Beg Algebra: Solving Quadratics by Factoring

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SOLVING QUADRATICS BY FACTORING
Okay, so solving an equation that just has x’s in it, even in the denominator, not too bad. You have to sort of work it through, get all the x’s to bubble onto the top and then work around, get all the x’s together and so forth. It’s great. But what if you make the equation that you're given a lot more exotic. In fact, what if there are powers on the x’s? So now I want to talk to you about the fact that it really is hip to be square. In particular, I want us to think about quadratic type equations.
Now quadratic equations sound sort of intimidating. But really, all it means is that there are x’s now that, instead of just appearing naked, actually now have a power of 2 on them. So let’s take a look at a very simple example. How about x2, you see, there’s the power of 2, equals 25.
Okay, if x2 = 25, now what I want to do is figure out what’s the right value for x to make these two things equal? Well, you might look at this and say, “Oh, yeah, I know, it’s just 5.” Well, if you plug in and check, you'll see that you're absolutely right. 52 = 25. So it looks like we’re done. But actually, that’s not the only solution, because what happens if for x I put in -5? Well, if I take -5 and square it, that -5 • -5, and I’d still get 25, because those negative signs would drop out. So, in fact, now we see there are two answers. Maybe there are three answers that we just don’t even know what the third one is yet. So when you have powers in the variables, you actually may get more than one answer.
Now, how can we see that for sure and how can we make sure that, in fact, we found all the answers, namely, +5 and -5, and there’s not a third answer somewhere else that we can find later? Well, here’s a great way to solve all quadratic type problems, and that is to create something that equals zero. In particular, bring everything over to one side and then have, on the other side, just a lone zero. Now, how would I do that here? Well, here I would just take this 25, I would want to move it to the other side so I’d subtract it, and so I’d see x2 - 25 = 0. Great!
Now, what would I do with this? Well, you see, instead of trying to solve it or take square roots or do things like that, what I’m going to do is factor, and this is the key thing for you to remember. And let me show you why this is going to be a good idea. You see, there are people that might say, “You know what? I know this stuff, I remember it, I saw it before.” And here’s what I’m going to do. I'm just going to take square roots of both sides and I’m just going to say the following: I'm just going to say, “x = 5,” and be done with it. The problem is that is not quite right, but this is a very, very common mistake. In fact, this is number two on my top ten list of classic mistakes. It’s the quadratic mistake. The thing to remember is that when you have an exponent of 2, it means there may, in fact, be two answers. So just taking square roots of both sides ain’t going to cut it anymore. And what you’ve got to do is make sure you're going to capture all the solutions. So instead of the square root business, which, you know, it only works with these kind of problems, if it even works then. Instead, let’s use the factor method.
So the factor method is to make sure that you just have zero on the right and everything else over on the left, and then look at what you’ve got there and try to factor. And if you need help factoring or want a refresher course in factoring, just click one of those buttons down there and you can see all the factoring. There’s more factoring than you ever thought humanly possible. I ought to know, because I factored those things down there. Anyway, this, I see, is the difference of two perfect squares. And so I know how to factor that. It would just be (x + 5) • (x - 5). And I still have that equal to zero. And now here is the key fundamental fact about mathematics that’s going to save us. It’s going to rescue us out of this dilemma. If I have two numbers, and if their product is zero, the only way that can happen is if either the first number were to be zero or the second number were to be zero. I can’t take two numbers, neither of which is zero, and multiply them together and produce zero. Try it! 2 and 7, it doesn’t work. How about 4 and -3? It doesn’t work. You can’t do it. So if I have a product of two things that equals zero, either this is zero or that’s zero, and that’s it. Well, if this is zero, I could write that down, x + 5 = 0, that would lead me to the solution x = -5, or maybe x - 5 = 0. In that case, I would see that x would have to be 5. And now, you see, I'm able to find not only both solutions, but discover that’s all there is, because I factored and saw that these are the only possible ways to make this thing zero. And you can go back and check and see that 52 = 25, -52 = 25. So when you’ve got squares involved, the trick always, is shift everything over, have a zero on one side and factor the other.
Okay, let’s try a more exotic example. Suppose I give you 3x2 - 7x = 0. Well, now we’re in great shape, because, in fact, they handed this stuff on a silver platter. I already have the zero there, now my job is to factor. What technique would I use here? Well, I look at this and I say, “Gee, there’s a common factor of x. Let me just factor out that common factor of
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x and see what we’re left with.” If I factor that out, so there’s the common x, I’d have to have a 3x here, that’s the remaining stuff after I factor out an x, and here, if I take away that x, I just have the -7. And so, I factored it as x • (3x - 7). Okay, so now I’ve got a product of two things that actually come together to make zero. The only way that can happen is if either this thing is zero or that thing is zero. Well, that means that either x = 0 or what’s the other possibility? That this thing equals zero. So 3x - 7 = 0. And now, I’m going to go off and solve this linear equation. I would bring this -7 to the other side and it becomes a +7, or add 7 to both sides, if you will. And I see 3x = 7. And now I divide both sides by 3. And if I divide both sides by 3, the 3’s now here cancel and I see x = 37. And so I see two solutions, x = 0 is a solution, you can try that really easily and see. Plug in zero here and you see 0 - 0 = 0, but also 37 is a solution. And you can actually check that by plugging in 37 here and squaring it and getting 949 × 3 and seeing what you get when you take –7 × 37. Look what happens. I would see 3 × 949 and I subtract off 737. Look what happens. The 3 and the 9 have a little bit of cancellation together that produces a 3 downstairs, so I have a 349. But then –7 × 7 = 349. Something minus itself is zero. It checks! And that’s it.
Okay, let me give you a chance to do one of these. I'm having too much fun here, so I’m going to give you one and see if you can solve it. 2x2 - 5x - 3 = 0. Okay, see if you can find out what values of x would satisfy this equation. Try right now.
Okay, well how did you make out? Well, what I would do is I would try to factor this. It’s a trinomial so I’m going to try to factor it into two pieces. I’ll put a 2x here and an x here in the hopes that this is going to factor somehow nicely. This negative sign tells me that, in fact, the signs here are going to be opposite. But I just can’t plunk down plus or minus anywhere, because these things are different. There’s a 2 here and not a 2 here. So I've got to be really careful with how I place those down. But I want the product to actually be 3 and I want to combine them so, in fact, I get a 5 somehow. So 3 and 1 might be a good idea. If I put the 3 here, that would produce a 6, and then a 1 here actually might be pretty good, because then I could subtract off the 1 to get the 5. So let me try this: I'm going to put the 3 here and a 1 here. Now, how should I put down those signs now? This is a 6, this is going to produce a 1, and I’m going to want the thing to be negative, so I'm going to want that big quantity to be negative. So let me put a negative sign here and a positive sign here. Let’s check and make sure that I’m really okay.
2x • x = 2x2, great. The inside term is an x, the outside term is a -6x. -6x + x = -5x. So that checks. And the last times the last gives me a -3. I'm okay. So I've just factored that thing, and so now I have two things that multiply to give zero, so either this thing, 2x + 1 = 0, or the other possibility is that x - 3 = 0. And I can solve each of these individually and I'm going to see my two solutions forming here. If I bring this over to the other side, I’d see a -1 on the right, so I’d see 2x = -1. If I divide both sides by 2, I see that x = -21. So that’s one solution and what’s the other solution? I come back here and solve this and see x = 3. So this quadratic actually has two solutions. We found it by factoring. One is -21 and the other one is 3. If you want, you can check. Plug them both back in and see what you get.
Okay, let’s try one quick last one here. I’ll do this one for you, because that last one might have knocked you out. It knocked me out, by the way. I’m sort of exhausted. All right, here we go, x2 - 4x + 4 = 0. Again, since everything is already over, I can just try to factor. Let’s see what happens now. I put an x and an x, I see that I'm going to have the same sign, and that same sign is negative. So I'm going to have a negative negative. I need something that actually multiplies to give 4 and combines to give 4, and 2 and 2 work great. Check x2, -2x and -2x = -4x, and -2 • -2 = 4. They're great.
Well, look it turns out that this is actually a perfect square. So I have two solutions, either x - 2 = 0 or the same thing, x - 2 = 0. So in either case what I see is x = 2. Aha! That’s sort of fun. It turns out this is a quadratic equation that has only
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one solution. Something wrong? Absolutely not. Quadratic equations may have no solutions, they may have one solution or they may have two solutions. They will ever have more than two. And here’s an example of where this has exactly one solution, even though you might have heard on the streets, “Oh if it’s quadratic, it has to have two.” That’s just street talk. I'm telling you straight talk, when you have a quadratic equation, there might be no solutions, there may be one solution or they’re maybe two.
Okay, that’s it for quadratics.