# Absolute Value Inequality Help

• Jan 11th 2009, 10:50 PM
nathan02079
Absolute Value Inequality Help
How do I solve this? Been stuck on it forever!

$|x+2|\geq{x\over{x+2}}$
• Jan 11th 2009, 11:24 PM
Prove It
Quote:

Originally Posted by nathan02079
How do I solve this? Been stuck on it forever!

$|x+2|\geq{x\over{x+2}}$

Square both sides to remove the modulus.

Then you have

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

You should be able to solve this inequality now.
• Jan 11th 2009, 11:38 PM
Inequality
Quote:

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

Then you have

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

You should be able to solve this inequality now.

You need to be careful if you're squaring both sides of an inequality:

$0.5 > - 3$

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

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

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

• $x>-2$
• $x<-2$

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

• Jan 11th 2009, 11:43 PM
earboth
Quote:

Originally Posted by nathan02079
How do I solve this? Been stuck on it forever!

$|x+2|\geq{x\over{x+2}}$

1. The domain is $d = \mathbb{R} \setminus \{-2\}$

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

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

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

Therefore $x>-2$

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

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

Therefore $x \leq -4$

Thanks o_O!
EDIT: I've removed my mistake. Corrections in red!
• Jan 12th 2009, 01:27 AM
o_O
Quote:

Originally Posted by earboth
4. $-(x+2) \geq \dfrac x{x+2}\ ,\ x < -2$

$-x^2-5x-4 \ {\color{red}\leq} \ 0$

Small correction in red.

Reason being:

\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}