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        riddles >> easy >> Too many solutions?
        
(Message started by: BNC on Jan 29^th, 2003, 11:24am)

Title: Too many solutions?
Post by BNC on Jan 29^th, 2003, 11:24am

A quadratic equation has either 2, 1 or 0 unique real solutions.

Well, look at this equation (nice work, towr!):

http://www.ai.rug.nl/~towr/PHP/FORMULA/formula.php?md5=515cc33ce67830472a4e9990dc6f491
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]

Title: Re: Too many solutions?
Post by towr on Jan 29^th, 2003, 12:58pm

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..:
http://www.ai.rug.nl/~towr/PHP/FORMULA/formula.php?id=24

[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]

Title: Re: Too many solutions?
Post by James Fingas on Jan 29^th, 2003, 1:24pm

BNC,

That is a very clever puzzle! [hide] 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! [/hide]

Title: Re: Too many solutions?
Post by Icarus on Jan 29^th, 2003, 7:08pm

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? :-/

Title: Re: Too many solutions?
Post by lukes new shoes on Jan 30^th, 2003, 5:10am

the equation isnt actually a quadratic

Title: Re: Too many solutions?
Post by aero guy on Jan 30^th, 2003, 5:17am

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.

Title: Re: Too many solutions?
Post by Garzahd on Jan 30^th, 2003, 10:38am

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.

Title: Re: Too many solutions?
Post by James Fingas on Jan 30^th, 2003, 11:16am

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!"

Title: Re: Too many solutions?
Post by aero_guy on Jan 31^st, 2003, 12:04pm

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.

Title: Re: Too many solutions?
Post by cathy on Jul 14^th, 2003, 3:33pm

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.