*Let \(t\) be the amount of time it takes the first machine (Machine A) to stuff a batch of envelopes by itself.That means that it will take the second machine (Machine B) \(t 1\) hours to stuff a batch of envelopes by itself.*

Also, as the previous example has shown, when we get two real distinct solutions we will be able to eliminate one of them for physical reasons. We’ll start off by letting \(t\) be the amount of time that the first car, let’s call it car A, travels.

Since the second car, let’s call that car B, starts out two hours later then we know that it will travel for \(t - 2\) hours.

\[\begin25t\\ 20\left( \right)\end\] At this point a quick sketch of the situation is probably in order so we can see just what is going on.

In the sketch we will assume that the two cars have traveled long enough so that they are 300 miles apart. That means that we can use the Pythagorean Theorem to say, \[ = \] This is a quadratic equation, but it is going to need some fairly heavy simplification before we can solve it so let’s do that.

Now, we know that the distance traveled by an object (or car since that’s what we’re dealing with here) is its speed times time traveled.

So we have the following distances traveled for each car.Also, we are going to assume that you can do the quadratic formula work and so we won’t be showing that work.We will give the results of the quadratic formula, we just won’t be showing the work.\[t = \frac = 10.09998\hspace\hspace\hspace\,\,\,\,\,\,\,\,\,\,\,t = \frac = - 8.539011\] As with the previous example the negative answer just doesn’t make any sense.So, it looks like the car A traveled for 10.09998 hours when they were finally 300 miles apart.Also, as we will see, we will need to get decimal answer to these and so as a general rule here we will round all answers to 4 decimal places.So, we’ll let \(x\) be the length of the field and so we know that \(x 3\) will be the width of the field.From the stand point of needing the dimensions of a field the negative solution doesn’t make any sense so we will ignore it. The width is 3 feet longer than this and so is 10.2892 feet. In this case this is more of a function of the problem.Notice that the width is almost the second solution to the quadratic equation. For a more complicated set up this will NOT happen.In this section we’re going to go back and revisit some of the applications that we saw in the Linear Applications section and see some examples that will require us to solve a quadratic equation to get the answer.Note that the solutions in these cases will almost always require the quadratic formula so expect to use it and don’t get excited about it.

## Comments Applications And Problem Solving

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