Question 7179
First of all, it is a requirement and they will tested on it, and how they'll do has a significant factor in determining where they'll end up in life later. Secondly, it doesn't hurt to know.


Quadratics do play a role, for example, in area and construction problems.


Example: Suppose that you were told that a piece of land had a width that was 15 feet wider than its length, and that the area was 5800 square feet. Could you determine what the dimensions of that land were?


With that problem, you'll be ending up solving the equation {{{ L(L+15) = 5800 }}} or further worked out, {{{ L^2 + 15L - 5800 = 0 }}} which you would need to factor to get {{{ (L - 80)(L + 95) = 0 }}}.


I hate to say this but even if students are convinced that the concepts can be used in a given real life situation, they will still complain UNTIL they find the need to use it DIRECTLY in their own personal lives.


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If they're complaining about completing the square (another concept that many students find ridiculously tedious and ultimately unnecessary), here's a problem:


Suppose that you have a bus, and you're renting it to organizations. Your bus can seat 80 people. You decide to charge the first person $30.00. If that person brings another person, you charge both of them $29.75 each. If there are three people, you charge $29.50 each. In other words, You charge EVERYBODY $0.25 less for every person who joins in. The thing is, there will come a point when you've got enough people to ride that you'll start losing profit if you add more people to the deal. With your starting price and discount price per additional person, will you: A) maximize your profit on the 80th person who rides? B) Be losing profit before you fill your bus, or C) would not yet reach your maximum profit on the 80th passenger?


So the total charge is dependent on how many people there are, how much you charge them, and what type of incentive you give depending on how many people there are. So, if there's one person, your total profit is $30.00. If there are two people, that's 2 people * ($30.00 - 0.25(1)). If there's 3 people, your total charge would be 3 people * ($30.00 - 0.25(2)).... Pretty soon, you'll see that your profit will be {{{ P(x) = x(30 - 0.25(x-1)) }}} or if simplified, {{{ P(x) = -0.25x^2 + 30.25x }}}


We'll have to use completing the square to rewrite that function in vertex form {{{ P(x) = a(x - h)^2 + k }}}. Once rewritten in vertex form, it becomes easy to pick out (h, k) where h is how many people it will take to maximize your profit, and k, what that maximum profit value is when you have h people. I take it that you know how to do this (because I assume that you're an educator). After completing the square, you should get {{{ P(x) = -0.25(x - 60.5)^2 + 915.0625 }}}. The maximimum profit is about $915.06 when you have 60.5 people. Since you can't have 1/2 a person, you would maximize your profit on the 60th person and start losing profit on the 61st person. To answer the original question, you would (B) be losing profit before you fill your bus.


Now, the challenge you could give to your students is to find how much you should discount (we used $0.25 that maximized profit on the 60th person) per additional person so that you'll maximize your profit exactly on the 80th person on your bus.


Now, I call that extremely practical, and if they're selling candy or something for a fund raiser, they can use this same kind of deal to get people to buy more, and they can decide the number of items limit a customer can buy so they don't start losing profit in case people buy too many.