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Thread: Absolute Value Inequality Help

  1. #1
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    Absolute Value Inequality Help

    How do I solve this? Been stuck on it forever!

    $\displaystyle |x+2|\geq{x\over{x+2}}$
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  2. #2
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    Quote Originally Posted by nathan02079 View Post
    How do I solve this? Been stuck on it forever!

    $\displaystyle |x+2|\geq{x\over{x+2}}$
    Square both sides to remove the modulus.

    Then you have

    $\displaystyle (x + 2)^2 \geq \frac{x^2}{(x + 2)^2}$.

    You should be able to solve this inequality now.
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  3. #3
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    Inequality

    Quote Originally Posted by Prove It View Post
    Square both sides to remove the modulus.

    Then you have

    $\displaystyle (x + 2)^2 \geq \frac{x^2}{(x + 2)^2}$.

    You should be able to solve this inequality now.
    You need to be careful if you're squaring both sides of an inequality:

    $\displaystyle 0.5 > - 3$

    $\displaystyle \Rightarrow 0.25 > 9$? I don't think so.

    If $\displaystyle x = -1.5$, then $\displaystyle |x+2| = 0.5$ and $\displaystyle \frac{x}{x+2} = -3 \dots$ ???

    I think instead, you'll need to look at the signs of $\displaystyle (x+2)$ and $\displaystyle |x+2|$ in the cases

    • $\displaystyle x>-2$
    • $\displaystyle x<-2$

    and consider each one as a separate inequality. Have you tried this approach?

    Grandad
    Last edited by Grandad; Jan 11th 2009 at 11:12 PM. Reason: Simplified solution
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  4. #4
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    Quote Originally Posted by nathan02079 View Post
    How do I solve this? Been stuck on it forever!

    $\displaystyle |x+2|\geq{x\over{x+2}}$
    1. The domain is $\displaystyle d = \mathbb{R} \setminus \{-2\}$

    2. $\displaystyle |x+2| = \left\{ \begin{array}{l} x+2,\ x >-2 \\ -(x+2),\ x < -2\end{array}\right.$

    3. $\displaystyle x+2\geq\dfrac x{x+2}\ ,\ x > -2$

    $\displaystyle (x+2)^2-x\geq 0~\implies~\left(x+\frac32\right)^2+\dfrac74\geq 0$
    is true for all x because both summands at the LHS are positive.

    Therefore $\displaystyle x>-2$

    4. $\displaystyle -(x+2) \geq \dfrac x{x+2}\ ,\ x < -2$

    $\displaystyle -x^2-5x-4 \bold{{\color{red}\leq}} 0~\implies~(x+1)(x+4) \bold{{\color{red}\geq}} 0~\implies~x \geq -1\wedge x\geq -4~\vee~ x \leq -1\wedge x \leq -4$

    Therefore $\displaystyle x \leq -4$

    Thanks o_O!
    EDIT: I've removed my mistake. Corrections in red!
    Last edited by earboth; Jan 12th 2009 at 12:47 AM. Reason: corrections
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  5. #5
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    Quote Originally Posted by earboth View Post
    4. $\displaystyle -(x+2) \geq \dfrac x{x+2}\ ,\ x < -2$

    $\displaystyle -x^2-5x-4 \ {\color{red}\leq} \ 0$
    Small correction in red.

    Reason being:

    $\displaystyle \begin{aligned} -(x+2) & \geq \frac{x}{x+2} \\ -(x+2)^2 & \ {\color{red} \leq} \ x \qquad \text{Since (x+2) is negative, inequality sign changes} \\ -x^2 - 4x - 4 & \leq x \\ -x^2 - 5x - 4 & \leq 0 \\ & \ \ \vdots \end{aligned} $
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