 This video will talk about quadratic applications. If a number is added to three times its square, the result is 30. So let's go through and see what we know. If a number is added to three times its square, the result is 30. So let's see if we can translate that. A number would be x is added to, so something's in front of that, three times its square. So three times and its square means that it would be x squared. And then the result is 30. So we can now look at this and say, oh, it's not equal to 0. So we need to subtract the 30 from both sides. So 3x squared plus x and then minus the 30 would be equal to 0. And if I take my x factor again and we take our a times our c, that's going to give me negative 90. And I take my b and that is positive 1. Now you can see that we've got this negative 90 that we need to, we've done negative 90 before. One of them is going to be negative and one of them is going to be positive. The bigger number has to be positive and the difference has to be 1. So that was our 10 and our negative 9. But this time, a was not 1. So we have to do the four terms. 3x squared and minus 30. And then in the middle, we had our minus 9x and plus 10x. Remember we said that if you let the negative one come first when this third term is positive, nice things happen. So let's factor. Here we have 3x squared minus 9x. They both have 3 and x in common. It's the greatest common factor. I need one more factor of x to have 3x squared when I multiply. And 3 times negative 3 would give me negative 9. And x is already on the outside. So that one factors that way. And then I look at positive 10x and negative 30. Remember we like to take the same sign if possible. So let's see if a positive 10 gets us what we want. If I take a positive 10, I still need the x. So I have x on the inside and positive 10 times negative 3 would give me minus 30. And when I look at those two things in parentheses, they now match, which is what I wanted. So we do the greatest common factor again. Remember that's what we did here. So our greatest common factor is going to be the x minus 3. And the other factor is going to be the 3x. I've already taken care of this factor over here. So plus 10. This is my final factorization. But let's remember I didn't do it up here. But let's remember it's always equal to 0. So we say x minus 3 equals 0. If we finish it out, adding 3 to both sides, x would equal positive 3. And if we have 3x plus 10 is equal to 0, then we subtract the 10. 3x is going to be equal to a negative 10. And then we divide by 3. So x is going to be equal to negative 10 over 3. So it says find the numbers. So the numbers are either numbers 3 or negative 10 thirds. So let's look at this next problem. And the next problem is talking about the width of a rectangular picture is 7 inches less than the length. And then the area of the picture is 78 inches squared. So we know something about area. And we know the width is 7 inches less than the length. So let's fill in what we asked. The formula area is length times width. And then it asks us which dimension length or width are we making the comparison to? Well, width is being compared to the length. So we're comparing to length. So the next question says, OK, well, give it a variable. Well, why not L? Then we'll know it's length when we solve for L. Then it says represent the other one in comparison. So I'm going to come back up here and read this sentence again. The width is, that means equal, 7 inches less than. So remember, less than means switch the order. So let's say width is 7 inches less than. So minus 7 over here. But what is it less than? It's less than the length. So W equal L minus 7. So set up the equation for area. And then we want to solve. So the area it said was 78 inches squared. So area 78 is equal to, let's start out with just a formula, length times width. So if I fill in, I already have my 78, but my length is L. And my width over here is L minus 7. And I can solve. So 78 is equal to L squared minus 7L. And remember, we want it to be equal to 0 to be able to solve it. So we're going to subtract the 78 from both sides. So 0 is equal to L squared minus 7L. And then minus 78. So when I do my x factor, I have a negative 78 up here. And I have a negative 7 here. And I really don't know what the factors of negative 7 d8 are. So I'm going to come back in here, go to my y equal, clear that out, and say negative 78. I'm going to divide it by my x. And then I'll look at my table. And in my table, I want a difference of negative 7. And there it is, 6 and negative 13. And if I add those, I actually get negative 7. So I know my signs are exactly the same. So I have a negative 13 and positive 6. I can say 0 is equal to L plus 6, just be positive here, and L minus 13. And if I set them equal to 0, each one, L plus 6 equals 0. I would subtract 6 from both sides. And L would be equal to negative 6. And if I have L minus 13 equals 0, and I would add 13 to both sides, then that would tell me that L was equal to 13. So now I know that the length is 13, because the length of negative 6 doesn't make sense. So I know that the length is 13. Well, how do I find the width? We've done that before. We go back up and plug it into this equation. So width is equal to my 13 minus 7. And 13 minus 7 is 6. And remember, with applications, we really like the right sentences. So the length, it is in inches. So length is 13 inches. And width is 6 inches. All right, so our final problem. A rocket is launched straight up into the air for the ground with an initial speed of 64 feet per second. The height of the rocket in feet is given by the equation h is equal to negative 16 t squared plus 40 t, where t is time in seconds after the launch. t has to be greater than 0 because it's time. And then find the times when the rocket is ground level. So there's a bunch of interesting things in here. This is going to be a nice set of fact here. And t is in seconds. And height is in feet. And then we also have this wonderful equation here that we'll probably need. And believe it or not, the 64 feet per second really isn't important. So it asks me to draw a picture. So I've got this little rocket here. And it's going to go up in the air. But then, of course, it's going to come back down. And it's going to have that parabola look. And this is my ground. So we already have an equation. So what variable are we solving for? Well, it says find the times. So we are solving for t. Always go and look what they're trying to find. That'll tell you the variable you need. So we're solving for t. So a circle of places on this picture where your solution would be. And we want to know when the rocket is on ground level. Well, it started on the ground level, it looks like. And it's going to land on ground level. So it'd be those two. So how do we solve this equation? Well, the ground would equal what? How many feet is the ground from the ground? It's 0 feet. And we want to know when it's on the ground. So the ground is going to be equal to my equation. Because the height on the ground is 0. And then I have negative 16t squared plus 40t. And the nice thing about this one is it's already said equal to 0. So I don't have to push anything over to the side. And it's two terms. And it's not a difference of squares. So my only option there is that it might have a common factor. And it does. It looks like it has 16 would be either 4 times 4 or 8 times 2. And 40 would be 4 times 10 or 8 times 5. So it's going to be an 8. And since we have a negative here, let's take out a negative 8. And then we have got a t squared and a t. So we're going to take out a t. And then that leaves us with negative 8 times positive 2 would give me negative 16. And I need one more factor of t. And then negative 8 times negative 5 would give me positive 40. And then the t is already on the outside. So those are my two factors. Then I'm going to set equal to 0. On negative 8t equal to 0, dividing by negative 8, I'm going to end up with t equal 0. And then I do this one. 2t minus 5 is equal to 0. I add the 5 to both sides. So 2t is going to be equal to positive 5. This reminds me that it's positive. And then I divide by 2. And t is going to be equal to 5 over 2. If you don't like that, we could also say that that would be 2.5. And remember that these are in seconds. So to answer the question, the rocket is on the ground at 0 seconds and 2.5 seconds.