If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:4:19

Solving equations by graphing

Video transcript

let's say you wanted to solve this equation 2 to the x squared minus 3 power is equal to 1 over the cube root of 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 you would say this is 2 to the x squared minus 3 is equal to X to the I could rewrite this this is 1 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 3 power and so then I would get if I raise something to an exponent then raise that to an exponent I can just multiply the exponents so it would be 2 to the negative 3 x squared plus 9 power I just multiplied both of these terms times negative 3 is equal to X to the negative 1/3 to the negative 3 negative 1/3 times negative 3 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 2 of both sides and I'd get negative 3x squared plus 9 is equal to log base 2 of X but once again not having 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 could say hey well what if I had one function or one equation that was Y is equal to 2 X 2 to the x squared minus 3 I should say and then you had another that was y is equal to 1 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 1 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 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 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 two we have f of X or I can even say this is y is equal to f of X which is equal to 2 to the x squared minus 3 and then in this yellowish color I have Y is equal to G of X which is equal to 1 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 the tool like desmos you can keep zooming in in order to get a more and more precise answer in fact you can even scroll over this and even tell you where they intersect but even if we're trying to explain 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 one point five and each of these is a tenth so this is one point six 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 one point six six and if you were to actually find the exact solution you would actually find this awfully close to one point six six 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 one point six six