Solve It Like A Pro - The Polynomial Style! 😁

Añadido: 2026-05-07

[00:00:00]Hello everyone. In this video we're going to be solving an interesting problem. I don't know what to call it. Maybe a rational equation, a rational expression, or algebraic expression. We are given a plus three over a equals six and we're supposed to evaluate a squared plus 18 over a based on the values of a we get from the first equation.

[00:00:23]So let's go ahead and evaluate the value of the second expression here by doing two methods. First method, I'm going to find the a value.

[00:00:36]All right?

[00:00:37]>> [snorts] >> So how do you solve this? Multiply both sides by a.

[00:00:41]Typical, right? You get a squared plus three equals six a.

[00:00:45]And then bring the six a over.

[00:00:48]And then write this if you want to use the quadratic formula or write it as a squared minus six a plus nine equals six. Same thing. I kind of um subtracted three and then added nine to both sides or just added six to both sides. It's probably easier.

[00:01:03]And I got a perfect square.

[00:01:05]And then I square root both sides.

[00:01:07]I get this and that.

[00:01:11]And then I add three to both sides.

[00:01:14]I get this and that. Okay, great. So those are the a values and I'm supposed to evaluate an expression based on these. But which a value am I going to use?

[00:01:26]That is the million dollar question, right? So let's find out. So what is the expression we're trying to evaluate? It is x squared plus 18 over x.

[00:01:38]Cool.

[00:01:39]So what happens if I replace a with three plus root six? Let's find out.

[00:01:44]Three plus root six squared plus 18 over three plus root six. Let's simplify this expression. The first one is a perfect square, a plus b quantity squared. So we're going to go ahead and expand it, right? Do you hopefully you remember the formula? a plus b on probably shouldn't use a here. A good choice. Not x either. How about m plus n?

[00:02:09]Okay.

[00:02:11]m squared plus n squared plus two m n.

[00:02:15]That's my favorite version even though some people are going to switch these around. I like to keep the squares together. Same thing.

[00:02:20]So we get three squared, root six squared, plus two a b.

[00:02:26]Plus 18. Now this one can be simplified right?

[00:02:35]To make the what is it called?

[00:02:37]Rationalizing the denominator. Okay, that's what it is.

[00:02:42]So when you multiply these two things from difference of two squares you get nine minus six which is three? Nine minus six that's already a hard question. Okay. Now what do you get from here?

[00:02:54]We get the following.

[00:02:55]18 three goes into 18 six times so they can simplify. Nine plus six is 15 so we get 15 plus six root six plus now I got a six times the three minus root six so I'm going to go ahead and distribute that. Hopefully you can see.

[00:03:12]18 minus six root six. And magically or mathematcially these two terms cancel out, disappear, and yay! We get 33.

[00:03:23]Awesome. Now you might be wondering what would happen if you used the different value of x like the conjugate the three minus root six. Let's give it a try. Why not, right?

[00:03:35]Why not?

[00:03:37]So if you replace x with three minus root six you're going to go through pretty much the same steps.

[00:03:43]So you're going to square this nine plus six. You know the drill, come on. Minus six root six. And this guy would be multiplied by its conjugate this time it's the other one. And obviously when you multiply these together you're going to get 18 times three plus root six and this shows you why it's not always a good idea to distribute. Just wait. Nine minus six is three. Three goes into 18 six times. This is 15 as before. Pretty much the same thing except for some sign changes. 18 plus six root six. Notice that the roles are reversed. One of them is negative the other one is positive. That's why this works in both cases. And that's kind of interesting like we have a variable expression but it seems to be a constant for this particular a value. That's so that's kind of interesting, isn't it?

[00:04:30]Okay. Now I'm going to go ahead and present the second method. So our first method was basically to solve the first equation. It turns into a quadratic. We find the radical values. We plug it into this one which is easy. We take care advantage of x what is it called?

[00:04:48]Conjugates, you know, simplify and we get a numerical answer.

[00:04:53]Right? And in both cases we got the same so most probably we did it right. Okay, hopefully. So let's go ahead and take a look at the second method. And what is really cool about the second method is because it's the second method. You know the second method is almost always the coolest one, right? Okay, great. So let me rewrite the equations. a plus three over a equals six. And from here we're supposed to evaluate a squared plus 18 over a.

[00:05:22]Awesome.

[00:05:24]Okay, I really love the second method by the way. I can't tell you enough about it.

[00:05:28]So here's what we're going to do. And I don't know if it got your attention but 18 is kind of three times six. I don't know if they're related. And for example, is this going to work for a plus k over a is equal to m and then a squared plus k m over a? Is this going to have a constant value? You can test it. I haven't tested it but I have a feeling it might. And math problems you should always start with a feeling like, okay, I feel like this is going to be good.

[00:05:55]>> [snorts] >> Anyways, so here's what I going to do.

[00:05:58]What we're going to do. I talked too much again.

[00:06:01]So I'm going to multiply both sides by a. Okay. Same same idea.

[00:06:08]But instead of trying to solve it I'm going to isolate a squared. And I'm pretty sure there's probably quite a few different branches off of this but I'm just going to keep a squared as a linear function. Make sense? Now I'm going to go back to my expression that I'm trying to evaluate. That is a squared plus 18 over a.

[00:06:31]And so this is my formula.

[00:06:34]Obviously I can replace a squared with something but what am I going to replace a with?

[00:06:39]The answer is nothing.

[00:06:42]Okay, why? Because you don't have to.

[00:06:45]Okay, one way to look at it is don't worry about the a replace a squared with six a minus three and solve it. Second idea is make a common denominator, get the a cubed, and then obtain a cubed as a linear function of a and then do it that way. You see there's a lot of paths. Okay.

[00:07:03]So but I'm not going to make a common denominator. So I'll just replace a squared directly which is six a minus three plus 18 over a. I do this because it everything falls into place nicely. And then I'll make a common denominator. How about that? You're like, ah this is cheating. No, it's not. Because look.

[00:07:20]If you make a common denominator you're going to multiply this by a, right? Like this and like that.

[00:07:27]Six a squared minus three a over a. So we can add them and this is divided by a.

[00:07:32]So what am I going to do with this though? Like I got this, okay.

[00:07:36]Well, I can replace a with something again, right? What is a? a squared?

[00:07:41]I have a squared so I I can replace a squared with six a minus three. Six times six a minus three minus three a plus 18 divided by a.

[00:07:51]And then I get 36 a minus 18 minus three a plus 18. This time the numbers cancel out instead of the variables. And we get 18 cancels out. 33 a over a. Obviously a is not zero and the answer is 33.

[00:08:09]And this brings us to the end of this video. Thanks for watching. I hope you enjoyed it. Please let me know. Don't forget to comment, like, and subscribe.

[00:08:15]I'll see you tomorrow.

[00:08:16]Probably tomorrow or next time with another video. Until then be safe, take care, and bye-bye.