I need help about this:

I a,b,c are real numbers such that , proof that system

don't have any solution in set of real numbers.

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- Jan 26th 2013, 07:38 AMdarencecyclical system of equations
I need help about this:

I a,b,c are real numbers such that , proof that system

don't have any solution in set of real numbers. - Jan 26th 2013, 08:54 PMchiroRe: cyclical system of equations
Hey darence.

Can you show you re-arrange each equation and show that the discriminant of each equation is the above? (i.e. discriminant = (b-1)^2 - 4ac)? - Jan 27th 2013, 12:23 AMdarenceRe: cyclical system of equations
i.e In first equation, move y to left side. Now, we have a quadratic equation of x. Her discriminant is (b-1)^2 - 4ac, and similar for all equations. How to continue ?

- Jan 27th 2013, 12:25 AMchiroRe: cyclical system of equations
Well since they are < 0 it means you have complex solutions. The reason is you have a negative number in the square root sign which means complex numbers (it would be >= for real number).

- Jan 27th 2013, 12:30 AMdarenceRe: cyclical system of equations
OK, but it is not solution of my problem.

ps. Discriminant will be (b-1)^2 - 4a(c-y). So, what to do ? - Jan 27th 2013, 01:10 AMchiroRe: cyclical system of equations
You have constraints on all values and you need to modify the set of three simultaneous equations ever so slightly.

- Jan 27th 2013, 08:18 AMdarenceRe: cyclical system of equations
I think that it is posible to resolve by prooving that x=y=z (I dont know how to ) , and than will be (b-1)^2-4ac<0 (it is supposed by problem). So, equation(s) dont have real solutions.

Sorry for my bad english. I wish you understood my idea. - Jan 27th 2013, 05:27 PMjohngRe: cyclical system of equations
If your question were phrased as f(f(f(x))) = x where f is a quadratic function, I think you wouldn't have any problem. Anyway here's a solution:

Attachment 26729 - Jan 27th 2013, 08:22 PMearthboyRe: cyclical system of equations
after a full day,my idea:

your equations:

summing them up we get

rearranging the terms,

or,

now by completing the squares we try to make the expression in the brackets a square.In general case:

, where and

using the same principles we get,

simplifying we get,

or,

dividing both sides by ,

now, (given) also therefore, and thus the__right hand side of the equation is a negative number__, while the__left hand side is the sum of squares and should always be positive__.

This is not possible for any real number solution, and hence we conclude no real solution exist.(Rock) - Jan 28th 2013, 04:04 AMdarenceRe: cyclical system of equations
Thank you boys! That's it!

@**johng**

Can you attach here a book where you find this solution ? - Feb 1st 2013, 08:30 AMdarenceRe: cyclical system of equations
One more problem:

In set of positive real numbers, solve system:

x^{y}=z

y^{z}=x

z^{x}=y - Feb 6th 2013, 12:40 PMdarenceRe: cyclical system of equations

Any idea ?

I found that because and are greater than zero. How to continue ? - Feb 21st 2013, 11:48 PMearthboyRe: cyclical system of equations
I think you should make new thread about new problems, unless the thread is about problem solving marathons...(Speechless)