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. 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.
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.
Example 1 We are going to
fence in a rectangular field and we know that for some reason we want to field
to have and enclosed area of 75 ft2. We also know that we want the width of the
field to be 3 feet longer than the length of the field. What are the dimensions
of the field?
Solution
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. Now, we also know that area of a rectangle is length times width
and so we know that,
x (x + 3) = 75
Now, this is a quadratic equation so let’s first write it in
standard form.
x2 + 3x = 75
x2 + 3x – 75 = 0
Using the quadratic formula gives,
Now, at this point, we’ve got to deal with the fact that there are
two solutions here and we only want a single answer. So, let’s convert to
decimals and see what the solutions actually are.
So, we have one positive and one negative. From the stand point of
needing the dimensions of a field the negative solution doesn’t make any sense
so we will ignore it.
Therefore, the length of the field is 7.2892 feet. The width is 3
feet longer than this and so is 10.2892 feet.
Notice that the width is almost the second solution to the
quadratic equation. The only difference is the minus sign. Do NOT expect this
to always happen. In this case this is more of a function of the problem. For a
more complicated set up this will NOT happen.
Now, from a physical standpoint we can see that we should expect to
NOT get complex solutions to these problems. Upon solving the quadratic
equation we should get either two real distinct solutions or a double root.
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.
Let’s work another example or two.
Example 2 Two cars start
out at the same point. One car starts out driving north at 25 mph. Two hours
later the second car starts driving east at 20 mph. How long after the first
car starts traveling does it take for the two cars to be 300 miles apart?
Solution
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.
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.
distance
of car A : 25t
distance
of car B : 20 (t – 2)
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.
So, we have a right triangle here. That means that we can use the
Pythagorean Theorem to say,
(25t)2 + (20(t − 2 ))2 + 20 = (300)2
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.
625t2
+ (20t – 40)2 = 90000
625t2
+ 400t2 – 1600t + 1600 = 90000
1025t2
– 1600t + 88400 = 0
Now, the coefficients here are quite large, but that is just
something that will happen fairly often with these problems so don’t worry
about that. Using the quadratic formula (and simplifying that answer) gives,
Again, we have two solutions and we’re going to need to determine
which one is the correct one, so let’s convert them to decimals.
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, even though the problem didn’t ask for it, the second car
will have traveled for 8.09998 hours before they are 300 miles apart. Notice as
well that this is NOT the second solution without the negative this time,
unlike the first example.
Example 3 An office has
two envelope stuffing machines. Working together they can stuff a batch of envelopes in 2 hours. Working separately it will take the second machine 1 hour
longer than the first machine to stuff a batch of envelopes. How long would it
take each machine do stuff a batch of envelopes by themselves?
Solution
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.
The word equation for this problem is then,
We know the time spent working together (2 hours) so we need to
work rates of each machine.
Here are those computations.
Note that it’s okay that the work rates contain t. In fact
they will need to so we can solve for it! Plugging into the word equation
gives,
Using the quadratic formula gives,
Converting to decimals gives,
Again, the negative doesn’t make any sense and so Machine A will
work for 3.5616 hours to stuff a batch of envelopes by itself. Machine B will
need 4.5616 hours to stuff a batch of envelopes by itself. Again, unlike the
first example, note that the time for Machine B was NOT the second solution from the quadratic without the minus sign.
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