College Algebra: Solve Quadratic Geometry Problems

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So now we know in some sense how to solve quadratic equations using the various methods of factoring or foiling or completing the square or using the quadratic formula. We can solve quadratic equations, but sometimes we'll be given a word problem and when we set up the problem, that is converting the words to, you know, unknown variables and so forth, we'll actually end up with some sort of equation that will be quadratic. There will be x's and squared powers and so forth. So then, well, how do you deal with that? Well, the thing to do is, of course, is to carefully convert the word problem to the math stuff and then solve the math stuff. Just like we did before with the linear equations. So it's this time again, once again, for another installment of solving "real world" problems. So again, we put the quotes on the "real world".
Let's start right now by taking a look at the following. So here's some garbage you can see here. If I take away this, some dirty socks, and of course, under the dirty socks, viola, is a dorm room. No surprise there. So, in fact, this brings up an interesting question so you can see the big mega speakers, sort of futon couch, coffee table, some lamps and there is the dorm room, a rectangular dorm room. Let's take a look at the question. So the perimeter of this rectangular dorm room is 34 feet. So that's the length around the whole thing, 34. And we're told that its area is 60 square feet. The question is what is the dimension of the room so what is the length and the width of the room.
Well, how can we figure this out? Well, let's think about this for a second together. We don't know the length and the width. Those are unknowns. So let me give them names. Let me call the width w and you can call them x and y, of course, if you want. I'm just going to call them w and I'll call the length l. I'm just naming these unknowns; notice that I'm doing this or, you know, we're doing this together. It's not given to us in the problem. But now I can talk about these things in a mathematical way so I can take all the words there and convert them to some sort of mathematical fact. So, for example, the first thing there says that the perimeter is equal to 34. So what does that mean? It means that the length around is 34. So let's add that up. Well, that's an l plus a w plus another l, so that's 2l, plus another w so that's 2w. So I see 2l + 2w and that has to equal 34. So that first sentence there can be converted to 2l +2w = 34. Now, in fact, actually I'm going to do a little more work, but let me just before I get too involved, let me just realize that I can divide everything through by 2 and simplify this a little bit. If I do this and factor out the 2 and divide, I would see l plus w equals half of 34, which would be 17. So, in fact, the first sentence that we're given leads to this. Not exactly, but after cancellation. So there's the first sentence. And what else are we told? We're also told that the area of this dorm room is 60 square feet. And how do you compute area? Well, area is base times height so that would be l times w would equal 60. So we know that l x w = 60. So those are the two facts that were given. In fact, maybe I should be fair and mark this. These are the two facts that I've just gleaned from the question. And now what's the question asking? They're asking for what are the dimensions. So what is l and what is w. So what I need to figure out is what these things are. I can't solve this because I have two equations, but I also have two things that I don't know. I have two unknowns. So what should I do?
Well, one technique is just to take a look at one of these equations and try to get one of the variables to disappear and become the first variable. So if I could have this equation just have l's in it, then I could solve it. How can I do that? Well, one way of doing that is by looking at this equation and realizing I can solve this for w. Let me actually solve this for w. In fact, I can even do this in my head. All I've got to do is take that l and bring it to the other side and I would see w = 17 - l. So this equation can be written as follows. It's identical to w = 17 - l. I just took this and solved. But the beauty of that is that now in place of this w right here, I can insert its twin which is 17 - l. So what I'm going to do now is go back to this which has two things I don't know and I'm going to replace one of the unknown things by something that has the same unknown thing I already have. And if I do that what I would see is I would see the l times w which I'm now going to write as 17 - l = 60. And now what you notice is I have just one equation and one unknown and I can solve. In fact, you may say, geez, it's factored so I have two things factoring and so 1 is, in other words, no, no, no. Remember to solve a quadratic we have to have everything equal 0. So what I want to do, in fact, is pull everything over. So pull that 60 over which means I'm going to have to factor this using some sort of other method. Let me first distribute that 17, or that l rather, not only to the 17, but also to the second l and when I do that what I see is 17l - l^2 = 60. And if I bring everything over to the right hand side, what I see is the following. I see 0 on the left and on the right I'll see if I bring this -l^2 over, it becomes a +l^2 so I have l^2. If I bring the 17l over to this side, I see a -17l and then I still have that +60. And my hope is that this can be factored somehow. So let's see if I can factor this. So I have 0 equals, so I'm going to put an l and an l. The plus sign tells me they're both the same and this minus sign tells me they're both minuses. I need two numbers whose product is going to be 60, but they combine together to produce 17. How about 5 and 12? Because that would give me a -5l and a -12l combined to give a -17l and -5 x -12 is, in fact, 60. So this is looking good. We're running out of room here. So let me just actually cover some of the previous work that we have here. Let me just keep going. So now what do I see? So therefore, these two factors multiply together to give 0, so I solve this just like always with the quadratic. Either this equals 0 or this equals 0. If this equals 0, that means that l would equal 5 or the other possibility is that l - 12 is 0 which means l = 12. So it looks like I've got two solutions and I've got to find out, you know, which one is the right one.
Well, first of all, what would the w be? Well, remember what w is? I'm going to slide this down a little bit. w is 17 - l so if l = 5, then what would w be? Well, it would be 17 - 5 which is 12. On the other hand, if l = 12, what would w be? Well, w would be 5. So what you notice is we get sort of the same dimensions either way, but it brings in sort of an interesting philosophical question, you know, what distinguishes length from width, right? I mean, if I have a rectangle like this, a hundred-dollar bill, Now this looks like the length and that's the width. But, of course, I could hold it this way and now this is the length and this is the width. So, I mean, this is the width and this is the length. The point is, you know, we're getting the same answer. We're seeing this 5 by 12. So using the quadratic formula we see the answer is 5 by 12 and so there we see the room is 5 feet by 12 feet. It sounds like a good size dorm room, doesn't it? And there's a bonus question here. How many roommates live in the room? Obviously, if it's a 5 by 12, it's spacious. It's almost a suite. Obviously there'll be six students living there and, in fact, maybe we should let them alone and I'll put back the dirty socks so they can have their privacy. You can see, by the way, the big fancy speakers, the futon, the chair and so on.
All right, enjoy solving these problems with the quadratic.
Equations and Inequalities
Word Problems with Quadratics - Math Topics
Solving a Quadratic Geometry Problem Page [1 of 2]