Quadratic Equation Methods

Welcome to Algebra 2. This is section 3.1. We're solving quadratic equations today. So we spent a lot of time in chapter 2 talking about quadratic functions. Now we're going to see how to handle a situation where we have a quadratic function, but we're not just looking at the graph of it. We're trying to solve that equation. We'll get to what that means here in just a second. First, for your introduction. Um, Here, a lot of these matching introductions that we have, I know I say it every time, but what I want you to do is pause the video really quickly just on this first part and see if you can't match each graph with each given function and come up with, I don't know, some patterns or something that would lead you to choose each graph to go along with each function provided. So go ahead and pause the video. I'll check back with you in a few seconds. All right, well, hopefully you've got some ideas down on paper. So now we notice that these functions look very similar, like 1 and 4 are the same type of function, just negated, and they come in pairs like that as well, and the graphs show that. You can see that we have similar graphs that are opening it up or down from the same point, kind of three different pairs going on here. Well, for number 1, I would hope that we would remember that we can factor. And then if we look at the graph where that graph equals zero height, those will be the solutions of our factors. So here, this first one is telling us that x equals zero or two. So we're thinking it could be letter a or it could also be letter e. So we have to tie that in with how the function is looking. It's a positive x squared. So we should say letter e for that one. Number two, here we cannot, oh, we can factor this one, x minus 1, x minus 1. So now the only solution we get is x equals 1, and we can tell yet again that it's opening upward. So this should be letter D. Number three, this one is not factorable, but if we do our negative b over 2a, we get positive 2 over 2, x equals 1. and we know it has to open up, the only option left that's opening up is C. Okay, and then we can apply our reflections, and I would hope that we would say, okay, we used letter E for number one. That means A, the same zeros, but opening downward, should be the negated version of number one. So that choice fits. Number five, this is the negated version of number two. So number two was graph D, and graph... f is the one that shows d vertically flipped or negated. And that leaves us with option b for number 6. Alright, so the patterns that we found above here are the functions written as part of an equation. So now we've got quadratic equations to find those real solutions. Well, remember this was f of x, which means y equals... x squared minus 2x. So all we've done is we've replaced the y with 0. So we want to find what x values give us a height of 0. And those results we got from factoring are exactly what we're looking for. So here we've got x equals 0 and 2. Here we had no solutions. Oh, I'm going out of order. Here we had x equals 1. Here we had x equals 0 and 2. So even though it was a totally different equation, we still got the same solutions. That is possible. For number 5, this one was x equals 1 as well. And for number 6, we should have arrived at no solutions for that one either. So just referring back to those graphs, looking at those solutions from factoring the x-intercepts. So, what are some techniques that we can set up to solve these equations? And again, we're dealing with the quadratic function and the x-intercepts. If we have the function set equal to zero, so we have our function and we set it equal to zero, then the x-intercepts are the solutions. We can also use square roots. Since we're dealing with quadratics, we'll have a variable squared. And if we're lucky that there is no b term involved, then we can isolate it into something squared equals a number and then take the square root of both. sides to undo the square root. It's important to realize, and this is an easy thing to forget, when we take the square root of both sides, that the answer could be either positive or negative. And the reason is, if I take negative 2 times negative 2, I get 4. So really, I took negative 2 and I squared it, and I came up with positive 4. But if I asked you, what's the square root of 4? you would certainly say 2. Well, technically this negative 2 worked just as well, so we need to keep ourselves honest with remembering that plus or minus sign. Another method, we can factor a quadratic, and really this would be the ideal one to look for first. If we notice that we don't have perfect squares for this square roots possibility, that one jumps off the page at us. It's a good idea to try factoring first. It ends up being very, very fast. And we'll learn other techniques later on that will help out too. But factoring is a nice process to work through. The zero product property is what allows factoring to work as a solution. So let's just say we had two numbers, x times y equals zero. Well, if we wanted... To realize, or if we wanted solutions out of that, find out what x equals, find out what y equals, what could happen is, in order to make this product be zero, that would happen when x equals zero, or when y equals zero. So when we factor out a polynomial and get something like x minus 5 times x plus 3 equals zero, then what we're really saying is, oh, that would equal zero when x minus 5 equals zero, and it would it would create a product of 0 when x plus 3 equals 0. And then we can go about solving, find that x could equal 5 or x could equal negative 3. But the fact that they're multiplying together to equal 0 is why that process works. All right, so our first example, solve the equation by graphing it. So remember when it's by graphing, we're looking for x intercepts. We want our function set equal to zero, so number four is set up beautifully for us. If I wanted to graph this out, well, maybe I think the fastest approach would be negative b over 2a. So, whoops, x equals 1. I wanted an arrow there. So x equals 1 is our axis of symmetry. And if I plug in a 1, I get negative 1 plus 2. plus 3. So I have the point 1, 4. And we can reflect that across the Axis of symmetry. Now we want to see where this crosses the x-axis. So let's see what happens when x equals 2. So if I plug in a 2, I get negative 2 squared. That's negative 4 plus 4 plus 3. I end up with a 3. Oh boy, oh boy, oh boy. This wasn't looking right. It started to look like an absolute value function. I found my mistake. I was just moving a little too quickly. That'd be a lesson to you. is on the axis of symmetry. That's our vertex. Then we have and we can reflect that across. And now let's see what happens when x equals 3. I've got negative 9 plus 6 plus 3 equals 0. Hey, and that does give me a height of 0 for an output. And I can reflect that across the axis of symmetry. There's our graph, and it's opening downward like it should, and it's bending like it should. So the solutions to our equation are x equals negative 1 and positive 2. Let's try number eight. Change colors really quickly. So one thing I notice is that, hey, 3x squared, 6x, negative 3. I can divide both sides of that equation by 3. and make it simpler to work with. So I'm going to go ahead and do that. x squared equals 2x minus 1, and it's not set equal to 0, so I can't start graphing this thing just yet. I need to see the function on one side of the equation if I want to graph it. So I have my x squared. I'm going to subtract over 2x, add over the 1 to get it set equal to 0. All right, and this is actually one of the same equations we had at our initial, oh, in our introduction. So this one is actually factorable, but they say to do this by graphing. So let's go through the same approach we did in the last one. x equals negative b over 2a. So again, x equals 1. And let's see what happens when we plug that into The function here for work in simplified. So if we plug in a 1, what we get is 1 minus 2 plus 1, and the output is 0. Now it's important to realize that when I divided both sides by 3, what I was really doing is I was taking away an important part of that graph. It was 3x squared. So I want to be sure that when I draw my graph, it has that 3x squared. Sure, dividing by 3 made the algebra much easier, and it doesn't change our answer at all, but it does affect the graph. So let's assume that we didn't divide by 3. Because now that it's 3x squared, it's going to be 3 times as tall. You know, that's going to affect our graph quite a bit. So now we've got our vertex. Now we can take a look at the rest of the equation here. Let's see, if I plugged in a 2. Oh no, let's use 0 instead. 0 is always nice to plug in. If I plug in a 0 to the left side of that equation, I end up with negative 3. So at 0, oh, I'm sorry, plus 3, I end up with a positive 3. We can reflect that across our y-axis. You can see there's the 3 times as tall happening. And we would not have seen that had we used our simplified function over here. This is just fine if we're going to solve by factoring. But I probably should not have done that since they wanted us to graph. Instead of dividing by 3, I should have simply used this for negative b over 2a and then made my table from there. All right, last one to solve by graphing. I've got x squared. I want to get this set equal to 0, so I need to subtract over the 2x. And then there's also a plus 2. That's set equal to zero. Let's find our axis of symmetry. We're just going to graph this out. So x equals negative b over 2a. Again, x equals 1. I realize all my examples had x equals 1 as an axis of symmetry. It's kind of boring. All right, so there's our axis of symmetry. That means our vertex. And we're not asked to find this, but just so you know what I'm writing on screen. 1, and then if I plug in a 1, I get 1 minus 2 plus 2. Hmm. So the vertex is the point 1, 1. Well, let's see what that graph's doing, because if it's crossing the x-axis anywhere, those are going to be our solutions. We have x-intercepts. So let's try an x value of 0. I plug in zero, again a nice value to plug into the left side of this equation, zero minus zero plus two, and you can see our graph is actually bending away from the x-axis. So this is a situation where we have no solutions. Now there's one other thing. This is not part of your example, but one other thing that I did want to point out. This is going to be a hypothetical situation, but let's just say that we had plotted out our vertex, and we got some points from our table of values, and we don't have nice, clean outputs. that are leading us to a height of zero. So you'll see how the x-intercepts, they're not at a nice clean point. Well if we wanted the solutions out of that graph, what we would say is that there's a solution between negative eight and negative seven, and another solution between, this looks like, what, negative 5 and negative 4. So when we're graphing by hand, not a lot of precision required if those points don't come out clean, but we can specify where those solutions are happening. And that's always nice to kind of have on hand that you can check, too. You know, does your answer make sense, that type of thing. Alright, example two. Solve the equation using square roots. So I want to get to a point where it's something squared equals a number. Well, I can see here in number 18, here's my something squared. Let's get that set up to equal a number. So we'll start off by adding five to both sides. Dividing both sides by two. And now we can get to the point where we've got something squared equals a number. Now we can square root both sides to undo that square. Square root cancels out the square. What we have left over is just what's underneath. That's called the radicand. And it's going to be plus or minus the square root of 13 over 2. Now before I go any further, there's a concept called rationalizing the denominator. This root 13 over 2 is really equal to the square root of 13 over the square root of 2. And we don't like leaving square roots in the denominator. It's an interesting story about why that is the case. So what we do is we multiply that fraction by 1. And in this case, I'm calling that 1 root 2 over root 2. That way, when I multiply across the denominator, I'm taking root 2 times itself. I'm squaring that to undo the square root. So across the top I get root 13 times root 2, which is root 26, all over a regular 2. And now there's no more radical in our denominator. We can use like terms much easier that way. Okay, so I'm going to apply that. x plus 2 equals plus or minus root 26 over 2. And to finish solving for x, we would simply subtract 2 from both sides. And I'm going to call that 4 over 2. just to get a common denominator on the right-hand side. So x equals negative 4 over 2 plus or minus root 26 over 2, or we could call that x equals negative 4 plus or minus root 26 all over that common denominator of 2. It looks really bizarre, but we should not be alarmed if our solutions come out this way. This is just one of those cases where it comes out not entirely clean. Alright, the next one. Solve using square roots. Get a different color here. Ooh, purple. Alright. So, what we want to have is something squared equals a number. And if I were to subtract over this 1 fifth x squared, Combine it with the other x squared term on the other side, I get 2 equals 2 fifths x squared. Luckily, those fractions had common denominators, so we could add them pretty easily. Now, I don't just have something squared equals a number. I need to get rid of that 2 fifths. So I'm going to divide both sides by 2 fifths, and dividing by a fraction means multiplying by the reciprocal on the other side. And if I thought of that 2 as 2 over 1, across the top I get 10, across the bottom I get 2. We could even simplify those 2s. One's being multiplied into the numerator, one's being multiplied into the denominator. The division in the fraction undoes that multiplication. We have 5 equals x squared. Now we can square root both sides to get our answer. Square root cancels out the square. x equals plus or minus square root 5. And that's a pretty nice answer. We had to do a lot of work to clean up our answer on problem 18. Here we got to our solution, and it's in its proper form. Pretty quick one. Example 3, solve by factor. This is something that we should be pretty comfortable with. We worked on it in the algebra refresher. It's something that you've seen for a couple years now. So, let's get started. The function is part of an equation, so I want to keep this 0 equals. That's what makes our 0 product property work, if you remember that. So, here we want to factor. Basically, we're going to unfoil this thing. Since we have a leading coefficient of 1, I know that my factors will both be z's, just single variables leading off. That way, if I were to foil it back together, I would get back to where I started, a z squared, first times first equals the first term. Now I want to figure out what numbers multiply together to get positive 25, but also add up to get negative 10. And after a little bit of thought, negative 5 and negative 5 works. So now I've got the case where this could equal 0 or this could equal 0 to give me a solution. And it's going to be the same thing in either case, z equals 5. If I took either of those factors and set them equal to 0 to get my solution, I would add 5 to both sides to get there. All right, for the next one, 32, a squared minus 49 equals 0. You'll notice there is no b term in there, so it's like a 0a in the middle. And a is squared, and then this 49 at the end is also a perfect square. That is 7 squared. This is a special type of factoring situation called a difference of squares. And we can factor it just like we did before. You know, we can say, okay, it's a quadratic, so we've got our two factors set equal to zero. The leading coefficient is one, so I know it's going to be a something times a something. What multiplies together to get negative 49 but adds up to zero? Well, positive seven and negative seven do the trick. But if you remember this difference of squares factoring, You can quickly tell that, oh, it's just the square root of the first thing leading off the factors, the square root of the second thing, the 49, ending our factors, and then one's positive and one's negative. And if we solve out of the first factor, we get a solution of a equals negative 7, and out of our second factor, a equals positive 7. Now they deliberately asked us to factor this, so we have to see this step here in the middle. I know it's real tempting after the last section to add 49 to both sides and then square root both sides. And yes, that will get you the answer, but we're practicing our factoring techniques. So a little picky on the details, but we want to be sure we have all these skills in place. Alright, number 34. Here you can tell we've got all the pieces of a trinomial, a typical quadratic function. We've got a regular number, we've got y terms, and we've got y squared terms. We need to combine those together and get them set equal to zero. When we're factoring, it has to be set equal to zero. So I noticed I have positive 2y squared on the right, positive 1y squared on the left. So I'm going to move the y squared, start building this quadratic function on the right-hand side. So I'm going to subtract y squared from both sides. I'm going to subtract 28 from both sides. And I'm going to add y to both sides. Now I've got 0 equals 1y squared. plus 3y minus 28. All right, let's get that factored out. We are very fortunate yet again. Leading coefficient is 1, so I know my factors open up with y's. 1 multiplies to get negative 28, but adds up, combines to get a positive 3 plus 7 minus 4. And if we're ever skeptical about whether we're right or not, we can always try and run through foiling to see if we get back to where we started. The solutions coming out of those factors, y equals negative 7 or y equals positive 4. And factoring is one of those skills that I hope we are doing. just automatic width, because that makes a lot of math in the future much, much easier. All right, example four, find the zeros of the function. Pretty vague. They don't have us do this by graphing, though we could. I mean, if we wanted to make a thumbnail sketch here, we could factor it even a little bit. But what happens if we factor it? We get something that's not clean. We have a factor of an x-intercept of zero and an x-intercept of negative 11. Oh, never mind, this one works beautifully. So if we want to find out where this equals zero, yeah, we were able to factor that right away. That g of x, that output, it means the same thing as y. We're letting that equal zero. No height. Zero of the function. All right, and we quickly find out that our loose x, well, that's being multiplied, and it could produce a product of 0 if x equals 0. Or if x were to equal negative 11, that's what's coming out of that second factor to give us a solution. And then 52, this one's a little bit more interesting. If we do 4, you'll notice our lead coefficient is not 1. And I know we covered this in our algebra refresher, but we really can't refresh this enough. So if we have that situation, we're trying to factor it. We're saying, okay, let's let f of x equal 0. Let's factor this right-hand side out. 4 times 9 is 36. What multiplies to reach 36 but adds up to 12? Huh. Well, we've got 1 and 36, 2 and 18, 3 and 12, we've got 4 and 9, we've got 6 and 6. Ah, 6 and 6, that must be it. So once we find those, our magic numbers that multiply to get 36 and add up to be negative 12, we just need to realize that they would both be negative, of course. We divide them by... our leading coefficient of 4, and simplify them. So negative 3 halves, and we have the same thing over here, negative 3 halves. And to put this into a factor, it's really telling us x and negative 3 halves, x and negative 3 halves, Which, if it were in fully factored form, I mean, we can get our solutions right away. We know x equals positive 3 halves, and that's it. But if we wanted to continue factoring this, just to bolster our knowledge, you'll notice if we tried to FOIL these factors together, we would start off with 1x squared. But that's not what we started with. So what we have to do is, these denominators of 2, they actually move out in front of the x. That way, if we do first times first, we get 4x squared. If we do last times last, we get plus 9. And our inside numbers give us negative 6x. Our outside numbers give us minus 6 more x's. So that's how we successfully factor that completely. But to get the solutions, we don't have to do all of that work. It's just kind of a nice little pattern to follow. All right, example five. This word problem is pretty interesting. It kind of does an excellent job of setting up a real-world situation that if you were like a business owner or really, for any process that you're trying to accomplish, you can take data points and try and maximize your efficiency here. And a store is just a really good example of that. So an athletic store sells about 200 pairs of basketball shoes per month. And now it only sells that number of pairs of shoes when it charges $120 per pair. For each $2 increase in the price, the store sells two fewer pairs of shoes. How much should the store charge to maximize monthly revenue? Okay, let's think about this graphically. So what do we want our inputs to be? What do we want our outputs to be? you'll notice that The thing that we can control that's being changed is the increase in price. So let's let price be our x-axis, and then our y-axis will represent the revenue in a given month. And we could, I guess, have it be pairs of shoes. if we wanted. But what do we know? The chapter is about intercepts. If we charge zero dollars per pair, then we're not going to make any money. And that's why I think revenue makes a better y-axis, because we could be giving away shoes, but the store is not going to make any money. Then we have this graph that's peaking at some price, and then sales start to drop off. It becomes too expensive for the customers to buy shoes. So we know that x equals 0 is an x-intercept of this graph. And we also know that there's some other point where the price is so high that customers won't pay that. Let's see if we can't find the pattern to locate that point. So let's think. If we charge $120, then we will sell 200 pairs of shoes in a month. If we were to raise that to $122, we would only sell 198 pairs of shoes. If we raised it again to $124, we would only sell 196 pairs of shoes. And if we continue this process on, you'll notice that we can only increase the price, let's see, what is it going to be, 100 times. before we get to zero pairs of shoes being sold. So 100 times that we add $2 to the price, that adds $200 total. $320 means we sell zero pairs of shoes. So here we've got x equals 320. Now because of the symmetry of a parabola, we know that Thank you. Halfway between those two points, or at x equals 160, that's where our vertex is going to be. That's where our maximum will occur. So let's look at our function. Our revenue function based on a certain price per pair of shoes is equal to, we'll use intercept form here, x times, well I should, sorry, I should do. is equal to um Sum number a times x times x minus 320. And this loose x is really just representing the factor x minus 0 from our first x-intercept. So now we can actually find the function. This beautifully ties in what we learned last chapter. Our revenue is going to be 120 times 200, right? They tell us somewhere over here on the graph, we don't know where that point is, when the price is 120. The revenue is 120 times 200, $24,000 made that month. So the revenue, 24,000, equals some variable a times x minus 320. And we know that that x value is 120. So here, why don't I go ahead and erase that. Use a little bit of foresight here. All right, so we got a times 120 times 120 minus 320. a times 120. times that's negative 200. And what we end up with is 24,000 equals A times negative 24,000. If we divide both sides by negative 24,000, we find A equals negative 1. And it makes sense that A should be negative because our parabola is opening downward, and we're finding a maximum, so that should all kind of lead us to believe that we're setting this up correctly. So again, our revenue function is negative x times x minus 320, since we know a value is negative 1. Then we want to find how much should the store charge to maximize its monthly revenue. Well, we found that to be $160 per pair. Now let's figure out what that monthly revenue would be. So now we need to plug in 160. We need to find R of 160. That's negative 160 times 160 minus 320. 20, which is 160. Oh, sorry, negative 160. Yeah, we don't want to end up with a negative revenue. That's a bad business. Yeah, it should be a negative times a negative. So our revenue at $160 per payer is $256,000. $25,600 per month. But again, this is not the only way to do this problem. I just saw using the x-intercepts as being the most beneficial, and for me it made it easiest to do by hand. A calculator can also do a nice job with this, as long as you know what direction you're headed in. That would definitely help with this part of the problem where we're trying to find out what that other intercept is. But yeah, a nice little problem. I like how it ties together. Our second difference idea, knowing what inputs and outputs should be, and analyzing a function with real life inputs. Very nice. Example 6. You drop a seashell into the ocean from a height of 40 feet. Write an equation that models the height h in feet of the seashell above the water after t seconds has gone by. How long is the seashell in the air? Well, we don't have enough information right now to work with this. It's kind of working off the assumption that our y value in height is equal to negative 16t squared plus y initial. And I could have used h's there, representing the initial height, the starting height. And the textbook gives you this equation in one of its little examples that it provides. But it is assumed that you know. this type of setup for dropping an object on Earth. And we saw something like this in, what was it, Chapter 1, when we dealt with the, maybe it was Chapter 2, dropping the object on Earth and on the Moon. We've seen this before. So our initial height, we know that that initial height is 40 feet, so we can go ahead and plug that right into our function. What they want to know is how long is the seashell in the air? So let's see what that function would look like. Well, it starts off at 40 feet, and then as time goes by, here's our time ticking away on the clock, as time goes by that object is falling faster and faster towards the ocean. What we need to do is find the zero. So we want to find where is the height, where is the y value zero. in this equation. And you can notice this is one that we can solve using square roots, so let's add 16t squared to both sides of the equation. Isolate the square term, that means dividing both sides by 16. And then we want to take the square root of both sides. If you're sitting there thinking, Mr. Hartz, Mr. Hartz, you can simplify this. Yes, I do realize that it's something that can totally be simplified. But if I'm going to square root a fraction, it's helpful to have the denominator be a perfect square. And 16 is a perfect square. No, but you know what? I can divide the top and bottom cleanly by 4, which will give me another perfect square denominator. So let's simplify it like this. I know we could go further, but let's leave it here. If we square root both sides of that equation, we find that t equals the square root of 10 over the square root of 4. And remember, plus or minus. But look at our graph. Yeah, our parabola, if we graphed the entire thing, would go negative and positive. But since we're talking about time going by, we only want the positive result. We can't go back in time. So t equals root 10 over 2. seconds until that lands. And we could use a calculator to find a decimal if need be, but this is an exact answer, which are definitely preferable. Okay, that does it for section 3.1. Kind of a long lesson, plenty of examples, but really nice to be working algebraically, not as much tedious graphing as before.

We're trying to solve that equation. We'll get to what that means here in just a second. First, for your introduction.

Um, Here, a lot of these matching introductions that we have, I know I say it every time, but what I want you to do is pause the video really quickly just on this first part and see if you can't match each graph with each given function and come up with, I don't know, some patterns or something that would lead you to choose each graph to go along with each function provided. So go ahead and pause the video. I'll check back with you in a few seconds.

All right, well, hopefully you've got some ideas down on paper. So now we notice that these functions look very similar, like 1 and 4 are the same type of function, just negated, and they come in pairs like that as well, and the graphs show that. You can see that we have similar graphs that are opening it up or down from the same point, kind of three different pairs going on here.

Well, for number 1, I would hope that we would remember that we can factor. And then if we look at the graph where that graph equals zero height, those will be the solutions of our factors. So here, this first one is telling us that x equals zero or two.

So we're thinking it could be letter a or it could also be letter e. So we have to tie that in with how the function is looking. It's a positive x squared. So we should say letter e for that one.

Number two, here we cannot, oh, we can factor this one, x minus 1, x minus 1. So now the only solution we get is x equals 1, and we can tell yet again that it's opening upward. So this should be letter D. Number three, this one is not factorable, but if we do our negative b over 2a, we get positive 2 over 2, x equals 1. and we know it has to open up, the only option left that's opening up is C.

Okay, and then we can apply our reflections, and I would hope that we would say, okay, we used letter E for number one. That means A, the same zeros, but opening downward, should be the negated version of number one. So that choice fits. Number five, this is the negated version of number two.

So number two was graph D, and graph... f is the one that shows d vertically flipped or negated. And that leaves us with option b for number 6. Alright, so the patterns that we found above here are the functions written as part of an equation. So now we've got quadratic equations to find those real solutions.

Well, remember this was f of x, which means y equals... x squared minus 2x. So all we've done is we've replaced the y with 0. So we want to find what x values give us a height of 0. And those results we got from factoring are exactly what we're looking for.

So here we've got x equals 0 and 2. Here we had no solutions. Oh, I'm going out of order. Here we had x equals 1. Here we had x equals 0 and 2. So even though it was a totally different equation, we still got the same solutions.

That is possible. For number 5, this one was x equals 1 as well. And for number 6, we should have arrived at no solutions for that one either.

So just referring back to those graphs, looking at those solutions from factoring the x-intercepts. So, what are some techniques that we can set up to solve these equations? And again, we're dealing with the quadratic function and the x-intercepts. If we have the function set equal to zero, so we have our function and we set it equal to zero, then the x-intercepts are the solutions.

We can also use square roots. Since we're dealing with quadratics, we'll have a variable squared. And if we're lucky that there is no b term involved, then we can isolate it into something squared equals a number and then take the square root of both.

sides to undo the square root. It's important to realize, and this is an easy thing to forget, when we take the square root of both sides, that the answer could be either positive or negative. And the reason is, if I take negative 2 times negative 2, I get 4. So really, I took negative 2 and I squared it, and I came up with positive 4. But if I asked you, what's the square root of 4?

you would certainly say 2. Well, technically this negative 2 worked just as well, so we need to keep ourselves honest with remembering that plus or minus sign. Another method, we can factor a quadratic, and really this would be the ideal one to look for first. If we notice that we don't have perfect squares for this square roots possibility, that one jumps off the page at us.

It's a good idea to try factoring first. It ends up being very, very fast. And we'll learn other techniques later on that will help out too. But factoring is a nice process to work through. The zero product property is what allows factoring to work as a solution.

So let's just say we had two numbers, x times y equals zero. Well, if we wanted... To realize, or if we wanted solutions out of that, find out what x equals, find out what y equals, what could happen is, in order to make this product be zero, that would happen when x equals zero, or when y equals zero. So when we factor out a polynomial and get something like x minus 5 times x plus 3 equals zero, then what we're really saying is, oh, that would equal zero when x minus 5 equals zero, and it would it would create a product of 0 when x plus 3 equals 0. And then we can go about solving, find that x could equal 5 or x could equal negative 3. But the fact that they're multiplying together to equal 0 is why that process works. All right, so our first example, solve the equation by graphing it.

So remember when it's by graphing, we're looking for x intercepts. We want our function set equal to zero, so number four is set up beautifully for us. If I wanted to graph this out, well, maybe I think the fastest approach would be negative b over 2a.

So, whoops, x equals 1. I wanted an arrow there. So x equals 1 is our axis of symmetry. And if I plug in a 1, I get negative 1 plus 2. plus 3. So I have the point 1, 4. And we can reflect that across the Axis of symmetry. Now we want to see where this crosses the x-axis. So let's see what happens when x equals 2. So if I plug in a 2, I get negative 2 squared.

That's negative 4 plus 4 plus 3. I end up with a 3. Oh boy, oh boy, oh boy. This wasn't looking right. It started to look like an absolute value function.

I found my mistake. I was just moving a little too quickly. That'd be a lesson to you. is on the axis of symmetry.

That's our vertex. Then we have and we can reflect that across. And now let's see what happens when x equals 3. I've got negative 9 plus 6 plus 3 equals 0. Hey, and that does give me a height of 0 for an output. And I can reflect that across the axis of symmetry. There's our graph, and it's opening downward like it should, and it's bending like it should.

So the solutions to our equation are x equals negative 1 and positive 2. Let's try number eight. Change colors really quickly. So one thing I notice is that, hey, 3x squared, 6x, negative 3. I can divide both sides of that equation by 3. and make it simpler to work with. So I'm going to go ahead and do that.

x squared equals 2x minus 1, and it's not set equal to 0, so I can't start graphing this thing just yet. I need to see the function on one side of the equation if I want to graph it. So I have my x squared.

I'm going to subtract over 2x, add over the 1 to get it set equal to 0. All right, and this is actually one of the same equations we had at our initial, oh, in our introduction. So this one is actually factorable, but they say to do this by graphing. So let's go through the same approach we did in the last one. x equals negative b over 2a.

So again, x equals 1. And let's see what happens when we plug that into The function here for work in simplified. So if we plug in a 1, what we get is 1 minus 2 plus 1, and the output is 0. Now it's important to realize that when I divided both sides by 3, what I was really doing is I was taking away an important part of that graph. It was 3x squared.

So I want to be sure that when I draw my graph, it has that 3x squared. Sure, dividing by 3 made the algebra much easier, and it doesn't change our answer at all, but it does affect the graph. So let's assume that we didn't divide by 3. Because now that it's 3x squared, it's going to be 3 times as tall.

You know, that's going to affect our graph quite a bit. So now we've got our vertex. Now we can take a look at the rest of the equation here.

Let's see, if I plugged in a 2. Oh no, let's use 0 instead. 0 is always nice to plug in. If I plug in a 0 to the left side of that equation, I end up with negative 3. So at 0, oh, I'm sorry, plus 3, I end up with a positive 3. We can reflect that across our y-axis. You can see there's the 3 times as tall happening. And we would not have seen that had we used our simplified function over here.

This is just fine if we're going to solve by factoring. But I probably should not have done that since they wanted us to graph. Instead of dividing by 3, I should have simply used this for negative b over 2a and then made my table from there.

All right, last one to solve by graphing. I've got x squared. I want to get this set equal to 0, so I need to subtract over the 2x.

And then there's also a plus 2. That's set equal to zero. Let's find our axis of symmetry. We're just going to graph this out. So x equals negative b over 2a. Again, x equals 1. I realize all my examples had x equals 1 as an axis of symmetry.

It's kind of boring. All right, so there's our axis of symmetry. That means our vertex.

And we're not asked to find this, but just so you know what I'm writing on screen. 1, and then if I plug in a 1, I get 1 minus 2 plus 2. Hmm. So the vertex is the point 1, 1. Well, let's see what that graph's doing, because if it's crossing the x-axis anywhere, those are going to be our solutions. We have x-intercepts.

So let's try an x value of 0. I plug in zero, again a nice value to plug into the left side of this equation, zero minus zero plus two, and you can see our graph is actually bending away from the x-axis. So this is a situation where we have no solutions. Now there's one other thing. This is not part of your example, but one other thing that I did want to point out.

This is going to be a hypothetical situation, but let's just say that we had plotted out our vertex, and we got some points from our table of values, and we don't have nice, clean outputs. that are leading us to a height of zero. So you'll see how the x-intercepts, they're not at a nice clean point. Well if we wanted the solutions out of that graph, what we would say is that there's a solution between negative eight and negative seven, and another solution between, this looks like, what, negative 5 and negative 4. So when we're graphing by hand, not a lot of precision required if those points don't come out clean, but we can specify where those solutions are happening.

And that's always nice to kind of have on hand that you can check, too. You know, does your answer make sense, that type of thing. Alright, example two.

Solve the equation using square roots. So I want to get to a point where it's something squared equals a number. Well, I can see here in number 18, here's my something squared. Let's get that set up to equal a number.

So we'll start off by adding five to both sides. Dividing both sides by two. And now we can get to the point where we've got something squared equals a number.

Now we can square root both sides to undo that square. Square root cancels out the square. What we have left over is just what's underneath. That's called the radicand.

And it's going to be plus or minus the square root of 13 over 2. Now before I go any further, there's a concept called rationalizing the denominator. This root 13 over 2 is really equal to the square root of 13 over the square root of 2. And we don't like leaving square roots in the denominator. It's an interesting story about why that is the case. So what we do is we multiply that fraction by 1. And in this case, I'm calling that 1 root 2 over root 2. That way, when I multiply across the denominator, I'm taking root 2 times itself. I'm squaring that to undo the square root.

So across the top I get root 13 times root 2, which is root 26, all over a regular 2. And now there's no more radical in our denominator. We can use like terms much easier that way. Okay, so I'm going to apply that. x plus 2 equals plus or minus root 26 over 2. And to finish solving for x, we would simply subtract 2 from both sides.

And I'm going to call that 4 over 2. just to get a common denominator on the right-hand side. So x equals negative 4 over 2 plus or minus root 26 over 2, or we could call that x equals negative 4 plus or minus root 26 all over that common denominator of 2. It looks really bizarre, but we should not be alarmed if our solutions come out this way. This is just one of those cases where it comes out not entirely clean.

Alright, the next one. Solve using square roots. Get a different color here. Ooh, purple.

Alright. So, what we want to have is something squared equals a number. And if I were to subtract over this 1 fifth x squared, Combine it with the other x squared term on the other side, I get 2 equals 2 fifths x squared. Luckily, those fractions had common denominators, so we could add them pretty easily.

Now, I don't just have something squared equals a number. I need to get rid of that 2 fifths. So I'm going to divide both sides by 2 fifths, and dividing by a fraction means multiplying by the reciprocal on the other side.

And if I thought of that 2 as 2 over 1, across the top I get 10, across the bottom I get 2. We could even simplify those 2s. One's being multiplied into the numerator, one's being multiplied into the denominator. The division in the fraction undoes that multiplication. We have 5 equals x squared.

Now we can square root both sides to get our answer. Square root cancels out the square. x equals plus or minus square root 5. And that's a pretty nice answer.

We had to do a lot of work to clean up our answer on problem 18. Here we got to our solution, and it's in its proper form. Pretty quick one. Example 3, solve by factor.

This is something that we should be pretty comfortable with. We worked on it in the algebra refresher. It's something that you've seen for a couple years now.

So, let's get started. The function is part of an equation, so I want to keep this 0 equals. That's what makes our 0 product property work, if you remember that.

So, here we want to factor. Basically, we're going to unfoil this thing. Since we have a leading coefficient of 1, I know that my factors will both be z's, just single variables leading off.

That way, if I were to foil it back together, I would get back to where I started, a z squared, first times first equals the first term. Now I want to figure out what numbers multiply together to get positive 25, but also add up to get negative 10. And after a little bit of thought, negative 5 and negative 5 works. So now I've got the case where this could equal 0 or this could equal 0 to give me a solution.

And it's going to be the same thing in either case, z equals 5. If I took either of those factors and set them equal to 0 to get my solution, I would add 5 to both sides to get there. All right, for the next one, 32, a squared minus 49 equals 0. You'll notice there is no b term in there, so it's like a 0a in the middle. And a is squared, and then this 49 at the end is also a perfect square.

That is 7 squared. This is a special type of factoring situation called a difference of squares. And we can factor it just like we did before. You know, we can say, okay, it's a quadratic, so we've got our two factors set equal to zero.

The leading coefficient is one, so I know it's going to be a something times a something. What multiplies together to get negative 49 but adds up to zero? Well, positive seven and negative seven do the trick. But if you remember this difference of squares factoring, You can quickly tell that, oh, it's just the square root of the first thing leading off the factors, the square root of the second thing, the 49, ending our factors, and then one's positive and one's negative.

And if we solve out of the first factor, we get a solution of a equals negative 7, and out of our second factor, a equals positive 7. Now they deliberately asked us to factor this, so we have to see this step here in the middle. I know it's real tempting after the last section to add 49 to both sides and then square root both sides. And yes, that will get you the answer, but we're practicing our factoring techniques.

So a little picky on the details, but we want to be sure we have all these skills in place. Alright, number 34. Here you can tell we've got all the pieces of a trinomial, a typical quadratic function. We've got a regular number, we've got y terms, and we've got y squared terms. We need to combine those together and get them set equal to zero.

When we're factoring, it has to be set equal to zero. So I noticed I have positive 2y squared on the right, positive 1y squared on the left. So I'm going to move the y squared, start building this quadratic function on the right-hand side. So I'm going to subtract y squared from both sides. I'm going to subtract 28 from both sides.

And I'm going to add y to both sides. Now I've got 0 equals 1y squared. plus 3y minus 28. All right, let's get that factored out. We are very fortunate yet again. Leading coefficient is 1, so I know my factors open up with y's.

1 multiplies to get negative 28, but adds up, combines to get a positive 3 plus 7 minus 4. And if we're ever skeptical about whether we're right or not, we can always try and run through foiling to see if we get back to where we started. The solutions coming out of those factors, y equals negative 7 or y equals positive 4. And factoring is one of those skills that I hope we are doing. just automatic width, because that makes a lot of math in the future much, much easier. All right, example four, find the zeros of the function.

Pretty vague. They don't have us do this by graphing, though we could. I mean, if we wanted to make a thumbnail sketch here, we could factor it even a little bit. But what happens if we factor it? We get something that's not clean.

We have a factor of an x-intercept of zero and an x-intercept of negative 11. Oh, never mind, this one works beautifully. So if we want to find out where this equals zero, yeah, we were able to factor that right away. That g of x, that output, it means the same thing as y. We're letting that equal zero. No height.

Zero of the function. All right, and we quickly find out that our loose x, well, that's being multiplied, and it could produce a product of 0 if x equals 0. Or if x were to equal negative 11, that's what's coming out of that second factor to give us a solution. And then 52, this one's a little bit more interesting.

If we do 4, you'll notice our lead coefficient is not 1. And I know we covered this in our algebra refresher, but we really can't refresh this enough. So if we have that situation, we're trying to factor it. We're saying, okay, let's let f of x equal 0. Let's factor this right-hand side out.

4 times 9 is 36. What multiplies to reach 36 but adds up to 12? Huh. Well, we've got 1 and 36, 2 and 18, 3 and 12, we've got 4 and 9, we've got 6 and 6. Ah, 6 and 6, that must be it. So once we find those, our magic numbers that multiply to get 36 and add up to be negative 12, we just need to realize that they would both be negative, of course.

We divide them by... our leading coefficient of 4, and simplify them. So negative 3 halves, and we have the same thing over here, negative 3 halves. And to put this into a factor, it's really telling us x and negative 3 halves, x and negative 3 halves, Which, if it were in fully factored form, I mean, we can get our solutions right away. We know x equals positive 3 halves, and that's it.

But if we wanted to continue factoring this, just to bolster our knowledge, you'll notice if we tried to FOIL these factors together, we would start off with 1x squared. But that's not what we started with. So what we have to do is, these denominators of 2, they actually move out in front of the x.

That way, if we do first times first, we get 4x squared. If we do last times last, we get plus 9. And our inside numbers give us negative 6x. Our outside numbers give us minus 6 more x's.

So that's how we successfully factor that completely. But to get the solutions, we don't have to do all of that work. It's just kind of a nice little pattern to follow.

All right, example five. This word problem is pretty interesting. It kind of does an excellent job of setting up a real-world situation that if you were like a business owner or really, for any process that you're trying to accomplish, you can take data points and try and maximize your efficiency here.

And a store is just a really good example of that. So an athletic store sells about 200 pairs of basketball shoes per month. And now it only sells that number of pairs of shoes when it charges $120 per pair. For each $2 increase in the price, the store sells two fewer pairs of shoes. How much should the store charge to maximize monthly revenue?

Okay, let's think about this graphically. So what do we want our inputs to be? What do we want our outputs to be?

you'll notice that The thing that we can control that's being changed is the increase in price. So let's let price be our x-axis, and then our y-axis will represent the revenue in a given month. And we could, I guess, have it be pairs of shoes.

if we wanted. But what do we know? The chapter is about intercepts.

If we charge zero dollars per pair, then we're not going to make any money. And that's why I think revenue makes a better y-axis, because we could be giving away shoes, but the store is not going to make any money. Then we have this graph that's peaking at some price, and then sales start to drop off.

It becomes too expensive for the customers to buy shoes. So we know that x equals 0 is an x-intercept of this graph. And we also know that there's some other point where the price is so high that customers won't pay that.

Let's see if we can't find the pattern to locate that point. So let's think. If we charge $120, then we will sell 200 pairs of shoes in a month. If we were to raise that to $122, we would only sell 198 pairs of shoes. If we raised it again to $124, we would only sell 196 pairs of shoes.

And if we continue this process on, you'll notice that we can only increase the price, let's see, what is it going to be, 100 times. before we get to zero pairs of shoes being sold. So 100 times that we add $2 to the price, that adds $200 total.

$320 means we sell zero pairs of shoes. So here we've got x equals 320. Now because of the symmetry of a parabola, we know that Thank you. Halfway between those two points, or at x equals 160, that's where our vertex is going to be.

That's where our maximum will occur. So let's look at our function. Our revenue function based on a certain price per pair of shoes is equal to, we'll use intercept form here, x times, well I should, sorry, I should do.

is equal to um Sum number a times x times x minus 320. And this loose x is really just representing the factor x minus 0 from our first x-intercept. So now we can actually find the function. This beautifully ties in what we learned last chapter.

Our revenue is going to be 120 times 200, right? They tell us somewhere over here on the graph, we don't know where that point is, when the price is 120. The revenue is 120 times 200, $24,000 made that month. So the revenue, 24,000, equals some variable a times x minus 320. And we know that that x value is 120. So here, why don't I go ahead and erase that.

Use a little bit of foresight here. All right, so we got a times 120 times 120 minus 320. a times 120. times that's negative 200. And what we end up with is 24,000 equals A times negative 24,000. If we divide both sides by negative 24,000, we find A equals negative 1. And it makes sense that A should be negative because our parabola is opening downward, and we're finding a maximum, so that should all kind of lead us to believe that we're setting this up correctly.

So again, our revenue function is negative x times x minus 320, since we know a value is negative 1. Then we want to find how much should the store charge to maximize its monthly revenue. Well, we found that to be $160 per pair. Now let's figure out what that monthly revenue would be.

So now we need to plug in 160. We need to find R of 160. That's negative 160 times 160 minus 320. 20, which is 160. Oh, sorry, negative 160. Yeah, we don't want to end up with a negative revenue. That's a bad business. Yeah, it should be a negative times a negative.

So our revenue at $160 per payer is $256,000. $25,600 per month. But again, this is not the only way to do this problem.

I just saw using the x-intercepts as being the most beneficial, and for me it made it easiest to do by hand. A calculator can also do a nice job with this, as long as you know what direction you're headed in. That would definitely help with this part of the problem where we're trying to find out what that other intercept is. But yeah, a nice little problem.

I like how it ties together. Our second difference idea, knowing what inputs and outputs should be, and analyzing a function with real life inputs. Very nice.

Example 6. You drop a seashell into the ocean from a height of 40 feet. Write an equation that models the height h in feet of the seashell above the water after t seconds has gone by. How long is the seashell in the air?

Well, we don't have enough information right now to work with this. It's kind of working off the assumption that our y value in height is equal to negative 16t squared plus y initial. And I could have used h's there, representing the initial height, the starting height. And the textbook gives you this equation in one of its little examples that it provides.

But it is assumed that you know. this type of setup for dropping an object on Earth. And we saw something like this in, what was it, Chapter 1, when we dealt with the, maybe it was Chapter 2, dropping the object on Earth and on the Moon.

We've seen this before. So our initial height, we know that that initial height is 40 feet, so we can go ahead and plug that right into our function. What they want to know is how long is the seashell in the air? So let's see what that function would look like.

Well, it starts off at 40 feet, and then as time goes by, here's our time ticking away on the clock, as time goes by that object is falling faster and faster towards the ocean. What we need to do is find the zero. So we want to find where is the height, where is the y value zero. in this equation.

And you can notice this is one that we can solve using square roots, so let's add 16t squared to both sides of the equation. Isolate the square term, that means dividing both sides by 16. And then we want to take the square root of both sides. If you're sitting there thinking, Mr. Hartz, Mr. Hartz, you can simplify this.

Yes, I do realize that it's something that can totally be simplified. But if I'm going to square root a fraction, it's helpful to have the denominator be a perfect square. And 16 is a perfect square. No, but you know what? I can divide the top and bottom cleanly by 4, which will give me another perfect square denominator.

So let's simplify it like this. I know we could go further, but let's leave it here. If we square root both sides of that equation, we find that t equals the square root of 10 over the square root of 4. And remember, plus or minus. But look at our graph.

Yeah, our parabola, if we graphed the entire thing, would go negative and positive. But since we're talking about time going by, we only want the positive result. We can't go back in time.

So t equals root 10 over 2. seconds until that lands. And we could use a calculator to find a decimal if need be, but this is an exact answer, which are definitely preferable. Okay, that does it for section 3.1.

Kind of a long lesson, plenty of examples, but really nice to be working algebraically, not as much tedious graphing as before.

Transcript for:Quadratic Equation Methods

Transcript for:
Quadratic Equation Methods