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  • - [Instructor] Let's say you wanted to solve this equation,

  • two to the x squared minus three power

  • is equal to one over the cube root of x.

  • Pause the video and see if you can solve this.

  • Well you probably realize that this is not so easy to solve.

  • The way that I would at least attempt to tackle it is,

  • you would say this is two to the x squared minus three

  • is equal to x to the,

  • I could rewrite this, this is one over x to the 1/3,

  • so this is x to the negative 1/3 power.

  • Maybe I can simplify it by raising both sides

  • to the negative three power.

  • And so then I would get,

  • if I raise something to an exponent,

  • then raise that to an exponent,

  • I could just multiply the exponents.

  • So it would be two to the negative three x squared

  • plus nine power.

  • I just multiplied both of these terms times negative three,

  • is equal to x to the negative 1/3 to the negative three.

  • Negative 1/3 times negative three is just 1,

  • so that's just going to be equal to x.

  • So it looks a little bit simpler, but still not so easy.

  • I could try to take log base two of both sides

  • and I'd get negative three x squared plus nine

  • is equal to log base two of x.

  • But once again not havin' an easy time solve this.

  • And the reason why I gave you this equation

  • is to appreciate that some equations

  • are not so easy to solve algebraically.

  • But we have other tools, we have things like computers,

  • we can graph things and they can

  • at least get us really close

  • to knowing what the solution is.

  • And the way that we can do that

  • is we can say hey, well what if I had one function

  • or one equation, that was y is equal to two x,

  • two to the x squared minus three I should say.

  • And you had another that was y is equal

  • to one over the cube root of x.

  • And then you could graph each of these

  • and then you could see where they intersect.

  • Because where they intersect

  • that means two to the x squared minus three

  • is giving you the same y as one over the cube root of x.

  • Or another way to think about it is they're going

  • to intersect at an x value,

  • where these two expressions are equal to each other.

  • And so what we could do is we could go

  • to a graphing calculator, or we could go to a site

  • like Desmos and graph it and at least try

  • to approximate what the point of intersection is,

  • and so let's do that.

  • So I graphed this ahead of time on Desmos.

  • So you can see here, this is our two sides of our equation

  • but now we've expressed each of them as a function.

  • Right here in blue we have two, we have f of x,

  • or I could even say this is y is equal to f of x which

  • is equal to two to the x squared minus three.

  • And then in this yellowish color I have y

  • is equal to g of x which is equal

  • to one over the cube root of x,

  • and we can see where they intersect.

  • They intersect right over there.

  • And we're not going to get an exact answer,

  • but even at this level of zoom

  • and on a tool like Desmos you can keep zooming in

  • in order to get to get a more and more precise answer.

  • In fact you can even scroll over this

  • and it'll even tell you where they intersect.

  • But even if we're trying to approximate,

  • just looking at the graph.

  • We can see that the x value, right over here

  • it looks like it is happening at around,

  • let's see this is 1.5, and each of these is a tenth.

  • So this is 1.6 and then it looks like

  • it's about two thirds of the way to the next one.

  • So this looks like it's about,

  • I'll say this is approximately 1.66.

  • And if you were to actually find the exact solution,

  • you would actually find this awfully close to 1.66.

  • So the whole point here is, is that,

  • even when it's algebraically difficult

  • to solve something you could set up

  • or restate your problem or reframe your problem

  • in a way that makes it easier to solve.

  • You can set this up as, hey let's make two functions,

  • and then let's graph them and see where they intersect

  • And the x value where they intersect well

  • that would be a solution to that equation,

  • and that's exactly what we did right there.

  • We're saying that hey, the x value,

  • the x solution here is roughly 1.66.

- [Instructor] Let's say you wanted to solve this equation,

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B1 中級

グラフで方程式を解く|代数2|カーンアカデミー (Solving equations by graphing | Algebra 2 | Khan academy)

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    林宜悉 に公開 2021 年 01 月 14 日
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