going from this to this difficult, you might
- want to review multiplying polynomial expressions or
- multiplying binomials, or actually, the special case
- where we square binomials.
- But the general view, it's this squared, which is that.
- And then you have minus 2 times the product of these 2.
- The product of those two is minus 12x or negative 12x.
- 2 times that is negative 24x, and then that squared.
- So this is what our equation has I guess we could say
- simplified to, and let's see what happens if we subtract x
- from both sides of this equation.
- So if you subtract x from both sides of this equation, the
- left-hand side becomes zero and the right-hand side
- becomes 4x squared minus 25x plus 36.
- So this radical equation his simplified to just a standard
- quadratic equation.
- And just for simplicity, not having to worry how to factor
- it and grouping and all of that, let's just use the
- quadratic formula.
- So the quadratic formula tells us that our solutions to this,
- x can be negative b.
- Negative 25.
- The negative of negative 25 is positive 25 plus or minus the
- square root of 25 squared.
- 25 squared is 625, minus 4 times a, which is 4, times c,
- which is 36, all of that over 2 times 4, all of that over 8.
- So let's get our calculator out to figure out what
- this is over here.
- So let's say so we have 625 minus-- let's see., this is
- going to be 16 times 36.
- 16 times 36 is equal to 49.
- That's nice.
- It's a nice perfect square.
- We know what the square root of 49 is.
- It's 7.
- So let me go back to the problem.
- So this in here simplified to 49.
- So x is equal to 25 plus or minus the square root of 49,
- which is 7, all of that over 8.
- So our two solutions here, if we add 7, we get x is equal to
- 25 plus 7 is 32, 32/8, which is equal to 4.
- And then our other solution, let me do that
- in a different color.
- x is equal to 25 minus 7, which is 18/8.
- 8 goes into 18 two times, remainder 2, so this is equal
- to 2 and 2/8 or 2 and 1/4 or 2.25, just like that.
- Now, I'm going to show you an interesting
- phenomena that occurs.
- And maybe you might want to pause it after I show you this
- conundrum, although I'm going to tell you why this
- conundrum pops up.
- Let's try out to see if our solutions actually work.
- So let's try x is equal to 4.
- If x is equal to 4 works, we get the principal root of 4
- should be equal to 2 times 4 minus 6.
- The principal root of 4 is positive 2.
- Positive 2 should be equal to 2 times 4, which is 8 minus 6,
- which it is.
- This is true.
- So 4 works.
- Now, let's try to do the same with 2.25.
- According to this, we should be able to take the square
- root, the principal root of 2.2-- let me make my radical a
- little bit bigger.
- The principal root of 2.25 should be equal to 2 times
- 2.25 minus 6.
- Now, you may or may not be able to do this in your head.
- You might know that the square root of 225 is 15.
- And then from that, you might be able to figure out the
- square root of 2.25 is 1.5.
- Let me just use the calculator to verify that for you.
- So 2.25, take the square root.
- It's 1.5.
- The principal root is 1.5.
- Another square root is negative 1.5.
- So it's 1.5.
- And then, according to this, this should be equal to 2
- times 2.25 is 4.5 minus 6.
- Now, is this true?
- This is telling us that 1.5 is equal to negative 1.5.
- This is not true.
- 2.5 did not work for this radical equation.
- We call this an extraneous solution.
- So 2.25 is an extraneous solution.
- Now, here's the conundrum: Why did we get 2.25 as an answer?
- It looks like we did very valid things the whole way
- down, and we got a quadratic, and we got 2.25.
- And there's a hint here.
- When we substitute 2.25, we get 1.5 is
- equal to negative 1.5.
- So there's something here, something we did gave us this
- solution that doesn't quite apply over here.
- And I'll give you another hint.
- Let's try it at this step.
- If you look at this step, you're going to see that both
- solutions actually work.
- So you could try it out if you like.
- Actually, try it out on your own time.
- Put in 2.25 for x here.
- You're going to see that it works.
- Put in 4 for x here and you see that they both work here.
- So they're both valid solutions to that.
- So something happened when we squared that made the equation
- a little bit different.
- There's something slightly different about this equation
- than that equation.
- And the answer is there's two ways you could think about it.
- To go back from this equation to that equation, we take the
- square root.
- But to be more particular about it, we are taking the
- principal root of both sides.
- Now, you could take the negative square root as well.
- Notice, this is only taking the principal square root.
- Going from this right here-- let me be very clear.
- This statement, we already established that both of these
- solutions, both the valid solution and the extraneous
- solution to this radical equation,
- satisfy this right here.
- Only the valid one satisfies the original problem.
- So let me write the equation that both of them satisfy.
- Because this is really an interesting conundrum.
- And I think it gives you a little bit of a nuance and
- kind of tells you what's happening when we take
- principal roots of things.
- And why when you square both sides, you are, to some
- degree, you could either think of it as losing or gaining
- some information.
- Now, this could be written as x is equal
- to 2x minus 6 squared.
- This is one valid interpretation of this
- equation right here.
- But there's a completely other legitimate interpretation of
- this equation.
- This could also be x is equal to negative 1
- times 2x minus 6 squared.
- And why are these equal interpretations?
- Because when you square the negative 1, the negative 1
- will disappear.
- These are equivalent statements.
- And another way of writing this one, another way of
- writing this right here, is that x is equal to-- you
- multiply negative 1 times that.
- You get negative 2x plus 6 or 6 minus 2x squared.
- This and this are two ways of writing that.
- Now, when we took our square root or when we-- I guess
- there's two ways you can think about it.
- When we squared it, we're assuming that this was the
- only interpretation, but this was the other one.
- So we found two solutions to this, but only 4 satisfies
- this interpretation right here.
- I hope you get what I'm saying because we're kind of only
- taking-- you can imagine the positive square root.
- We're not considering the negative square root of this,
- because when you take the square root of both sides to
- get here, we're only taking the principal root.
- Another way to view it-- let me rewrite
- the original equation.
- We had the square root of x is equal to 2x minus 6.
- Now, we said 4 is a solution.
- 2.25 isn't a solution.
- 2.25 would've been a solution if we said both of the square
- roots of x is equal to 2x minus 6.
- Now you try it out and 2.25 will have a
- valid solution here.
- If you take the negative square root of 2.25, that is
- equal to 2 times 2.25, so that is equal to 4.5 minus 6, which
- is negative 1.5.
- That is true.
- The positive version is where you get x is equal to 4.
- So that's why we got two solutions.
- And if you square this-- maybe this is an easier way to
- remember it.
- If you square this, you actually get this equation
- that both solutions are valid.
- Now, you might have found that a little bit
- confusing and all of that.
- My intention is not to confuse you.
- The simple thing to think about when you are solving
- radical equations is, look, isolate radicals, square, keep
- on solving.
- You might get more than one answer.
- Plug your answers back in.
- Answers that don't work, they're extraneous solutions.
- But most of my explanation in this video is really why does
- that extraneous solution pop up?
- And hopefully, I gave you some intuition that our equation is
- the square root of x.
- The extraneous solution would be valid if we took the plus
- or minus square root of x, not just the principal root.

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