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  1. #1
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    root

    please help with it:

    square root(x+2-x^2)<square root(x^2-3x+2)+x
    Last edited by agripa; January 1st 2008 at 05:25 AM.
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  2. #2
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    Quote Originally Posted by agripa View Post
    please help with it:

    square root(x+2+x^2)<square root(x^2-3x+2)+x
    Square both sides:

    x^2+x+2 + x^2 -3x + 2 - 2(x-1)\sqrt{x^2 - 4} < x^2
    x^2 - 2x + 4 < 2(x-1)\sqrt{x^2 - 4}
    ((x-1)^2 + 3)^2 < 4(x-1)^2(x^2 - 4)

    Then it is ugly, very ugly.... I wonder if there is a neat solution.
    Soroban is answering it anyway, so let us see...
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  3. #3
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    Quote Originally Posted by agripa View Post
    please help with it:
    square root(x+2+x^2)<square root(x^2-3x+2)+x
    The original post by agripa had a pdf file of questions from Lithuania, but it has been deleted.
    The problem printed there was: \sqrt {x + 2 - x^2 }  < \sqrt {x^2  - 3x + 2}  + x. Note that agripa has made a sign error in recopying the problem. The question appearing on the test paper is a concept question.

    \sqrt {x + 2 - x^2 } is only defined in the interval \left[ { - 1,2} \right].
    Looking at the graphs of both sides, one can easily see that the only solution is x \in [ - 1,0).
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Plato View Post
    The original post by agripa had a pdf file of questions from Lithuania, but it has been deleted.
    The problem printed there was: \sqrt {x + 2 - x^2 }  < \sqrt {x^2  - 3x + 2}  + x. Note that agripa has made a sign error in recopying the problem. The question appearing on the test paper is a concept question.

    \sqrt {x + 2 - x^2 } is only defined in the interval \left[ { - 1,2} \right].
    Looking at the graphs of both sides, one can easily see that the only solution is x \in [ - 1,0).
    But lets say this mistaken question is posed, then do you have a solution?

    Thanks
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  5. #5
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    Quote Originally Posted by agripa View Post
    please help with it:

    square root(x+2+x^2)<square root(x^2-3x+2)+x
    \sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x

    x^2 + x + 2 < (x^2 - 3x + 2) + x^2 + 2x\sqrt{x^2 - 3x + 2}

    x^2 + x + 2 < 2x^2 - 3x + 2 + 2x\sqrt{x^2 - 3x + 2}

    -x^2 + 4x < 2x\sqrt{x^2 - 3x + 2}

    Do NOT cancel that x! (If x is negative then we would be dividing by a negative number. Not kosher here as that changes the < to a >.)

    (-x^2 + 4x)^2 < 4x^2(x^2 - 3x + 2)

    x^4 - 8x^3 + 16x^2 < 4x^4 - 12x^3 + 8x^2

    -3x^4 + 4x^3 + 8x^2 < 0

    3x^4 - 4x^3 - 8x^2 > 0

    x^2(3x^2 - 4x - 8) > 0

    x^2 \left (x^2 - \frac{4}{3}x - \frac{8}{3} \right ) > 0

    The factoring here is a bit rough but can be done using the quadratic formula:
    x^2 \left ( x - \frac{1 + 2 \sqrt{7}}{3} \right ) \left ( x - \frac{1 - 2 \sqrt{7}}{3} \right ) > 0

    The critical points here are x = 0, \frac{1 \pm 2 \sqrt{7}}{3}, so we have four intervals on the real line to test in the original (that is to say, all the way back to the beginning!) inequality:
    x \in \left ( - \infty, \frac{1 - 2\sqrt{7}}{3} \right ) \implies \sqrt{x^2 + x + 2} > \sqrt{x^2 - 3x + 2} + x (Nope!)

    x \in \left ( \frac{1 - 2\sqrt{7}}{3}, 0 \right ) \implies \sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x (Check!)

    x \in \left ( 0, \frac{1 + 2\sqrt{7}}{3} \right ) \implies \sqrt{x^2 + x + 2} > \sqrt{x^2 - 3x + 2} + x (Nope!)

    x \in \left ( \frac{1 + 2\sqrt{7}}{3}, \infty \right ) \implies \sqrt{x^2 + x + 2} < \sqrt{x^2 - 3x + 2} + x (Check!)

    So the solution set for x is
    \left ( \frac{1 - 2\sqrt{7}}{3}, 0 \right ) \cup \left ( \frac{1 + 2\sqrt{7}}{3}, \infty \right )

    (Note that there is a place where the domain of x is restricted. As this falls outside of the derived solution set I'm not going to worry over it.

    (The graph is for the equivalent problem: \sqrt{x^2 + x + 2} - ( \sqrt{x^2 - 3x + 2} + x) < 0. We are looking for where the expression is negative.)

    -Dan
    Attached Thumbnails Attached Thumbnails root-buttugly.jpg  
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