I need help about this:

I a,b,c are real numbers such that $\displaystyle (b-1)^2-4ac<0$ , proof that system

$\displaystyle ax^2+bx+c=y\\ ay^2+by+c=z \\ az^2+bz+c=x$

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

Printable View

- Jan 26th 2013, 06:38 AMdarencecyclical system of equations
I need help about this:

I a,b,c are real numbers such that $\displaystyle (b-1)^2-4ac<0$ , proof that system

$\displaystyle ax^2+bx+c=y\\ ay^2+by+c=z \\ az^2+bz+c=x$

don't have any solution in set of real numbers. - Jan 26th 2013, 07:54 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 26th 2013, 11:23 PMdarenceRe: 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 26th 2013, 11:25 PMchiroRe: 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 26th 2013, 11:30 PMdarenceRe: 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, 12:10 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, 07:18 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, 04:27 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, 07:22 PMearthboyRe: cyclical system of equations
after a full day,my idea:

your equations:

$\displaystyle ax^2+bx+c-y=0$

$\displaystyle ay^2+by+c-z=0$

$\displaystyle az^2+bz+c-x=0$

summing them up we get $\displaystyle ax^2+bx+c-y+ay^2+by+c-z+az^2+bz+c-x=0$

rearranging the terms, $\displaystyle ax^2+bx-x+c+ay^2+by-y+c+az^2+bz-z+c=0$

or, $\displaystyle \bigg(ax^2+(b-1)x+c \bigg)+\bigg(ay^2+(b-1)y+c \bigg)+\bigg(az^2+(b-1)z+c \bigg)=0$

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

$\displaystyle am^2+bm+c=a(m-h)^{2}+k$, where $\displaystyle h=\frac{-b}{2a}$ and $\displaystyle k=c-\frac{b^2}{4a}$

using the same principles we get,

$\displaystyle \bigg(a(x+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)+\bigg(a(y+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)+\bigg(a(z+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)=0$

simplifying we get,

$\displaystyle a[(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2]+3(\frac{4ac-(b-1)^2}{4a})=0$

or,$\displaystyle a[(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2]=-3(\frac{4ac-(b-1)^2}{4a})$

dividing both sides by $\displaystyle a$,

$\displaystyle (x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2=-\frac{3}{4}.\frac{4ac-(b-1)^2}{a^2}$

now, $\displaystyle (b-1)^2-4ac<0$ (given) $\displaystyle \implies 4ac-(b-1)^2>0$ also $\displaystyle a^2>0$ therefore, $\displaystyle \frac{4ac-(b-1)^2}{a^2}>0$ 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, 03:04 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, 07:30 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, 11:40 AMdarenceRe: cyclical system of equations
$\displaystyle x^3-3x^2+2x=y^2 \\ y^3-3y^2+2y=x^2$

Any idea ?

I found that $\displaystyle x,y \in [0,1]\cup [2,+\infinity )$ because $\displaystyle x^2$ and $\displaystyle y^2$ are greater than zero. How to continue ? - Feb 21st 2013, 10:48 PMearthboyRe: cyclical system of equations
I think you should make new thread about new problems, unless the thread is about problem solving marathons...(Speechless)