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Algebra - Notes

Internet Explorer 10 & 11 Users : If you have been using Internet Explorer 10 or 11 to view the site (or did at one point anyway) then you know that the equations were not properly placed on the pages unless you put IE into "Compatibility Mode". I believe that I have partially figured out a way around that and have implimented the "fix" in the Algebra notes (not the practice/assignment problems yet). It's not perfect as some equations that are "inline" (i.e. equations that are in sentences as opposed to those on lines by themselves) are now shifted upwards or downwards slightly but it is better than it was.

If you wish to test this out please make sure the IE is not in Compatibility Mode and give it a test run in the Algebra notes. If you run into any problems please let me know. If things go well over the next week or two then I'll push the fix the full site. I'll also continue to see if I can get the inline equations to display properly.
Preliminaries Previous Chapter   Next Chapter Graphing and Functions
Quadratic Equations : A Summary Previous Section   Next Section Equations Reducible to Quadratic in Form

 Application of Quadratic Equations

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 the field to have an 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?



So, we’ll let x be the length of the field and so we know that  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,



Now, this is a quadratic equation so let’s first write it in standard form.



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?



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  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.



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,



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, 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 to stuff a batch of envelopes by themselves?



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)  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,




So, to solve we’ll first need to clear denominators and get the equation in standard form.



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.

Quadratic Equations : A Summary Previous Section   Next Section Equations Reducible to Quadratic in Form
Preliminaries Previous Chapter   Next Chapter Graphing and Functions

Algebra (Notes) / Solving Equations and Inequalities / Applications of Quadratic Equations    [Notes] [Practice Problems] [Assignment Problems]

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