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Math Help - cyclical system of equations

  1. #1
    Junior Member darence's Avatar
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    cyclical system of equations

    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.
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  2. #2
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    Re: 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)?
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  3. #3
    Junior Member darence's Avatar
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    Re: 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 ?
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  4. #4
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    Re: 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).
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  5. #5
    Junior Member darence's Avatar
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    Re: 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 ?
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  6. #6
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    Re: cyclical system of equations

    You have constraints on all values and you need to modify the set of three simultaneous equations ever so slightly.
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  7. #7
    Junior Member darence's Avatar
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    Re: 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.
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  8. #8
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    Re: 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:

    cyclical system of equations-mhfcomposition3.png
    Thanks from darence
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  9. #9
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    Re: cyclical system of equations

    Quote Originally Posted by darence View Post
    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.
    Thanks from darence
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  10. #10
    Junior Member darence's Avatar
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    Re: cyclical system of equations

    Thank you boys! That's it!

    @johng
    Can you attach here a book where you find this solution ?
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  11. #11
    Junior Member darence's Avatar
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    Re: cyclical system of equations

    One more problem:
    In set of positive real numbers, solve system:
    xy=z
    yz=x
    zx=y
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  12. #12
    Junior Member darence's Avatar
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    Re: cyclical system of equations

    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 ?
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  13. #13
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    Re: cyclical system of equations

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