Question about SDMB's "No homework questions" policy re math questions specifically

The BBQ Pit
1

I’ve wondered about this for awhile now, and since this is the place to put these questions -

I really think the SDMB policy on homework questions, and specifically math homework questions is a bit too restrictive. I’m reasonably good at financial and investment math, but when I was in high school and college there were advanced algebra and probablity questions that would spin me out. It would have been a God send to have a ready resource that would assist me and point me toward a solution.

A lot of people are probably thinking “Lazy students! Just work harder”, but that’s not the way it is. In getting your head wrapped around abstract math concepts a good example and guide is like gold, and unfortunately a lot of people at the lower end of the math intuitive curve are intimidated in large groups from asking “I don’t get it” questions. It’s abjectly humiliating.

Beyond this, I’ve seen a few math questions sneak under the wire here, and the level of guidance offered is truly superlative. It’s head and shoulders above a lot of the classroom examples offered, and the SDMB respondents who offer help don’t seem to mind. Some math snobs might disdain the simple algebra help requests, but many math mavens here are glad to help.

Invariably these “please help me” threads often get shut down with stern mod warnings about not asking the SDMB for help with [thunder crash]“HOMEWORK QUESTIONS”.[/thunder crash]

Why are we so anal about this? We answer questions about goat porn for Gods sake, but lord help you if you’ve got a homework question! Most people aren’t asking us to write term papers. It’s typically 1:00 AM and they’ve got a stumper they need some help with. Why not offer a helping hand? Limit it to registered members, or maybe set a forum for it, but to ban these questions entirely seems mean spirted, and to some degree, against the basic “Fighting Ignorance” ethos of this board.

2

I personally think it’s silly. We answer all kinds of science questions that could easily be on a homework assignment. We discuss topics that could easily be an essay for an English or History class.

As someone who had poor school experience with math and science, I enjoy reading threads about them to try to catch up a little bit.

3

It is my understanding that “Do My Homework For Me” threads are not allowed, but it’s perfectly permissible to ask for help getting on the right track, especially if you post your work to show you’ve made an effort on your own, and if you are up front about stating that you’re asking for help with homework. If this is about the thread being discussed in ATMB, then I have to say that samclem fumbled the ball quite badly in that case. That thread should never have been locked.

4

What’s the reason behind the rule? Is it to prevent the board from becoming flooded with homework requests?

5

This equation was in GQ in a thread that was closed down by Samclem:



    1         +        3        =       4
----------         ----------          ---
  x[sup]2[/sup] - 4             x + 2              5

Here’s the thing: I haven’t had to work such a problem since high school, which was over two decades ago. I simply don’t remember the “rules” for solving it. I e-mailed Samclem for a solution, since he’s the one who closed the thread; but I haven’t heard back from him.

I am not a student, and this is not homework for me.

But I’d still like to know the solution.

I know that the denominators must be the same. There was something about multiplying the right side by one ((x-2)/(x-2)) and I came up with this:



    1               3             4(x-2)
------------  + -----------  =  ------------
(x+2)(x-2)        x+2             5(x-2)

Where do I go from there? Or was that the wrong way to start? It seems to me that it’s just the same thing as the original equation. :confused:

As I said, it’s been a long time since I’ve needed to solve such an equation, and it’s never come up in real life. I think enough time has passed since that thread was posted that the OP must have solved it one way or another.

Obviously the solution is x=3 [1/(3[sup]2[/sup]-4) + 3/(3+2 = 4/5, or 1/5 + 3/5 = 4/5]. But what are the steps to solving the equation?

6

What about those of us who are asked to research goat porn as a homework assignment? :confused:

7

Hint: multiply both sides by (x -2). This will give you a common denominator on the left. I’m sure you can work it out from there.

8

The answer is also x = 3/4.

To solve this, you wanna get the x’s up top. So multipy both sides by (x^2 - 4). This give you:

1 + 3 ( x - 2 ) = (4/5) (x^2 - 4 )

Distribute everything to get:

1 + 3x - 6 = (4/5) x^2 - (16/5)

Combine terms and take out the (1/5) to make it easier:

(1/5) (4x^2 - 15x + 9) = 0

From here, it’s really obvious. Get rid o’ the (1/5) and factor the quadratic to get:

( 4x - 3 ) ( x - 3 ) = 0

9

I so agree with the OP. I’m an older college student and I am struggling with a few subjects. Math was one (I got an A in my summer college algebra class, finally, and I worked my ass off to get it), this semester it is chemistry. I take advantage of the extra help offered at school - I spend time in the chemistry help room and also see my professor when I have specific questions. Due to a severe shortage of tutors at my school I am not able to have one right now, but I think I am making every reasonable effort to learn this stuff. Unfortunately the help room and my prof are not available 24/7. On the weekends and later at night when I have questions, the SD is so enticing as a resource. But I know the policy and don’t want my threads shut down. :frowning: I don’t want someone to outright give me the answers, I need to know this stuff myself on tests.

The way I see it is: If I am having trouble figuring out the problems (Which are not handed in or graded in my chem class, the prof highly recommends we do them, but it is optional) I ask someone to explain the concepts. If the chem help room is open, I’ll ask there (and oddly enough, the students who work in the help room are more apt to outright give the answers without much explanation as to WHY, so I don’t understand how helpful that is, or else they tend to explain things too far above the level I am currently at, making me even more confused). If the prof is available, I’ll ask him. If neither resource is open/available, I’d like to be able to ask here. Some concepts I need to have explained to me more than once, or presented in a different way, before I fully understand them.

10

Okay:



    x-2          3x-6          4x-8
----------  + ----------  = ----------
(x+2)(x-2)        x+2           5

Getting rid of x-2 in the first bit and adding the second gives me:



     1           3x-6          4x-8
----------  + ----------  = ----------
   x+2            x+2           5

Or:

   3x-5          4x-8
----------  = ----------
   x+2             5

Multiplying by the denominators gives me:

15x -25 = 4x[sup]2[/sup] - 16

So

15x - 4x[sup]2[/sup] = 9

-4x[sup]2[/sup] + 15x - 9 = 0

Multiply both sides by -1 is

4x[sup]2[/sup] - 15x +9 = 0

That’s not right. I’ve gone wrong somewhere.

11

How can x be 3 and 3/4?

If I replace x with 3, I get 1/(3*3)-4, or 1/5; and 3/(3+2), or 3/5. 1/5 + 3/5 = 4/5, or .2 + .6 = .8

But if I replace x with .75, I get -.290909(repeating) + 1.090909(repeating) = .8, or .8 = .8

How does that happen? :confused:

12

The original equation:

1/(x[sup]2[/sup] - 4) + 3/(x + 2) = 4/5

Put both parts of the left-hand side over a common denominator. Since (x + 2) divides (x[sup]2[/sup] - 4), that is the common denominator. Normally, you’d pick the least common multiple of the two denominators, but if one divides the other, that second one is the LCM.

1/(x[sup]2[/sup] - 4) + (3x - 6)/(x[sup]2[/sup] - 4) = 4/5

Combine terms:

(3x - 5)/(x[sup]2[/sup] - 4) = 4/5

Cross-multiply:

5(3x - 5) = 4(x[sup]2[/sup] - 4)

Expand:

15x - 25 = 4x[sup]2[/sup] - 16

Combine terms again:

4x[sup]2[/sup] - 15x + 9 = 0

You could try to factor that, but it’s easier to just use the quadratic formula:

x = (15 + sqrt(225 - 144))/8

x = (15 + sqrt(81))/8

x = (15 + 9)/8

So the two solutions we’ve got are 3 and 3/4. Sometimes you can take steps that introduce new roots, but we didn’t do any of those here, so both of those should be solutions to the original equation.

As far as homework questions go, the SDMB has never had a clear policy, either stated or applied. It would be nice to have the rules.

13

Think of it this way:

When you’re solving an equation, you’re not finding out what x is. You’re finding out what values x could stand for that would make the equation a true statement. In this case, if you substitute 3 for x, the equation is true. If you substitute 3/4 for x, the equation is true. There’s nothing else you could substitute for x to make the equation true.

14

How do you get from 4x[sup]2[/sup] - 15x + 9 = 0 to your equation?

Also: How would you factor 4x[sup]2[/sup] - 15x + 9 = 0?

I’m sorry if I’m being obtuse, but it’s been a long time. I’ve lost the skills for solving this sort of thing. (And it was a struggle in high school just to get a B.)

That is true, as has been shown; but I’ve always thought of it as being “find out what x is”.

15

The equation I used is the quadratic formula, which is the solution to the general equation ax[sup]2[/sup] + bx + c = 0. In that case, x = (-b + sqrt(b[sup]2[/sup] - 4ac))/2a.

As far as factoring goes…well, nobody really does that outside of high school algebra. It requires some creativity and some luck, whereas the quadratic formula is literally a no-brainer, given that computers can solve equations that way.

Yeah, a lot of people think that. It’s a mistake, but you can probably blame it on your teachers.

16

That’s at least partially because it’s easier to lecture on “find out what X is” than “find all the possible values for X.” When you’re just getting into algebra, the only times you’ll generally have more than one possible X are with an absolute value or a < or > sign (eqtns with powers higher than 1 usually don’t make their ways into beginning algebra until students are comfortable just with 3x+4=7), so at least initially it’s easier to deal with “figure out what X is” rather than “figure out what values X could be.”

There’s a new method being used now for factoring, involving using a square. It still involves some guess work and luck, but for my money it’s easier on most students than the old method, which didn’t work unless you had more than a little luck and skill. Usually I won’t even bother factoring something by hand if the first term has a coefficient other than 1.

17

Two administrators both said basically what I did, in this thread:

18

Interesting. It’s probably the opposite of the method I’ve been using for multiplication.

19

I’m not following you. Can you go step-by-step, using 4x[sup]2[/sup] - 15x + 9 = 0? That is, I assume you’re dividing the left side of the equation with “something”, and then subtracting everything except x from both sides.

Wait…

I’ve just found this site.

I’ve never seen this formula. Is it a “standard” formula, like the Pythagorean Theorem? That is, if you have a quadratic equation, is the solution always



    -b± sqrt(b[sup]2[/sup] - 4ac)
x = ------------------
           2ac

?

This is not something that I was ever taught in high school. Although I went through trig and pre-calc (and it was a struggle, as I said, to earn a decent grade) we always used factoring.

So let me try to figure this out without looking at what you posted earlier:

x = (15 ± sqrt(15[sup]2[/sup] - 4(4 * 9)) / 2(4)

x = ((15 ± sqrt(225 - 144)) / 8

x = (15 ± sqrt(81)) / 8

x = (15 ± 9) / 8

x = 24 / 8 = 3
x = 6 / 8 = .75

20

That’s a pretty sloppy policy IMO, and quite frankly it forces busy mods to work out the nature of a question’s nuances, or just adopt a “no homework questions period” policy, and they seem to be choosing the latter in many cases.