Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Applications Air travel. A turb jet flies 50 miles an hour faster than a super prop plane. If a turbo jet goes 2000 miles in three hours less time than it takes the Super prop to go 2800, find the speed of each plane. So the first thing we're going to do is we're going to call one of these lines turbo and one super. So the distance for the turbo was 2000 and the other one 2800. We don't know the time for the turbo. I'm just going to call it T turbo and T super. And the rate we know that the turbo went 50 faster than the Super. So if we call X the Super, we know that the turbo is 50 faster. So now we can solve for T turbo. Time of the turbo is 2000 / 50 + X because rate times time equal distance and time of the Super is going to be 2800 / X. And we were told that the turbo takes three hours less time. So if you thought about this, if it took three hours less time, let's say it took two hours. So the other one had to take 5 S in order to get these balance. We're going to take the time of the turbo, which was less amount of time and add two of the three to get the time of the Super to keep those balanced. So the time of the turbo 2000 / 50 + X +3 is going to equal 2800 / X. Looking at our common denominator, we get 50 + X * X and we know that X isn't going to equal -50 or zero as that would give us a zero in the denominator. So when we multiply, we get 2000X plus 150X plus three X ^2 equaling 2800 * 50, which is going to be 140000 plus 2800 X. So I just multiplied each and every term by 50 + X * X and I went ahead and distributed. So the 50 + X cancels in this first one, leaving us 2000 X 50 + X * X If we thought about distributing their fifty X + X ^2 times the three and then the XS cancel for this last term 2800 times that. Now to solve this, we're going to take everything to one side so that we get zero on the other side. So three X ^2 -, 650 X -140,000, and now we're going to use quadratic formula that says X equal the opposite of B plus or minus square root b ^2 -, 4 * a * C all over 2A. And if I grab my calculator and plug it in, I get -400 thirds. And 3:50, we're talking about rates. So -400 thirds didn't make any sense. So if X is 350, then 50 + X is 400, and that's mph. The next one talks about the navigation of the Hudson River at a rate of three mph. So if we're going upriver and downriver, going upriver we're going against the current. So X -, 3 and downriver we're going with it. So the down is pushing us along or +3. We go 60 miles an hour up and 60 miles an hour down. So we have time of up and time of down South. Time of up is going to be 60 / X -, 3 and time it down is going to be 60 / X + 3 and then it tells us the total time. So the time up plus the time down is going to equal 9. So our common denominator here X -, 3 X +3. When we distribute, we get sixty X + 180 + 60 X -180 equaling nine X ^2 -, 81. If I take everything to one side, we get nine X ^2 -, 120 X -81. If I plug that into my calculator quadratic formula first, sorry, and then the calculator X equals the opposite of b ± sqrt b ^2 -4 * a * C all over 2A. So when I compute this, I'm going to see that X is going to give me 20 ± sqrt 427 thirds. Well, that doesn't really give us much of an answer. So we're actually going to do approximates here. And if I put this into my calculator, I get approximately -.22 and 13.55. Well, negative .22 doesn't make any sense because we're talking about a rate. So it's about 13.55 mph or about 13 1/2 mph. The next one we're going to let you try. We're going to have a work problem. Two pipes are connected to the same tank. Working together, they can fill the tank in two hours. The larger pipe working alone can fill the tank in three hours less time. So let's let X - 3 be larger tank, which means X's smaller tank, and we're going to get one job in X -, 3 hours. They're working together, so we're going to add one job in X hours equals one job in two hours. Our common denominators 2 ** -3. So 2X plus two X - 6 = X ^2 - 3 X. Taking everything to one side, we get X ^2 -, 7 X +6 quadratic formula X equal opposite of B plus or minus the square root b ^2 -, 4 * a * C all over 2A. When we compute this out, we can see that X is going to be one or six. Now we have X, but we also have X -, 3. I can't take three away from 1:00 to get a number that's positive. So if I think about X -, 3 with six, if I take 3 away from six, I get three. So this number didn't work. So one is going to take six and the other is going to take three. The smaller take takes six hours. The bigger tank takes three hours. Now this one actually could have been factored, but I wanted to practice the quadratic formula. The next two examples are going to be solving for a variable using the quadratic formula. So the first thing I'm going to do is I'm going to multiply to get rid of the fraction. I always, I'm going to get rid of the fraction. I'm going to take everything to one side and then I'm going to pay attention. And with the quadratic formula, my a here is going to be 1, my B is going to be -3, and my C is going to be -2 N. It's the whole thing without the K because we're solving for K. So K equals the opposite of b ± sqrt b ^2 - 4 * a * C all over 2A. So K is going to be 3 ± sqrt 9 + 8 N all over 2. And that's as far as I can go. The 9 + 8 NI can't combine. They're not like terms. Can't take the square root of either of them because it's not a single monomial. This next one we need to have zero on one side, so Pi r ^2 - π S I'm going to put the R at the end because that π S is going to all be the coefficient on the R term minus a. So here for the quadratic, my a is π, my B is negative π S and my C is negative a. So we're solving for R the opposite of B plus or minus the square root of B squared -4 * a * C all over 2A. So R is going to be positive π S plus or minus the square root π ^2 s ^2 + 4 a π all over 2π. I can't combine the things underneath the square root because they're not like terms, and I can't take the square root of a portion of it because it's not a monomial. So that's our final answer. Thank you and have a wonderful day.