cyclical system of equations

**darence** · January 26th 2013, 06:38 AM

I need help about this:

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

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

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

**chiro** · January 26th 2013, 07:54 PM

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)?

**darence** · January 26th 2013, 11:23 PM

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 ?

**chiro** · January 26th 2013, 11:25 PM

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).

**darence** · January 26th 2013, 11:30 PM

OK, but it is not solution of my problem.

ps. Discriminant will be (b-1)^2 - 4a(c-y). So, what to do ?

**chiro** · January 27th 2013, 12:10 AM

You have constraints on all values and you need to modify the set of three simultaneous equations ever so slightly.

**darence** · January 27th 2013, 07:18 AM

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.

**johng** · January 27th 2013, 04:27 PM

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:

$cyclical system of equations-mhfcomposition3.png$

**earthboy** · January 27th 2013, 07:22 PM

Originally Posted by darence

I need help about this:

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

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

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

after a full day,my idea:
your equations:
$ax^2+bx+c-y=0$
$ay^2+by+c-z=0$
$az^2+bz+c-x=0$
summing them up we get $ax^2+bx+c-y+ay^2+by+c-z+az^2+bz+c-x=0$
rearranging the terms, $ax^2+bx-x+c+ay^2+by-y+c+az^2+bz-z+c=0$
or, $\bigg(ax^2+(b-1)x+c \bigg)+\bigg(ay^2+(b-1)y+c \bigg)+\bigg(az^2+(b-1)z+c \bigg)=0$
now by completing the squares we try to make the expression in the brackets a square.In general case:
$am^2+bm+c=a(m-h)^{2}+k$ , where $h=\frac{-b}{2a}$ and $k=c-\frac{b^2}{4a}$
using the same principles we get,
$\bigg(a(x+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)+\bigg(a(y+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)+\bigg(a(z+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)=0$
simplifying we get,
$a[(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2]+3(\frac{4ac-(b-1)^2}{4a})=0$
or, $a[(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2]=-3(\frac{4ac-(b-1)^2}{4a})$
dividing both sides by $a$ ,
$(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2=-\frac{3}{4}.\frac{4ac-(b-1)^2}{a^2}$
now, $(b-1)^2-4ac<0$ (given) $\implies 4ac-(b-1)^2>0$ also $a^2>0$ therefore, $\frac{4ac-(b-1)^2}{a^2}>0$ 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.

**darence** · January 28th 2013, 03:04 AM

Thank you boys! That's it!

@johng
Can you attach here a book where you find this solution ?

**darence** · February 1st 2013, 07:30 AM

One more problem:
In set of positive real numbers, solve system:
x^y=z
y^z=x
z^x=y

**darence** · February 6th 2013, 11:40 AM

$x^3-3x^2+2x=y^2 \\ y^3-3y^2+2y=x^2$

Any idea ?
I found that $x,y \in [0,1]\cup [2,+\infinity )$ because $x^2$ and $y^2$ are greater than zero. How to continue ?

**earthboy** · February 21st 2013, 10:48 PM

I think you should make new thread about new problems, unless the thread is about problem solving marathons...

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