Solving equations by graphing | Algebra 2 | Khan academy

Let's say you wanted to solve this equation: (2^{x^2 - 3} = \frac{1}{\sqrt[3]{x}}). Pause this 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 to say this is (2^{x^2 - 3} = x^{-\frac{1}{3}}). I could rewrite this: (1) over (x^{\frac{1}{3}}) is (x^{-\frac{1}{3}}). Maybe I can simplify it by raising both sides to the negative (3) power.

So then I would get: if I raise something to an exponent, then raise that to an exponent, I can just multiply the exponents. It would be (2^{-3(x^2 - 3)}). I just multiplied both of these terms times (-3), which is equal to (x^{-\frac{1}{3}}^{-3}). Negative (\frac{1}{3}) times negative (3) is just (1), so that's just going to be equal to (x).

It looks a little bit simpler, but still not so easy. I could try to take (\log_2) of both sides, and I’d get: (-3x^2 + 9 = \log_2{x}). But once again, I’m not having an easy time solving this.

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.

The way that we can do that is we could say, “Hey, what if I had one function, or one equation, that was (y = 2^{x^2 - 3})?” I should say, and then you had another that was (y = \frac{1}{\sqrt[3]{x}}).

Then you could graph each of these and see where they intersect. Because where they intersect, that means (2^{x^2 - 3}) is giving you the same (y) as (\frac{1}{\sqrt[3]{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.

So what we could do is go to a graphing calculator or a site like Desmos and graph it to at least try to approximate what the point of intersection is. So let's do that. I graph 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 (2^{x^2 - 3}). We can even say this is (y = f(x)), which is equal to (2^{x^2 - 3}). In this yellowish color, I have (y = g(x)), which is equal to (\frac{1}{\sqrt[3]{x}}).

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 to get a more and more precise answer.

In fact, you can even scroll over this and it can 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 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).

It looks like it's about two-thirds of the way to the next one, so this looks like approximately (1.66). If you were to actually find the exact solution, you'd find this awfully close to (1.66).

So the whole point here 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.”

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).”