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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

• 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,

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