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The nasty easy equations

by Hobin
Tags: equations, nasty
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Hobin
#1
Mar14-12, 11:34 AM
P: 194
If you laugh when reading this, you lose.

So, I was solving a bunch of extremely easy equations, and encountered this one I wanted to simplify (i.e. group into terms):
[itex]x^{3}-3x^{2}-4x+12[/itex]
Having an inflexible mind, I immediately decided that I should somehow be able to solve it using the sum of cubes formula. As you can most likely guess, this didn't work so well. It's an incredibly simple equation, and when you realize that it's just a matter of grouping the first two and the last two terms, it takes only a few seconds to solve.

The problem is, of course, that sometimes I simply 'forget' to use technique A of solving a problem because you fixate on using technique B.

How often do you people experience these annoyances? And, more importantly, what is your prefered method of dealing with them? Me, I often just bruteforce an equation through all possible algebraic options, but I suspect it can be done more efficiently. I also suspect that getting it 'right' faster may simply be a result of experience (or 'mathematical maturity', as some like to call it).

PS: Yes, you can correlate my mood to the amount of stupid little mistakes I make in math.
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tiny-tim
#2
Mar14-12, 11:59 AM
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Hi Hobin!
Quote Quote by Hobin View Post
[itex]x^{3}-3x^{2}-4x+12[/itex]

I should somehow be able to solve it using the sum of cubes formula.
I don't understand.

There's no general way of solving a cubic other than the really complicated one.
If an exam question asks you to solve a cubic, you can be sure that at least root is easy to find

just try 1, 2, 3, until you find one that works!
Hobin
#3
Mar14-12, 12:07 PM
P: 194
Quote Quote by tiny-tim View Post
I don't understand.
The point was that I could apply some general formulae to the previous equations (along the lines of [itex]A^{3}+B^{3} = (A+B)(A^{2}-AB+B^{2})[/itex]), and I immediately jumped to the (wrong) conclusion that it would be possible to force-fit this equation, too. Unfortunately, this wasn't the case, and resulted in a lot of wasted time. The easiest method of simplification here was to put the first two and the last two terms in separate groups: for example, [itex]a = x^{3}-3x^{3}[/itex] and [itex]b = 4x-12[/itex], then the equation becomes [itex]a - b[/itex]. This was much easier to simplify, and this is the kind of thing that frustrates me because it turns out to be so easy.

Thus, my question was if people experience this kind of thing - you think it's easy to solve an equation or a problem, but it turns out that whatever you throw at it, it doesn't work. And in the end, of course, you realize that you simply forgot a very little something you should've thought of minutes ago.

Mark44
#4
Mar14-12, 02:56 PM
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P: 21,216
The nasty easy equations

Quote Quote by Hobin View Post
The point was that I could apply some general formulae to the previous equations (along the lines of [itex]A^{3}+B^{3} = (A+B)(A^{2}-AB+B^{2})[/itex]), and I immediately jumped to the (wrong) conclusion that it would be possible to force-fit this equation, too. Unfortunately, this wasn't the case, and resulted in a lot of wasted time. The easiest method of simplification here was to put the first two and the last two terms in separate groups: for example, [itex]a = x^{3}-3x^{3}[/itex] and [itex]b = 4x-12[/itex], then the equation becomes [itex]a - b[/itex]. This was much easier to simplify, and this is the kind of thing that frustrates me because it turns out to be so easy.

Thus, my question was if people experience this kind of thing - you think it's easy to solve an equation or a problem, but it turns out that whatever you throw at it, it doesn't work. And in the end, of course, you realize that you simply forgot a very little something you should've thought of minutes ago.
But you don't have a sum of cubes, so the formula above doesn't apply.

The equation you want to factor is fairly simple to factor by grouping. The first two terms have a factor of x - 3, and so do the third and fourth terms.
Hobin
#5
Mar14-12, 03:07 PM
P: 194
Quote Quote by Mark44 View Post
But you don't have a sum of cubes, so the formula above doesn't apply.

The equation you want to factor is fairly simple to factor by grouping. The first two terms have a factor of x - 3, and so do the third and fourth terms.
I know! That's why it was so hard to solve.

What I'm asking is if you people often experience that are you trying very hard to use an incorrect method to solve a problem, only to find out later that it would've been much easier (i.e. not downright impossible) if you weren't being such a fool and actually looked at the equation properly.
dodo
#6
Mar14-12, 03:35 PM
P: 688
A quick one is that from[tex]x(x^2-3x-4)=-12[/tex]you can see that [itex]x=3[/itex] is a solution. But that was sheer luck. :)
Hobin
#7
Mar14-12, 03:40 PM
P: 194
Quote Quote by Dodo View Post
A quick one is that from[tex]x(x^2-3x-4)=-12[/tex]you can see that [itex]x=3[/itex] is a solution. But that was sheer luck. :)
Heh. I bet most of us have sometimes passed an exam by writing down similar answers: get the correct answer on a calculator, then mess around with the equation a bit and use the 'it is obvious that'-rule to make it seem like you understood what to do.
SHISHKABOB
#8
Mar14-12, 04:24 PM
P: 614
Quote Quote by Hobin View Post
Heh. I bet most of us have sometimes passed an exam by writing down similar answers: get the correct answer on a calculator, then mess around with the equation a bit and use the 'it is obvious that'-rule to make it seem like you understood what to do.
my math professors never let me use calculators on exams :(
Hobin
#9
Mar14-12, 05:26 PM
P: 194
Quote Quote by SHISHKABOB View Post
my math professors never let me use calculators on exams :(
That's true, too! I remember a few classes where that wasn't allowed. Mostly, these were the introductory let's-see-if-you-can-do-algebra-and-trigonometry-classes. On the other hand, graphing calculators were banned from exams much more often.
willem2
#10
Mar14-12, 07:25 PM
P: 1,395
According to the rational root theorem.

http://en.wikipedia.org/wiki/Rational_root_theorem

all rational roots will be integers that divide 12.
Mark44
#11
Mar14-12, 07:42 PM
Mentor
P: 21,216
Quote Quote by Hobin View Post
What I'm asking is if you people often experience that are you trying very hard to use an incorrect method to solve a problem, only to find out later that it would've been much easier (i.e. not downright impossible) if you weren't being such a fool and actually looked at the equation properly.
Sure, I think that most of us have spent some time trying to solve a problem a certain way, only to find out later that there is a much simpler way of going about it. Experience is probably the best teacher in cases like that, when you have learned a number of techniques of doing something.

I had an instructor in a class I took one time who said, "If the only tool you have is a hammer, everything looks like a nail." With more experience, you learn different techniques, and pros and cons of each.

In an unrelated area, I like to work crossword puzzles, especially the ones in the NY Times. In a given week the puzzles start off easy on Monday, and get trickier and tricker later in the week. The Friday and Saturday puzzles are usually quite a bit harder than the ones on Monday and Tuesday. A strategy that works for me is to not get locked into a single meaning for a clue - try to think is there some other meaning? For example, does "Lead object" (hokey example) mean an object made out of lead metal, or does it mean an object that is in front?

I think there's a connection here with doing math - don't get locked into a single interpretation of what the problem is saying. It could be that your interpretation is incorrect, so you'll probably struggle more at finding a solution.


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