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Math Help - inequality on the set of real numbers

  1. #1
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    inequality on the set of real numbers

    Solve the following inequality on the set of real numbers


    x^2-3\sqrt{x^2+3}\leq1
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by perash View Post
    Solve the following inequality on the set of real numbers


    x^2-3\sqrt{x^2+3}\leq1
    x^2-3\sqrt{x^2+3}\leq1 \Rightarrow (x^2 - 1)^2 \leq 9(x^2 + 3)
    A little bit of algebra leads us to,(that is bringing everything to the LHS and factorising)
     (x-\sqrt{13})(x+\sqrt{13})(x^2+2) \leq 0
    \forall x \in \mathbb{R}, x^2 + 2 > 0
    So,
     (x-\sqrt{13})(x+\sqrt{13})\leq 0
    Since exactly one of them is non-positive,
    x \in [-\sqrt{13},\sqrt{13}]
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  3. #3
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    Hello, perash!

    Solve on the set of real numbers: . x^2-3\sqrt{x^2+3}\:\leq\:1

    We have: . x^2-1 \:\leq \:3\sqrt{x^2+3}

    Square: . x^4-2x^2 + 1 \:\leq \:9x^2+27\quad\Rightarrow\quad x^4 - 11x^2 - 26 \:\leq \:0

    . . which factors: . (x^2+2)(x^2-13)\:\leq\:0


    Since (x^2+2) is always positive, (x^2-13) must be negative.

    We have: . x^2-13 \:\leq \:0\quad\Rightarrow\quad x^2\:\leq\:13\quad\Rightarrow\quad |x| \:\leq\:\sqrt{13}


    Therefore: . -\sqrt{13}\;\leq\: x \:\leq\:\sqrt{13}


    Drat . . . too slow again!
    .
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, perash!

    [size=3]
    We have: . x^2-1 \:\leq \:3\sqrt{x^2+3}

    Square: . x^4-2x^2 + 1 \:\leq \:9x^2+27\quad\Rightarrow\quad x^4 - 11x^2 - 26 \:\leq \:0
    I always get confused when I need to square an inequality.

    x^2 - 1 \leq 3\sqrt{x^2 + 3}

    If x \in (-1, 1) then when we square this we shouldn't we get
    (x^2 - 1) \geq 9(x^2 + 3)
    since the LHS is negative? So don't we have to do the problem separately on (-1, 1) and \mathbb{R} - (-1, 1)?

    -Dan

    Edit: This question has been resolved. See here.
    Last edited by topsquark; December 31st 2007 at 09:01 AM.
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    Lord of certain Rings
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    Quote Originally Posted by topsquark View Post
    since the LHS is negative? So don't we have to do the problem separately on (-1, 1) and \mathbb{R} - (-1, 1)?

    -Dan
    I will try to answer this question,
    First we note RHS is always positive for this problem,so if the LHS is negative then, LHS \leq 0 \leq RHS , so it will always hold.

    The point is almost all steps in the inequality reduction are equivalent.So there is no reason it will be wrong.
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