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The nasty easy equations 
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#1
Mar1412, 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. 


#2
Mar1412, 11:59 AM

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P: 26,148

Hi Hobin!
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 … 


#3
Mar1412, 12:07 PM

P: 194

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. 


#4
Mar1412, 02:56 PM

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P: 21,311

The nasty easy equations
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. 


#5
Mar1412, 03:07 PM

P: 194

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. 


#6
Mar1412, 03:35 PM

P: 688

A quick one is that from[tex]x(x^23x4)=12[/tex]you can see that [itex]x=3[/itex] is a solution. But that was sheer luck. :)



#7
Mar1412, 03:40 PM

P: 194




#8
Mar1412, 04:24 PM

P: 614




#9
Mar1412, 05:26 PM

P: 194




#10
Mar1412, 07:25 PM

P: 1,398

According to the rational root theorem.
http://en.wikipedia.org/wiki/Rational_root_theorem all rational roots will be integers that divide 12. 


#11
Mar1412, 07:42 PM

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P: 21,311

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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