Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Application problems of quadratics canoeing during the first part of a canoe trip. Tim covered 60 kilometers at a certain speed, so his distance was 60 at a certain speed. He then traveled 24 kilometers at a speed that was 4 kilometers an hour slower, so X -, 4. The total trip was 8 hours. Well, if we know distance equals rate times time, we can think of time being distance divided by rate. So in this box here I'm going to put 60 / X and in this one I'm going to put 24 / X -, 4 these two times total. So one of the times and the other time we're going to add them together because they're the total time is going to equal 8. So now we need to solve this and we're going to solve it by getting our common denominator of X * X -, 4, keeping in mind that X could never be 0 or 4 because that would give us a zero in the denominator. So when I multiply everything through by the common denominator, we're going to get 60 * X - 4 + 24 * X equaling 8X times X - 4. If you don't see what's happened there, we multiplied each and every term by X * X -, 4. And what'll happen in these first terms is the X cancels leaving us 60 * X - 4 and the next 1, the X minus fours cancel, leaving 24 * X. And then the last one, nothing cancels. So we have 8X times X -, 4. We're going to then distribute. So sixty X - 240 + 24 X equal eight X ^2 - 32 X. So if we combine our like terms, we get 84 X -240 equal eight X ^2 -, 32 X. We're going to take everything to one side. So if we take everything to one side, we're going to get 0 equaling eight X ^2 -, 116 X plus 240. Now I like to keep my numbers as small as possible. All those coefficients are divisible by a four. So I'm going to get 0 equal two X ^2 -, 20 nine X + 60. From here that might factor, but that seems like it might be a lot of work. Hence the section is about quadratic formula. So I could just plug it into the quadratic formula X equal the opposite of B plus or minus the square root b ^2 -, 4 * a * C all over 2A. And the nice thing about the quadratic formula is it always works even if the original equation didn't factor. So we get X equal 29 plus or minus. The square root -29 ^2 is 841 and 4 * 2 * 60 is 480 / 2. So X equal 29 ± sqrt 361 / 4. So that's 29 + 19 / 4 plus or minus. So 29 + 19 is going to be 48 / 4 or 12, and 29 - 19 is 10 / 4 or five halves. Now that's what X equals. But we also need to know X - 4. So if X equal 12, then X - 4 would have to equal 8. If X equal 5 halves, then X -, 4 would equal five halves -4 or -3 halves. That one doesn't make any sense because it's a negative. So our answers are going to be 12 kilometers per hour on first part and eight kilometers per hour on the second. So Tim covered or Tim traveled 12 kilometers per hour in the first part and 8 kilometers per hour on the second part. Now we actually do need to check this. So does 60 / 12 + 24 / 8 equal 860 / 12 is 5, 24 / 8 is 3. So it does. Check this next example. Air travel. A turbo jet flies 50 kilometers or 50 mph faster than a super prop. So X + 50 and X, we're going to let this be the turbo and this one be the Super prop. The turbojet goes 2000 miles and the Super prop goes 2800. So once again, if distance equals rate times time, then the time equals distance divided by rate. So we're going to have 2000 / X + 50, and we're going to have 2800 / X. Now this time it tells us that the turbojet goes 2003 hours less time than it takes the Super prop to go 2800. So the time for the turbo is less in order to make them balanced. If this is less, I'm going to add to it to get it to equal the bigger. I could have also subtracted from the Super prop would have done the same thing. Our common denominator X * X + 50. Remembering X can never equal 0 or -50 because we'd have zero in the denominator there. So when I multiply through by the common denominator, we're going to get 2000X plus three X * X + 50 equal 2800 * X + 50. So 2000 X plus three X ^2 + 150 X equal 2800X plus 140,000. Taking everything to one side, we're going to get three X ^2 -, 650 X -140,000 equaling 0. Plugging that into my quadratic formula, it may factor, it may not, but with the quadratic formula, doesn't matter if it factors. So the opposite of b ± sqrt b ^2 - 4 * a * C all over 2A. If I plug and chug that through I get X = 650 ± sqrt 2 one 02500 over six. The square root of that big number is really just 1450, so I'm going to get X = 650 + 1450 / 6, which is really just 350. Or I'm going to have X = 650 -, 1450 / 6, which is -400 thirds. Now, because we're talking about a rate, this one doesn't make any sense because it's a negative. So 350 is our X. So our X + 50 would have to be 400 to do our check. 2000 / 400 + 3. Does that equal 2800 / 350? Well, 2000 / 400 is 5 and 2800 / 350 is 8, so it does indeed check. So the speed of each plane, the speed of turbo is the turbo. Turbo was 400 mph and the speed of super prop is 350 mph. Looking at another example, we're going to look at the Hudson River flows at a rate of three mph. A patrol boat travels 60 miles upstream and returns. So if it returns, it's also going 60. So we have upstream and we have downstream. Well, if we're flowing with a river upstream, we're going against the current and downstream it's pushing us a longer, we're going faster. So it we're going with the current against is going to be -3 and width, it's going to be +3 the time again, distance equal rate time times. So time is distance over rate. So 60 / X - 3 and 60 / X + 3. It's a total time, so the time to go up plus the time to go down has to equal a total time of nine. The common denominator X - 3 X plus 3, where X can never equal 3 or -3. So we'd get 60 * X + 3 + 60 * X - 3 equal 9 * X - 3 X +3. So sixty X + 180 + 60 X -180 equal nine X ^2 - 81. Taking everything to one side, we're going to get 0 equaling nine X ^2 -, 120 X -81. I like, if at all possible, to keep my coefficients as small as possible. So I'm going to divide everything through by a three, and I'm going to get three X ^2 -, 40 X -27. If we plug all of that into the quadratic formula, X equals the opposite of B plus or minus the square root b ^2 -, 4 * a * C all over 2A. So X is going to equal +40 ± sqrt 1924 all over six. So keep simplifying. Oh, sqrt 191 thousand 924 doesn't simplify. So if we just grabbed our calculators at this point to give ourselves approximations, those two answers would be our exact. An approximation would be 13.98 and -.64. Does negative make any sense? If we're talking about speed of a boat in Stillwater, no. So our answer is going to be about 14 mph. Now, this one's a little harder to check, and the check would be more of an approximate unless we use the exact. So if we wanted to check it, we'd get an approximate, not an exact. We would put 14 in and we would realize that this is really close. So 60 / 11 + 60 / 17 is close. There was some rounding errors going on there. So the speed of the boat and Stillwater is about 14 mph. The next example is going to be a work problem. We're going to have two pipes are connected to the same tank working together. They can fill the tank in two hours. So one tank in two hours. The larger pipe working alone can fill it in three hours less time than the smaller ones. So X -, 3 is larger pipe time, and plain old X is smaller pipe time. So 1 / 2 is going to equal 1 / X - 3 + 1 / X the common denominator 2X times X - 3, where X can't ever equal zero or three. So we're going to get X * X - 3 equaling two X + 2 * X - 3. If we distribute those out and take everything to one side, we're going to get X ^2 - 7, X plus 6 = 0. And that's one that we might see factors pretty easily -6 * -1 gives us 6, and they add to give us -7 So X = 6 and X equal 1. But we don't need just X, We also need X - 3. Well, if X is six, X -, 3 is 3. But if X is one, X - 3 = -2. Does that make any sense for time? And hopefully we're saying no. So our answer has got to be the six and the three. So 1/2 would equal 1 / 3 + 1 / 6. Getting a common denominator 3/6 equals 2/6 plus 1/6 so it checks. So the how long would the smaller one take? So the smaller pipe will take six hours. We also are going to look at solving equations for variables. So if we wanted to solve for S, we would divide each side by 6, and then to get rid of that square, we're going to square root. Because these are actual formula equations, we're going to look at the positives here. If we wanted to solve for R, we're going to cross multiply to start to get rid of our denominators. Then we're going to divide by F to get the R by itself. And to get rid of the R-squared, we're going to square root. So R is going to equal the square root of GM1M2 all over F. This next one we're going to cross multiply to get rid of our fraction. Then we're going to take everything to one side. This one, we're not going to be able to just factor. Because we have AK squared and AK. So we're going to actually use quadratic formula here. If we're using the quadratic formula, we realize our a is going to be one. It's the coefficient on the K ^2. Our B is going to be -3. It's the coefficient on the K Our C is going to have a variable in it. It's going to be a -2 N So K is going to equal the opposite of B plus or minus the square root b ^2 -, 4 * a * C all over 2A. When we simplify that up, we get 3 plus or minus the square root 9 + 8 N all over 2. And that's our answer. Now, once again, these are formula problems. So we really want the plus, SO3 plus square roots of 9 + 8 N all over 2. If we wanted to be safe, we could always give plus and minus on all of them. And then look at what the actual situation is as to whether we need the positive or the negative on #8. We're going to do the same thing. We're going to take everything to one side, so 0 equal π r ^2 + π Rs minus a. We're going to use quadratic formula, so the coefficient on the R-squared term is π. The coefficient on the plane R term is π S and our constant's going to be negative a. So R equal the opposite of B plus or minus the square root of that b ^2 -, 4 * a * C all over 2A. So we get negative Pi s ± sqrt π ^2 s ^2 + 4 Pi a all over 2π. We can't pull out a sqrt π ^2 s ^2 because it's not a monomial. We can't do a sqrt 4 because it's not a monomial. So that is our final answer We're going to talk about next. An object being tossed downward with an initial velocity of V naughty will travel a distance of S meters, where s = 4.9 T squared plus V naughty t + t and T is measured in seconds. So a ring is dropped from a helicopter at an altitude of 75 meters, so our S is going to be 75. Approximately how long does it take the ring to reach the ground? So we're trying to find the time. If it's just being dropped, it's not being thrown. It's going to have a zero velocity and 0 * t is really just going to give us 0. So we get 75 equal 4.9 T squared. We'd have 75 / 4.9 equaling t ^2 and we would square root to get T equal sqrt 75 / 4.9. If we put that into our calculator, we'd get approximately 3.9 seconds. Part B says a coin is tossed downward at initial velocity of 30 meters per second from an altitude of 75 S. Our height is now 75 four point nine t ^2 and our velocity is 30 going downward. And the original equation said that V naughty was being tossed downward. So we're going to take everything to one side and then we're going to use the quadratic formula. So T equal the opposite of b ± sqrt b ^2 -4 * a * C all over to A. If we grab our calculators and put that in, we'd get approximately 1.9 seconds. The last part of this equation says approximately how far will an object fall. So now we're looking for the distance if it's falling for two seconds with an initial downward velocity of 20. So S equal 4.9 * 2 ^2 + 20 * 2. If we grab our calculators, we get 59.6 meters. Thank you and have a wonderful.