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Topic: Too many solutions? (Read 1822 times) |
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BNC
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Too many solutions?
« on: Jan 29th, 2003, 11:24am » |
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A quadratic equation has either 2, 1 or 0 unique real solutions.
Well, look at this equation (nice work, towr!):
Assume a<b<c
BUT:
x=a; x=b and x=c are all uniqe solutions!
[edit]
Hi!
Can't seem to link to the formula (#28 on towr's database). Can anyone help?
[/edit]
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| « Last Edit: Jan 29th, 2003, 10:57pm by BNC » |
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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towr
wu::riddles Moderator
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Re: Too many solutions?
« Reply #1 on: Jan 29th, 2003, 12:58pm » |
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You forgot a 'b' at the end of the url..
Maybe an easier alternative is to use formula?id=28
(Which makes you wonder why I didn't put that under image at the first place)
for the time being..:
[edit 25/2/03]the reason for using an md5 as index is when the database disappears and all formulas get renumbered (it's now 24)
hmmzz.. For some strange reason using the md5 doesn't work for this formula when linked from anywhere on this board.. It works for other formulae, and from other places.. peculiar[/edit]
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| « Last Edit: Feb 25th, 2003, 2:48am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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James Fingas
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Re: Too many solutions?
« Reply #2 on: Jan 29th, 2003, 1:24pm » |
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BNC,
That is a very clever puzzle! It makes me wonder if there are even more solutions, other than x=a, x=b, x=c ... so much for the foundations of mathematics!
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Doc, I'm addicted to advice! What should I do?
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Icarus
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Re: Too many solutions?
« Reply #3 on: Jan 29th, 2003, 7:08pm » |
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Very clever indeed! But James, I don't follow your hidden remark. The "wonder" is true but what does this say about the foundations of mathematics? Or am I being dense? 
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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lukes new shoes
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Re: Too many solutions?
« Reply #4 on: Jan 30th, 2003, 5:10am » |
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the equation isnt actually a quadratic
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aero guy
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Nice hint James. I "figured it out" by noting that the equation is AT MOST quadratic. I assume that if I bothered to reduce it you would get a nifty little answer.
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Garzahd
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Re: Too many solutions?
« Reply #6 on: Jan 30th, 2003, 10:38am » |
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Sure enough. If you make a common denominator, you get
(x-a)(x-b)(a-b) + (x-b)(x-c)(b-c) - (x-a)(x-c)(a-c) = (a-b)(b-c)(a-c)
(this reduction becomes much simpler if you remember that (a-b) = -(b-a). )
Anyway, check the coefficients of x2 in the expansion of the left side: a-b + b-c - (a-c) = 0.
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James Fingas
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Re: Too many solutions?
« Reply #7 on: Jan 30th, 2003, 11:16am » |
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I wasn't going to hide my remark, but then I thought:
"What if somebody tries another value for x? Then for sure they will figure it out!"
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Doc, I'm addicted to advice! What should I do?
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aero_guy
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Re: Too many solutions?
« Reply #8 on: Jan 31st, 2003, 12:04pm » |
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that is the nifty part of the problem: they make it very easy to try a, b, or c, but you get into a bunch of algebra with anything else.
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cathy
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If you expend the equation, you will realize that all the unknows cancel out. So, this are infinite number of solutions. not only does x=a, x=b, x=c satify this equation, x can be anything and still satify this equation.
A simply example so that you can understand this easier.
x+a = x +a
how do you solve this equation? you cancel out the x on both sides and realize that no matter what x is, the equation will always be satified. x = a, x = b, x = c..... does matter.
even though this is a first order equation, it still had infinite number of solutions.
The same theory goes for the complicated equation in this question. It just takes a little more time to realize that you can cancel out all the x-terms.
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