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Thread: Solving inequality

  1. #1
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    Solving inequality

    See this example:

    $\displaystyle \frac{x - 1}{x + 1} \geq \frac{2x}{x - 1} $

    We multiply both sides by $\displaystyle (x + 1) (x - 1)$

    and get:

    $\displaystyle (x - 1)^2 \geq 2x(x +1)$

    Criteria 1:

    x < -1
    $\displaystyle (x - 1)^2 \geq 2x(x +1)$
    $\displaystyle x^2 - 2x + 1 \geq 2x^2 + 2x$
    $\displaystyle x^2 + 4x -1 \leq 0$

    $\displaystyle x_{1,2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$

    $\displaystyle x \in [-2-\sqrt{5}, -2+\sqrt{5}]$ but since $\displaystyle x < -1 $ we get $\displaystyle x \in [-2-\sqrt{5}, 1)$

    Criteria 2:
    $\displaystyle x \in (-1, 1)$

    $\displaystyle (x - 1)^2\leq 2x(x +1)$ why less or equal??

    from that we get: $\displaystyle x^2 +4x -1 \geq 0 \Rightarrow x \in [-2+\sqrt{5},1)$ (why not (-1, -2+\sqrt{5}] ?)

    Criteria 3:
    x > 1
    $\displaystyle x^2 + 4x - 1 \leq 0 \Rightarrow no solution$

    All together:

    $\displaystyle x \in [-2-\sqrt{5},-1) U [-2+\sqrt{5},1)$
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  2. #2
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    If $\displaystyle x \in (-1,1)$, then $\displaystyle (x - 1)(x + 1)$ is negative.
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  3. #3
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    short and simple. thanks
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