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# 二次公式的證明 (英): 二次公式的證明

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- In the last video, I told you that if you had a quadratic
- equation of the form ax squared plus bx, plus c is
- equal to zero, you could use the quadratic formula to find
- the solutions to this equation.
- And the quadratic formula was x.
- The solutions would be equal to negative b plus or minus
- the square root of b squared minus 4ac, all
- of that over 2a.
- And we learned how to use it.
- You literally just substitute the numbers a for a, b for b,
- c for c, and then it gives you two answers, because you have
- a plus or a minus right there.
- What I want to do in this video is
- actually prove it to you.
- Prove that using, essentially completing the square, I can
- get from that to that right over there.
- So the first thing I want to do, so that I can start
- completing the square from this point right here, is--
- let me rewrite the equation right here-- so we have ax--
- let me do it in a different color-- I have ax squared plus
- bx, plus c is equal to 0.
- So the first I want to do is divide everything by a, so I
- just have a 1 out here as a coefficient.
- So you divide everything by a, you get x squared plus b over
- ax, plus c over a, is equal to 0 over a, which
- is still just 0.
- Now we want to-- well, let me get the c over a term on to
- the right-hand side, so let's subtract c
- over a from both sides.
- And we get x squared plus b over a x, plus-- well, I'll
- just leave it blank there, because this is gone now; we
- subtracted it from both sides-- is equal to negative c
- over a I left a space there so that we can
- complete the square.
- And you saw in the completing the square video, you
- literally just take 1/2 of this coefficient right here
- and you square it.
- So what is b over a divided by 2?
- Or what is 1/2 times b over a?
- Well, that is just b over 2a, and, of course, we are going
- to square it.
- You take 1/2 of this and you square it.
- That's what we do in completing a square, so that
- we can turn this into the perfect square of a binomial.
- Now, of course, we cannot just add the b over 2a squared to
- the left-hand side.
- We have to add it to both sides.
- So you have a plus b over 2a squared there as well.
- Now what happens?
- Well, this over here, this expression right over here,
- this is the exact same thing as x plus b over 2a squared.
- And if you don't believe me, I'm going to multiply it out.
- That x plus b over 2a squared is x plus b over 2a, times x
- plus b over 2a. x times x is x squared.
- x times b over 2a is plus b over 2ax.
- You have b over 2a times x, which is another b over 2ax,
- and then you have b over 2a times b over 2a, that is plus
- b over 2a squared.
- That and this are the same thing, because these two
- middle terms, b over 2a plus b over 2a, that's the same thing
- as 2b over 2ax, which is the same thing as b over ax.
- So this simplifies to x squared plus b over ax, plus b
- over 2a squared, which is exactly what we have written
- right there.
- That was the whole point of adding this term to both
- sides, so it becomes a perfect square.
- So the left-hand side simplifies to this.
- The right-hand side, maybe not quite as simple.
- Maybe we'll leave it the way it is right now.
- Actually, let's simplify it a little bit.
- So the right-hand side, we can rewrite it.
- This is going to be equal to-- well, this is
- going to be b squared.
- I'll write that term first. This is b-- let me do it in
- green so we can follow along.
- So that term right there can be written as b
- squared over 4a square.
- And what's this term?
- What would that become?
- This would become-- in order to have 4a squared as the
- denominator, we have to multiply the numerator and the
- denominator by 4a.
- So this term right here will become
- minus 4ac over 4a squared.
- And you can verify for yourself that that is the same
- thing as that.
- I just multiplied the numerator and the
- denominator by 4a.
- In fact, the 4's cancel out and then this a cancels out
- and you just have a c over a.
- So these, this and that are equivalent.
- I just switched which I write first. And you might already
- be seeing the beginnings of the quadratic formula here.
- So this I can rewrite.
- This I can rewrite.
- The right-hand side, right here, I can rewrite as b
- squared minus 4ac, all of that over 4a squared.
- This is looking very close.
- Notice, b squared minus 4ac, it's already appearing.
- We don't have a square root yet, but we haven't taken the
- square root of both sides of this
- equation, so let's do that.
- So if you take the square root of both sides, the left-hand
- side will just become x plus-- let me scroll down a little
- bit-- x plus b over 2a is going to be equal to the plus
- or minus square root of this thing.
- And the square root of this is the square root of the
- numerator over the square root of the denominator.
- So it's going to be the plus or minus the square root of b
- squared minus 4ac over the square root of 4a squared.
- Now, what is the square root of 4a squared?
- It is 2a, right?
- 2a squared is 4a squared.
- The square root of this is that right here.
- So to go from here to here, I just took the square root of
- both sides of this equation.
- Now, this is looking very close to the quadratic.
- We have a b squared minus 4ac over 2a, now we just
- essentially have to subtract this b over 2a from both sides
- of the equation and we're done.
- So let's do that.
- So if you subtract the b over 2a from both sides of this
- equation, what do you get?
- You get x is equal to negative b over 2a, plus or minus the
- square root of b squared minus 4ac over 2a, common
- denominator.
- So this is equal to negative b.
- Let me do this in a new color.
- So it's orange.
- Negative b plus or minus the square root of b squared minus
- 4ac, all of that over 2a.
- And we are done!
- By completing the square with just general coefficients in
- front of our a, b and c, we were able to derive the
- quadratic formula.
- Just like that.
- Hopefully you found that as entertaining as I did.

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