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Hi. My name is Rebecca Muller. During this session, we're going to investigate the different methods we can use to solve quadratic equations in one variable. Let's now review what a quadratic equation actually is. A quadratic equation in one variable, x, is an equation that can be written in the form ax squared plus bx plus c equals 0, where the values of a, b, and c are just constants and the value of a does not equal 0. So notice that this last part will tell us that we'll definitely have an x squared term in our equation. Now, in this session, we want to look at three different methods for solving these quadratic equations. The first one is going to be by the square root method. This is going to include a process called "completing the square." Then we'll move on and look at solving quadratic equations by factorization. And then finally, we'll look at solving quadratic equations by using the quadratic formula. So now let's begin with the idea of a square root method on an equation that looks something like this, if we have x squared equals the value of 16. A couple things to point out-- we have the variable x squared isolated on one side of the equation. There are no other variable terms in this equation. So I have a perfect square, x squared, equals a constant value of 16. Now, you probably can solve this by inspection. What numbers when squared would give us the value of 16? And I'm sure that what you're thinking right now is, I know that answer. That answer is the number 4. And you're partly correct. It turns out that we end up having two square roots for a number. There's going to be the positive value. That is, if I take the value of 4 and square it, we end up with 16. But it turns out that if I start off with negative 4 and square that, we also come up with 16. So there's a positive square root of 16 and a negative square root of 16. And what we can think of in the format of the equation when we solve it is to say that if we're solving for x, x is going to equal plus or minus, giving us both the positive and the negative, square root of 16. The symbolism with the radical sign if used without the plus minus just means the principal square root, which means the positive value. So the plus minus is important. Now, the square root of 16 would be the number which when squared gives us 16. And that's going to give us the value of 4. So we're going to end up with x equals plus or minus the value of 4. And of course, we saw those two solutions in this format. We're going to look at a second example where we're going to use, again, the square root method, which is just taking square root of both sides of the equation. I'm going to state it a little bit differently now. If I start off with the equation x minus 3 quantity squared equals the value of 24, then what I have is on one side of the equation a perfect square. This means that this involves all of my variables, it's a perfect square, and it equals a constant. I'm now going to use the square root method, which is to take the square root of both sides of this equation. On the left-hand side, when I take the square root of something squared, I'm just left with what we could call the base. That in this case is going to be the x minus 3. On the right-hand side, we need to have the plus or minus the square root of the value 24. So again, this process right here is taking the square root of both sides of the equation. Now, I'm going to go a little bit further with this one in that notice that I'd like to still solve for x. That is, I can add 3 to both sides of the equation to get there. So let's go and do that next. x equals 3 plus or minus the square root of 24. But it turns out that the square root of 24 is not in its simplest format. You might recall this from another algebra class you've had. If I can take a number, such as the square root of 24, and rewrite it so that under the radical I have a product of two numbers, one of which is a perfect square-- and I'll demonstrate that now-- I notice that 24 can be rewritten as 4 times the value of 6. 4 is a perfect square. That allows me in my next step to rewrite this as the square root of 4 multiplied times the square root of 6. Now I can simplify the square root of 4. That's going to give us the value of 2. And then I have that multiplied times the square root of 6. One more thing just to again emphasize a subtlety of this process-- when I take the square root of 16 over here, I had to put the plus or minus in front of it. This symbolism with just the square root sign means the principal square root. So you'll notice that right now I'm not writing plus or minus in front of that value because I'm going to have it already in my equation down here. In other words, at this point, you can just separate this into a product of two square roots. Take the square root of the perfect square. Square root of 4 is 2. 6 is not a perfect square nor can it be written as a product which has a perfect square in it. So I have to leave it as the square root of 6. And finally now, using this simplification, my final answer is going to be x equals 3 plus or minus 2 times the square root of 6. And so we end up really with two solutions. We have 3 plus 2 square root 6. And we have 3 minus 2 square root 6. And those are going to be the two solutions for this second equation. So now let's consider this equation. We have x squared plus 6x minus 6 equals 0. You know, we just now went over the square root method. And that required us to have a perfect square isolated on one side of the equation equal to a constant. So notice first of all, I see the x squared here. But notice I have another term here that has the 6x with it. So we don't want to try to use the square root method in this process in this format. However, we can do something called "completing the square" to put it into a format so that we can use the square root method. And I'm going to demonstrate that now. The first step in doing this completing the square method is to end up with only the variable terms on one side of the equation. So I'm going to begin by adding 6 to both sides of this equation. Rewriting it, we'll have x squared plus 6x. And I'm going to leave a little space here because you're going to see in a moment why. I'm going to end up adding something on there. I'm going to put equals. And when I add 6 to 0, I end up with 6. Now, I have only variable expressions on one side of the equation. The next thing to check in this process is whether or not the coefficient of my x squared term is equal to 1. In this example, it is already equal to 1. I'll end up doing a second example in a moment where it doesn't equal 1. And I'll demonstrate what to do in that circumstance. Right now, we have our variable expressions isolated on one side of the equation. We have a leading coefficient of 1. That is, the coefficient of x squared equals 1. Here's our process. We're going to take 1/2 of our middle coefficient. So that's 1/2 times 6. That value is equal to 3. We're going to take the value of 3. We're going to square that value. And we're going to add it to both sides of the equation. And I'm going to go ahead and simplify here. When I add 3 squared, that's the same thing as adding 9. So now, remember, this is an equation. As long as I do the same thing to both sides of an equation, I end up with an equivalent equation. So all I've done in this process is to add 9 to both sides of the equation. So what have I accomplished? Well, if I just rewrite this side of the equation, hopefully you'll spot it. What I have is x squared plus 6x plus 9 equals-- I'm going to go ahead and simplify. 6 plus 9 is 15. What I've done is I've created a perfect square. If you look at this format, x squared plus 6x plus 9, that is equivalent to the expression x plus 3 quantity squared. Check it out. Multiply that out and make sure that you see that it's equivalent. So let me go back a minute and just re-emphasize a few things. I take half of my middle coefficient, which is equal to 3 in this circumstance. I square it, and I add it to both sides of the equation. The expression that I end up with on the left-hand side is going to end up being a perfect square. That's going to be that x value plus the value of 3, which is, remember, that half of that middle coefficient. And that's in parentheses squared. This is the format within which we can use the square root method in that we can now take square roots of both sides of the equation. We'll end up with x plus 3 equals plus or minus the square root of 15. We can solve for x by subtracting 3 from both sides of the equation so that x is going to equal negative 3 plus or minus the square root of 15. We can look at our square root to see whether or not we could simplify any further. But when we think about the factorization of 15, we can think of it as 3 times 5 or 1 times 15, of course. There's not going to be a perfect square that is a factor of 15. So we've got both of the solutions to this quadratic equation given to us now, and that's in its simplest form. If you were going to write them separately, one of the solutions is negative 3 plus the square root of 15. And the other solution is negative 3 minus the square root of 15. However, most the time, you will see this notation with the plus minus in most textbooks and in most examples that you're going to see. It's time for a quick quiz. To complete the square on the expression x squared minus 5 x. What value should be added. Should it be A. 25, B. -5 halfs or C. 25 ports. Choose A, B, or C now. You're right the correct answer is 25 fourths. Sorry the answer is 25 fourths. To see why let's consider a process. we look at the coefficient of X which in this case is -5. we take that value of -5 and then we take half of that value within square the result. This number is going to be 25 divided by 4 which is port C. That is the value that we're going to take in order to complete the square on that expression. So I promised you a second example of working with completing the square, and here it is. The equation I want to work with is 2x squared minus 3x plus 1 equals 0. In this example, notice that my leading coefficient-- that is, the coefficient on x squared-- is equal to the value of 2. So it's going to require an extra step. The first step I'm going to use-- and actually, these two steps are interchangeable. But I'm just going to use the same process I used earlier-- is to isolate the values that have the variables in them. In other words, we have two terms that have a variable. We have a single term that is a constant. I'm going to start off by subtracting 1 from both sides of the equation. So that's going to look like 2x squared minus 3x equals negative 1. So again, this process is to isolate the variable expressions. Next, notice that I have a leading coefficient of 2. I need to have a leading coefficient of 1 in order to use the process that we described earlier for completing the square. So how can we accomplish that? Well, right now, 2 is being multiplied times x squared. If I divide that first term by 2, I will have a leading coefficient of 1. But you can't just decide to divide a single term by 2. We have to divide both sides of the equation by 2. When I divide the left-hand side of the equation, which has two terms in it, that basically means I'm going to divide each term independently by the value of 2. And division by 2 is really going to affect the coefficient. So I'm going to do that all in one fell swoop here. We're going to divide the first term by 2. That's going to give us our x squared value with a leading coefficient of 1, which is what we need. In my middle term, I have a coefficient of negative 3. I'm going to divide by 2. That would give us negative 3x divided by 2. And I'm going to rewrite that as negative 3/2 x. I'm going to go ahead and leave myself some space here because I'm going to come back and do the completion of the square on this step. But I'm going to go ahead and write on the right-hand side the value of negative 1 also divided by 2. So that's going to give me negative 1/2. So again, remember, isolate your variable expressions. And then divide out the leading coefficient in order to make sure you end up with in our next step a coefficient of 1 in front of the x squared. So now we're ready to complete the square. To complete the square, the process is to take 1/2 of the middle coefficient. I'm just going to write it in right here. It's going to be 1/2 times-- because taking 1/2 of something is multiplying by 1/2-- 1/2 times the negative 3/2. Now, that's going to end up giving us a negative 3/4. We're going to take this value and square it. We're going to add the negative 3/4 squared to both sides of the equation. And I'm going to go ahead and simplify on the right-hand side as I go. So if I take negative 3/4 and multiply times another negative 3/4-- well, first of all, a negative times a negative is a positive. To multiply 3/4 times 3/4, we multiply numerators, and we multiply denominators. That is, you end up with the numerator squared, which is going to give us a value of 9, that is. And we end up with the denominator squared, which is going to give us a value of 16. So once more, just going through the arithmetic, squaring gives us a positive value. 3 squared is going to give us a 9. 4 squared is going to give us a 16. Now, I'm going to do a fair amount of simplification in my next step. First of all, let's go ahead and write down what we're going to end up with when we have our perfect square. This expression-- because we have, quote, "completed the square"-- is going to be a perfect square where the first term is going to be an x. Because our middle coefficient is negative, we're going to have that minus right there in order to give us that negative in the middle term when we multiply out. The value that I have here is going to be the same as 1/2 of this, which was going to be minus 3/4. So again, it's really like adding the negative 3/4 inside of the set of parentheses. You know, if you're doing this and you get really good at it, you can actually just realize that when the middle term is negative, you're going to end up with a minus sign. When the middle terms are positive, you're going to end up with a plus sign. And in doing this step, where you take 1/2 times the middle coefficient, you can actually kind of disregard the negative as long as you've got it taken care of in this step. I hope that made sense just now. So in other words, you can simplify your work a little bit by just noting the minus sign moving down or the plus side moving down. On the right-hand side of that equation, I have an addition of two fractions. And of course, to add two fractions, we need a common denominator. So when I look at my denominators, I have a denominator of 2 and a denominator of 16. The common denominator will be 16. I can write negative 1/2 as negative 8/16. And then I'm to be adding that to 9/16. All right. Let's rewrite this once more. We're going to have x minus 3/4 quantity squared. And I'm going to go ahead and add the two fractions together. Negative 8/16 plus 9/16 is equal to 1/6. Great. Now we have a perfect square isolated on one side of the equation equal to a constant. We can now take the square root of both sides of the equation. That's going to give us x minus 3/4 equals plus or minus the square root of 1/16. Let's simplify. We're going to have x minus 3/4 equals plus or minus the square root of 1/16 is the square root of 1, which is 1, over the square root of 16, which is 4. Solving for x, we can add 3/4 to both sides of the equation. And now notice in this case I'm going to be able to simplify a little bit further. So I'm going to separate this into two parts. We're going to have x equals 3/4 plus 1/4. And I'm going to have x equals 3/4 minus 1/4. Here we're going to have 3/4 plus 1/4. That's going to be 4/4, so x equals 1. Over here, we're going to have 3/4 minus 1/4. That's going to be 2/4. And simplifying gives us x equals 1/2. So I end up with two solutions that are in the simplest forms, the values of 1 and the value of 1/2. Now, although the process of completing the square can be used to solve any quadratic equation, it's not necessarily the most efficient method that you're going to learn. On the other hand, it turns out that that process is important when you're using some other applications in math. So even though it may not be the one to go to when you're doing quadratic equations necessarily, it is a process that's important to learn how to use. So now I'm going to move on to another process, and that's going to be working with factorization. So to demonstrate this idea, let's just remember a few things that are pretty obvious. You know that if you multiply 0 times any number, the product is going to also be 0. If I tell you that I've got the number 3 and I'm multiplying times some unknown value, and that product is 0, what do we know that 3 has to be multiplied times? It has to be multiplied times 0. Likewise, in the other order, if I had some number multiplied by something like negative 1/2, and I told you the result was equal to 0, what would that number have to be? That number would have to be 0. So this can be summarized into the following property. It's called the "zero-product property." It basically just says that if a times b equals 0-- so you have two things equal to 0 when they're multiplied together-- then either a equals 0, b equals 0, or both of them are equal to 0, because of course 0 times 0 is also 0. So this is going to be the crux of the idea that we're going to use when we use a process to solve quadratic equations called "factorization." So I've written a quadratic equation on our board. We have x squared equals the value of x. What I want to do is demonstrate this factorization process so that we can use the zero-product property in order to solve this quadratic equation. Now, of course, that requires you to understand how to factor. So I'll be reviewing techniques in factorization as we're going along with these examples. The first thing that we need to do is to rewrite this equation set equal to 0 because that factorization process, that zero-product property, requires that we have something equal to 0. Let me stop for a moment and just point out that I cannot use the square root method on this example because I do not have both variable expressions on one side of the equation. Yes, x squared is a perfect square. But the square root method requires me to have a perfect squared equal to a constant. And of course, here is a variable expression. So again, we'll go with the zero-product property by starting off by setting this equal to 0. So we have x squared minus x equals 0 as our first step. What we'll concentrate on doing now is factoring the side of the equation that is set equal to 0. Here I notice I have only two terms. When I only have two terms, there are a number of things we can look for. But in this case, what I notice is the first thing you always look for in factoring, which is a common factor. I have a common factor of x in each term. So I'm going to factor out that common factor. We'll factor out an x, leaving me with x multiplied times x to give me x squared minus-- remember that when you factor an x out from your second term, we have to note that x times 1 would give me back the x that I have in my second term here. So one more time, we notice that we have a common factor of x. And that can be factored out. Now I notice that I have two things multiplied together. I'm just going to emphasize that here. I have x multiplied times x minus 1. In order for the product to be equal to 0, one of the two factors must equal 0, or both of them could equal 0 if they were the same factor. So at this point, I'm going to separate this into two equations. The first equation is going to say that simply, x could be equal to 0. Or the second equation will state that the second factor-- that is, x minus 1-- could be equal to 0. And I can solve this second equation to give me x equals 1. So what it says is that if I let x equal 0 or I let x equal 1 that I will have a solution to the original equation. We can easily check that. If we look at our original equation and I let x equal 0 in that equation, I'm going to have 0 squared equals 0. And of course, 0 squared is 0, so 0 equals 0. That checks. If I let x equal the value of 1 and I substitute it into the original equation, we have 1 squared equals 1. And of course, 1 squared is 1, so 1 equals 1. And that checks. So we have two solutions. x could equal 0, or x could equal the value of 1. Now, I'm going to do a second example that looks a lot like this one, but there's a difference. We have x squared equals the value of 25. I just want to point out that in this instance, I could use the square root property. That is, I could take square roots of both sides of this equation and solve for x because I do have a perfect square isolated equal to a constant. So just contrast this with the first example for a moment to make sure you get that idea. It's easy to get those confused. And you want to make sure that you're clear on when you can use the square root property and when you can't. Now, even though I can use the square root property to solve this second equation, I'm going to choose to do it using factorization. And to do that requires me to first set it equal to 0. So I'm going to have x squared minus 25 equals 0. Now, how do we factor x squared minus 25? This you should recognize as being a difference of two squares. And in a previous course, you've probably learned about the factorization of a difference of two squares. You're going to end up having x minus 5 times x plus 5. That is, if you have a format of a squared minus b squared, that can be factored as a minus b times a plus b. So the x that is being squared ends up being the first term in both of my factors. The 5 that is being squared ends up being the second term in both of my factors. And one time I'm adding, and one time I'm subtracting. Notice that if you're having trouble or you need to review this process, go ahead and spend time right now multiplying this out. Convince yourself that this is actually the factorization. At this point, I have two things multiplied together equal to 0. So I can say, well, either the first factor, x minus 5, equals 0, or the second factor, x plus 5, equals 0. Solving each of these equations independently gives us x equals 5 or x equals negative 5. Now, again, let's just kind of review a moment. Had you done this process using the square root property, you would have taken square root of both sides of the equation. And you would have said x equals plus or minus the square root of 25. So square root of 25-- we would think it was 5. So we would end up with the positive value of 5 and the negative value of 5. The beauty of using the factorization instead is that you don't make the mistake of forgetting to put the plus minus. You can see both solutions. And of course, it's easy to check those because we can see that we would end up having both of those square roots as our solutions. Next, I'm going to look at a series of equations in which I'm going to talk about the factorization process. The first one I want to start with is 2x squared minus 3x plus 1 equals 0. This is the same equation I used as the second example when I did completing the square. So just compare that process to what we're about to do, and you'll see how much more efficient factorization can be. Here I have the first term with the x squared. So I have it in the format ax squared. The middle term is multiplied times x, so that's ax squared plus bx. And the third term is a constant. So that's ax squared plus bx plus c equals 0. Now, this is a trinomial. You may remember that term. It's a polynomial expression with three terms. I'm going to take this and factor it into a product of two binomials. So there are a lot of different methods you might learn for doing this. I'm going to demonstrate what I think is the most efficient method for thinking about factoring a trinomial. First of all, because I have it written already set equal to 0, and I have it written in the standard format with the x squared minus the value times x plus the constant, I'm going to go ahead and just put a double set of parentheses here. What I know is that I'm going to end up with a linear expression in my first set of parentheses and a linear expression in my second set of parentheses because if you have x multiplied times an x, we're going to get an x squared. So I'm going to start off with this first term. How can I think about the factorization of 2x squared? Well, 2 is just going to be thought of as 2 times 1. x squared can be thought of as x times x. So I'm going to write down the factorization of 2x squared in the first terms that I see in each of my what are going to be linear factors. So that's going to be 2x multiplied times x. Now, again, I'm going to demonstrate what I think is the most efficient method for doing this. The next thing I'm going to do is look at my last term, my constant. It's just the value of 1. What I know is that the second term in each of my linear factors will have to multiply together to give me back the value of 1. And 1 is pretty easily factored when you're thinking about integer values as just 1 times 1. So I know this is going to be a 1, and this is going to be 1. Now, one thing you'll notice is I have not paid attention to the sign in the middle term yet. Nor have I paid attention to the fact that this is a positive one. I'm about to now consider all of that in my next step. What we're doing is what can be called the reverse of what you might have learned as the FOIL method. The FOIL method is where you learned how to multiply two binomials together-- First, Outer, Inner, Last. What I've looked at is that the first terms-- the 2x times x is going to give me the first term here. The L, the last terms-- 1 times 1 will give me the last term here. The middle term, the negative 3x, is going to be arrived at by looking at your inner and your outer terms. So I'm going to go ahead and write all of this down. And then we'll end up doing this in a lot quicker format in our next examples. When I multiply the inner terms, I notice I have one x. When I multiply my outer terms, I notice I have 2x times 1, which is going to be 2x. Now, if this factorization process is going to work, then I need to be able to combine 1x and 2x and end up with a negative 3x. And of course, the only way to make it a negative 3x is to make both of these negative. So if I add negative 1x and a negative 2x, I end up with negative 3x. Well, how can I make this inner term negative? The x is already positive. If I put a minus sign in front of this one, then now when I multiply together, I end up with a negative 1x. How can I make this outer term a negative 2x? This 2x is already positive. If I put a minus sign in front of the 1 on the second factor, I end up with a negative 2x. There's one thing to check at this point because it looks like I'm done. Have I maintained the correct sign for my constant? When I multiply negative 1 times negative 1, do I end up with the correct sign on the constant, which is a positive 1? And of course, we do. Two negatives multiplied together give us a positive. So what I know now is that I have the correct factorization for this trinomial. Now I can use the zero-product property. I can take my first factor, 2x minus 1, set it equal to 0. Or the second factor, x minus 1, could equal 0. To solve the first linear equation, we're going to add 1. So that's going to be 2x equals 1. And then we're going to divide by 2, so x equals 1/2. To solve the second equation, we're going to add 1 to both sides of the equation. And we end up with x equals 1. So our two solutions to the quadratic are the values 1/2 and 1. Let's look at a second example now. We're going to look at 8 minus 2x minus x squared equals 0. And I'm going to do this a little faster because you need to become more accomplished at doing factorization. But I'm going to say everything that you need to hear in order to understand the factorization. One thing is I notice that I don't have this in standard form. I have my x squared with a negative sign in front of it. Well, it turns out that in this first example, you were looking at descending powers of x. In the second example, you're looking at ascending powers of x. They're going up. And you can actually factor in this format just as easily as you can in that format. So I'm going to go ahead and take it as it is and look at the factorization as a product of two binomials. The 8 is going to be multiplied together to give me the first terms here. I'll end up multiplying the last terms together to give me x squared. Well, that part's pretty easy. What do I multiply together to end up with an x squared? It's going to be x times x. Once more, you notice I'm not paying attention to the sign-- the fact that it's negative-- yet. Now I need to look at factors of 8. Well, I can either use 1 times 8, or I could use 2 times 4. I'm looking now at my middle term in order to figure out which one I should use. Is there a way to somehow combine 8 and 1 and end up with 2? Well, not by addition. However, if I look at the other factorization of 8-- that is, 4 times 2-- I note that I could subtract those values and end up with a 2. So that's how I'm going to consider it. I'm going to say that I'm going to factor the 8 as 4 times 2. At this point, I need to consider my inner and my outer terms in that product to see whether or not I can end up with the correct signs-- the negatives in both cases here-- and whether I can end up with the minus 2x as my middle coefficient. So my inner term is 2x. My outer term is 4x. Can I combine 2x and 4x and end up with a negative 2x? The answer is yes if I make this a negative 4x plus 2x. How do I make this positive? I put a plus sign in front of that x so that these two multiply together to give me positive 2x. How do I make the outer term negative? I put a negative sign in front of this x so that when I multiply 4 times negative x, I get negative 2x. Last thing to check-- did I end up with the correct sign on my last term here? Yes, x times negative x is negative x squared. So this is the correct factorization. Now I can use my zero-product property. 4 plus x equals 0, or 2 minus x equals 0. Here I can subtract 4 from both sides of the equation, x equals negative 4. Here I can just add x to both sides of the equation, 2 equals x. And so I end up with two solutions. x could equal negative 4, or x could equal the value of 2. Let's do one more example, 6x squared plus 5x minus 6 equals 0. And again, I notice that this is a trinomial set equal to 0. I'm going to factor it as a product of two binomials. Now, in this case, I've got to really think about all the different combinations. But I'm going to point out something that's pretty important when you're working with these values in figuring out what's going to work best. You want to be not haphazard. You want to be kind of structured when you're working these. So when I look at the value of 6 as my coefficient of x squared, I know I could either use 1 times 6 or 2 times 3. To decide which one to use, I am going to glance at my middle term. I notice the value of 5 there. Now, when you're thinking about your factors of 6-- again, 1 times 6 or 2 times 3-- these two values are far apart. Those are extremes. They're going to give you larger values than the ones that are closer together, the 2 and the 3. So because the value of 5 is not particularly large compared to my outside values of 6, and when particularly large could be bigger than that product, then I'm going to use the ones that are closer together as my go-to. I'm going to put down 2x times 3x. Now, remember, I'm just going to simply look at my last term, my constant. It also happens to be a 6. For exactly the same reasoning, I am going to use the 2 and the 3. And here's the crux of what I want you to understand. If I put a 2 right there, that's going to be incorrect. Because if I put a 2 right there-- let me just write it down here. If I put 2x, and I put a 2-- let's just say it's a plus 2-- notice I have a common factor of 2. Well, I can't have a common factor of 2 because I didn't have a common factor to start with. So there is no way I can do this factorization using the number 2 here, which means if this factorization process is going to work, this has to be a 3, and that has to be a 2. Again, this is kind of an educated guessing type of process, but it works 90% of the time. That is, think about your extremes. Think about the ones that are closer together. Figure out whether this middle term is extreme or not. If not, go with the ones that are closer together. And then just use a process of elimination to come up with stuff. Here, my inner term is 9x. My outer term is 4x. Can I combine 9x and 4x and end up with a positive 5x? Yes-- plus 9 minus 4. Make this positive by putting a plus sign here. Make the outer negative by putting a minus sign there. Do I end up with a negative 6 as my constant? I do. So now take each factor independently-- 2x plus 3 equals 0 or 3x minus 2 equals 0-- and solve each for the value of x. Here I'll subtract 3. 2x equals negative 3. Divide by 2. x equals negative 3/2. Here I'll add 2. 3x equals 2. Divide by 3. x equals 2/3. And I come up with my two solutions. So I've gone through these three problems in succession, really being very detailed at the beginning, speeding it up at the end. But if you look back at it, these are really the same process over and over again. It's just a matter of becoming savvy and looking at how my number combinations are going to work in order to end up with your product that is going to give you back the original trinomial. And from that point on, it's pretty simple to come up with your solutions. So now that we've gone through a number of things dealing with factorization process for solving quadratics, it's time for you to try one on your own. What I'd like you to do is to try to solve the equation 2x squared plus 13x minus 24 equals 0. And we're going to use the factorization process in order to try to solve this. So pause the video now. Try the problem on your own. And then restart the video when you're ready to look at their solutions together. Ok. So let's see how we did. We have a trinomial set equal to 0. I'm going to go ahead and say, well, I know I'm going to have a product of two binomials. I'm going to take my first term, 2x squared, and I'm going to look at the factors that I can come up with. And that's pretty simple because there's really only one choice here. You're going to have those multiplied together. That's going to be 2x multiplied times x. Now, it doesn't really matter the order in which you write them down. But this is pretty much your only choice when you're working with integer values. Now I'm going to look directly at my last term. My last term is 24. I know that I'm going to multiply two numbers together in these last spots in order to end up with a product of 24. And I've got to think about the fact that there are a lot of options here. So I'm going to write some things down and talk about my thought process as I'm doing so. I want to be fairly structured, either by writing everything down or at least thinking about it. I'm going to start with 1 times 24. And then I'm going to go ahead and say, well, does 2 work? Yes-- 2 times 12. Does three work? Yes-- 3 times 8. Does 4 work? Yes-- 4 times 6. Does 5 work? No. And then I go to 6. And I notice, oh, look. 6 is already there. So I know that these are the four factorizations that I could use for the value 24. I'm going to now glance at my middle term. And what I'm worried about is only whether that coefficient is really big compared to my numbers on the outside, the 2 and the 24. It's not extremely large. So my rule of thumb is to start with the values that are closer together to see whether or not those would work. Now, when I look at the 4 and the 6, one thing to note is both of these are even. If I put an even number in this first factor, notice that I already have a 2x. I would have a common factor of 2. I cannot create a common factor that didn't already occur in the original format. And of course, we didn't have a common factor. So therefore, I know the 4, 6 combination is not going to work. So now let's move our way up. I have a 3 and an 8. Can I use this combination? Well, I can try it. But now let's think about what would have to go in this first blank that I have here. Could it be the value 8? No, 8 is an even number. I would have a common factor that didn't already appear. So if this is going to work, that would have to be the 3. So I'm going to write it down. I'm going to have the 3 and the 8. So what I know right now is that I've got a setup where I get back my first term, and I get back my constant. Can we end up with that positive 13x in my middle term by looking at my inner and my outer terms here? Let's do it. We end up with 3x if I look at the inner. The outer is going to be 16x. Ooh, this is looking good. Can I combine these and end up with a positive 13x? Yes. If I make this a positive 16 and this a minus 3x, those will combine to end up giving me back 13x. So I just put a minus sign here and a plus sign there. Last thing to check-- are those signs going to work out with my constant? My constant is actually a subtraction of 24, so a minus 24. When I multiply negative 3 times 8, I end up with a negative 24. So this factorization works. At this point, then I know that this is the factorization. Now I can use my zero-product property, set my first factor equal to 0. Either that equals 0, or the second factor is equal to 0. Let's solve each of these independently. I will add 3 to both sides of the equation-- so 2x equals 3-- and then divide by 2 to give me x equals 3/2. For the second equation, I can solve for x by subtracting 8 from both sides of the equation. So x is going to equal negative 8. And now I have the two solutions for that quadratic equation. I hope you did well on your example. Now, learning how to factor efficiently is going to be important to you in any algebra course. You're going to use it not only to solve quadratic equations, but in many other types of problems. So take time now to become proficient in that so that you can do better later on in the course. For now, we're going to move on to our third method for solving quadratic equations. Remember, we had the square root method, then we looked at factorization. This method is called using the quadratic formula. But we first have to see where the quadratic formula emanates from. Why does it work? Now we're going to derive what's called the "quadratic formula." And the way to do that is to take the general format of a quadratic equation-- that is, ax squared plus bx plus c equals 0-- and complete the square on that in order to solve for x. So let's run through that quickly. We're going to have a isolation of our x terms. We can do that by subtracting c from both sides of the equation. What we need to do next is to divide by a. And so we're going to end up having our ax squared divided by a plus bx divided by a equals negative c divided by a. That can be rewritten as x squared plus b over ax equals negative c over a. Now, what occurs next is we have to complete the square, which means we're going to take 1/2 of our middle coefficient. That's going to be 1/2 of b over a. And we're going to end up getting b divided by 2a. To complete the square means to take the square of that result and to add it to both sides of the equation now. So that will read x squared plus b over ax plus the quantity b over 2a squared equals negative c over a plus that same quantity squared-- again, both sides of the equation. Now, what are we going to do now? We know that we have a perfect square. So we're going to have x plus b over 2a quantity squared equals negative c over a plus b squared over 4a squared. Notice that in the last term, I've just gone ahead and apply the square to both the numerator and denominator. We need to combine the terms on the right. And that means we need to have a common denominator. That common denominator is going to be 4a squared. And so notice that I've taken the term negative c over a. And I need to multiply both numerator and denominator by 4a. So let's think about what that's going to look like in the next step. And here we have it. We're going to have that first term as negative 4ac divided by 4a squared plus the other term on the right-hand side, b squared over 4a squared. And now I'm just going to go ahead and combine those since I have a common denominator and rewrite the equation as x plus b over 2a quantity squared equals b squared minus 4ac divided by 4a squared. We can now continue by taking the square root of both sides of the equation. We then need to simplify the right-hand side. So I can take the square root of the numerator divided by the square root of the denominator. Notice that I already have the plus or minus in the problem. So when I take the square root of the denominator, even though there's a variable involved, I can still notice that I can just look at it as being 2a. And we take the term from the left-hand side, the b over 2a, and move it to the right-hand side. Notice that it becomes negative. So again, we've done two steps in one. We've taken the square root of the fraction that was on the right. And we have transferred b over 2a to the right-hand side of the equation, making it negative b over 2a. And finally, because we have the same denominator, we can now combine those to come up with the quadratic formula-- x equals negative b plus or minus the square root of b squared minus 4ac divided by 2a. Now, we don't want to have to complete the square every time we want to use quadratic formula. But what this allows us to do is take an equation in the format ax squared plus bx plus c equals 0 and solve for x by substituting in the values of a, b, and c. Let's use the following example to see how to use the quadratic formula. We have x squared plus 5x minus 7 equals 0. The first step is to identify what we have for our a, b, and c value. a is going to be the coefficient of x squared. So our a in this case is equal to 1. b will end up being the coefficient of our x term. So our b is equal to 5. And c is going to be the constant value, which is equal to negative 7. One thing to note is make sure you have this set equal to 0 before you attempt to identify your a, your b, and your c because the quadratic formula depends on you having this written in general form. Now, to recall the quadratic formula, we have x equals negative b plus or minus the square root of b squared minus 4 times a times c all divided by 2a. When you're trying to learn formulas, a good idea is every time you have to use it, just rewrite it. And that way, it's going to be ingrained in your memory. To proceed, we're going to now substitute in the values of our a, our b, and our c. So our x value is going to equal-- we're going to take the negative of our b value, or that is the opposite of the B value. So our b value is 5, which means we're going to end up with a negative 5 here plus or minus the square root of-- we're going to have b squared. So that's going to be 5 squared minus our 4 multiplied times our a value, which is 1, multiplied times our c value, which is negative 7. We'll then divide that whole thing by 2, multiplied times the a value, which is 1. And now it's just a matter of simplification. So let's do that by rewriting negative 5 plus or minus the square root of-- 5 squared is 25. We're going to have a negative of 4 times 1 times negative 7. When I'm doing this, I like to look at the signs first and figure out what I'm going to end up with. Will it be positive or negative? So I have a negative here, a positive here, and a negative there. So when I multiply a negative times a negative, I know I get a positive. And then that times another positive value remains positive. So this is going to be a plus sign. Then I can just concentrate on my numbers. I've got 4 times 1, which is 4 times 7, which is 28. So I've already determined the signs. And now I can just write down the numerical value. That's being divided by the value of 2. And then finally, we'll add the 25 plus the 28. And that is going to give us x equals negative 5 plus or minus the square root of 25 plus 28 is going to be 53. And that's divided by 2. Now, the last thing I consider is whether or not I can simplify the square root of 53. But it turns out I can't do that because 53 is a prime number. So our two solutions are negative 5 plus square root 53 all divided by 2, negative 5 minus square root of 53 all divided by 2. Now let's look at a second example. And I'm going to be concentrating now on the simplification process as we're doing this. We're going to have 2x squared plus x plus 3 equals 0. We're going to identify that a is equal to the value of 2, our b value is equal to 1, and our c value is equal to 3. Here's my quadratic formula. I'm going to use it and substitute in. We're going to have x equals the negative of our b value, which is going to be a negative 1, plus or minus the square root of b squared-- and I'm going to start demonstrating what I would normally do. I would go ahead and simplify this right now. b squared is 1 squared, which is 1, minus-- now I'm going to write everything out because in my experience, if you're going to mess it up by trying to do it mentally, it might happen right here. So go ahead and write out this portion. So let's go ahead and square the b, minus 4 times a, which is 2, times c, which is 3. And then that's all divided by 2 times a. I can go ahead and simplify that. Twice the number 2 is going to be 4. Now my next step-- I'm going to simplify underneath that radical symbol. So I'm going to have negative 1 plus or minus the square root of-- notice that this is going to be a minus under here. So it's going to be 1 minus 4 times 2 is going to be 8 times 3 is 24. And then that's divided by value 4. And so we have x equals negative 1 plus or minus the square root of 1 minus 24 is a negative 23. And then that's divided by 4. Now, at this point, when I look at the value that's underneath the square root, I see that it's a negative number. You know we can't take the square roots of negative numbers in the set of real numbers. So there are no real solutions to this equation because underneath the radical symbol, I ended up with a negative. And in fact, let's just go back a minute to this previous one. Remember that we talked about this was going to be a prime number, so you couldn't simplify any further. And we ended up with two solutions here. But what type of solutions are they? Well, they're two solutions. But notice that this is going to be irrational. We're not going to be able to simplify it any further. You can't write it as a fraction where we have a numerator and denominator that are both integer values. So we end up with two irrational solutions here. And we're going to look at next three more examples. And we're going to concentrate on the types of solutions that we come up with and the simplification process in order to review this with you. Now we're going to look at a few more examples. And in these examples, I'm going to concentrate on the simplification process and on identifying the types and number of solutions. Let's assume by now that you're pretty good at identifying the a, the b, and the c. Let's go right into using the quadratic formula on this first one. We have x squared plus 4x minus 7 equals 0. So we'll have x equals negative b, which will be negative 4, plus or minus the square root of b squared. So that's going to be 4 squared, which is 16, minus 4 times a, which is the coefficient of x squared. That's 1 times c, which is going to be our constant value, which is a negative 7. Divide that by 2 times the a value, which is 2 times 1, which is 2. So we have x equals negative 4 plus or minus the square root of-- we're going to have 16. We notice that we have a minus and a minus, so that's going to give us a plus. 4 times 1 times 7 is 28. And we divide that all by 2. Continuing, x equals negative 4 plus or minus the square root of-- adding 16 and 28 gives us a value of 44. And divide by 2. Now we need to notice that the square root of 44 can be simplified. We know that 44 is the same as 4 times 11. So we can think of that as x equals negative 4 plus or minus-- we can think of the square root of 4 as being the value of 2. And then the square root of 11 cannot be simplified. So again, this is 4 times 11. Square root of 4 is 2, and the 11 remains underneath the radical symbol. And then that's divided by 2. Now, at this point, what I note is that I have a common factor in my numerator. That common factor is the value 2. And notice that I have a 2 in the denominator that's going to be able to be a common factor between the numerator and denominator. That all together tells me that what I should do is factor out the 2 from the two terms in the numerator, leaving me with negative 2 plus or minus the square root of 11. In the denominator, I have the 2. 2 is now a common factor that will now divide out. And so a lot of times, you'll see divide by 2, we get a 1, divide by 2, we get a 1, like it was cancelling. And we end up with our solutions, x equals negative 2 plus or minus the square root of 11. Now, how many solutions do we have? Well, we can have negative 2 plus square root of 11 or negative 2 minus square root of 11. So we have two solutions. And we can't simplify that radical expression any further. So we're going to end up with two irrational solutions here. Now, one last time, let's use the quadratic formula in order to solve a quadratic equation. This time it's 3x squared plus x minus 2 equals 0. Every time I'm using this, we're coming up with a slightly different version of what could happen as far as the simplification process occurs. And that's my point of doing all of these examples. So let's see what happens in this last one. We're going to end up with x equals negative b, which is going to be in this case negative 1 since the coefficient of x is 1, plus or minus the square root of b squared, which is 1, minus 4 times our a value, which is 3, times our c value, which is negative 2. We'll divide that whole thing by twice our a value. That's 2 times 3, or 6. x then is going to equal negative 1 plus or minus the square root of-- that's 1. And then we have a negative and a negative and a positive. So that result will be a plus. 4 times 3 times 2 is 12 times 2, or 24. And we divide that by 6. So x equals negative 1 plus or minus the square root of 25 divided by 6. Now, in this case, the number underneath the radical symbol is a perfect square. So I can rewrite this as x equals negative 1 plus or minus the square root of 25, which is 5, divided by 6. Well, now I can simplify further. I'm going to have x equals negative 1 plus 5 divided by 6. So let's go ahead and take that to its conclusion. Negative 1 plus 5 is going to be a 4 divided by 6. And I'll get x equals 2/3 when I simplify. Or I could have that x equals negative 1 minus 5 divided by 6. So x would equal negative 6 divided by 6, x equals negative 1, giving us two solutions that are rational. We don't end up with a radical expression. So here-- two rational solutions. Notice that in all of these examples, the decision on the type of solutions had to do with the number that was underneath the square root symbol. That number would tell us whether or not we had two solutions-- and if I ended up with any number under there besides 0, that was going to be two solutions-- or if I had a rational solution because it could be simplified, or if I had irrational solutions because it couldn't be simplified, or if I ended up with no real solutions because it was a negative. For this reason, that radicand-- the number underneath the square root-- is called a "discriminant." It discriminates between the different types of solutions that you can have for a quadratic equation. It's time for another quick quiz. The discriminant is the value b squared minus 4 ac. The value allows you to discriminate between the number and the types of solutions for quadratic equation. What do we know about the solutions if the discriminant is equal to 28. Is the answer A there are 2 rational solutions. B. There are 2 irrational solutions. Or C. There's only one rational solution. Choose between A, B and C. Your correct the answer is B. There or to irrational solutions. The value of 28 is not a perfect square and therefore when we simplify we're not going to be able to end up without the square root symbol in our quadratic formula. Sorry the answer is B. There are two irrational solutions. The value of the discriminant is 28 which is not a perfect square. Therefore we're going to end up with a value left underneath the square root symbol once we simplify. So we've seen three different methods for solving quadratic equations. When I'm working with my students, here's what I typically tell them to do. Start off with the square root method if you have a perfect square already as part of the problem. If you have to go through the completing the square, I'm going to suggest that you use one of the other methods. So again, if you can easily isolate a perfect square, set it equal to a constant, that's the method to go with. But other than that, the second method I suggest is to use factorization. Set it equal to 0 and see whether or not you can factor it. Typically, if you're good at factorization, that's going to be a quicker method than our last process, which is using the quadratic formula. Also, there's this idea that you can end up with a lot more careless mistakes with quadratic formula than you can with factorization. So again, these are your three methods. Try them in this order whenever you're dealing with quadratic equations that you come across.

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