Absolute Value Inequality (another)

Question

Solve the following inequality

$$|a-2| > |a+4|$$

Here I separated it into cases as shown

$a<-4$

$$-(a-2) > -(a+4) \implies 2-a>-a-4 \implies 0>-6$$

Always true, so we get $\mathbb{R} \cap (-\infty , -4) = (-\infty ,-4)$

$-4<a<2$

$$-(a-2)>a+4 \implies 2-a>a+4 \implies a<-1$$

Taking interception $(-4,2)$ $\cap $ $(-\infty,-1)$

$a>2$

$$a-2 > a+4 \implies -2>4 $$

Always false, so no solution from there. Finally, I checked end points and noticed that they do not work in this inequality. However, I do not know how to proceed further. Could you assist me?

Regards

Answer 1

You have to distinguish the following cases: $$x\geq 2$$ then we get $$x-2>x+4$$ $$-4\le x<2$$ then we get $$x-2>-x+4$$ $$x<-4$$ then we have $$-x+2>-x-4$$

Answer 2

As commented, $\;a=4\;$ doesn't really fits in the inequality. You can now put your solution set as

$$(-\infty,-4)\cup\left((-4,-2)\cap(-\infty,-1)\right)=(-\infty,-4)\cup(-4,-1)=(-\infty,-1)\setminus\{-4\}$$

Answer 3

If you interpret the absolute values in terms of distance, it is immediate:

$|a-2|>|a+4|$ means $a$ is nearer to $-4$ than to $2$, so it is less than the arithmetic mean of $2$ and $-4$: $$a<\frac{2-4}2=-1$$

Other method: \begin{align} |a-2|>|a+4|&\iff(a-2)^2>(a+4)^2 \\ &\iff a^2-4a+4>a^2+8a+16\\ &\iff 12a+12<0\iff a<-1. \end{align}

Answer 4

$|(a+1)-3|>|(a+1)+3|;$

$x:=a+1$;

$|x-(+3)| >|x-(-3)|;$

Real number line :

Distance from a point $x$ to $(+3)$ is bigger than from $x$ to $(-3)$, i.e. $x<0.$

(Note : At $x=0$ both distances are equal, for $x <0$, $x$ is to the left of $0$ , closer to $(-3)$ ).

$x= a+1<0$, $a<-1$.