Absolute Value Inequality (another)

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

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

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

• With $\;a=4\;$ your inequality is $$2=|4-2|>|4+4|=8...\text{you still think this is true?}$$ – DonAntonio Jan 13 '19 at 13:18
• @DonAntonio Not really, thanks for pointing out. – Melz Jan 13 '19 at 13:19

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

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

• What does this mathematical symbol mean \? – Melz Jan 13 '19 at 13:23
• Set difference. Sometimes it is also used $\;A-B\;$ ... A simple minus sign. – DonAntonio Jan 13 '19 at 13:23
• The right answer seems to be $(-\infty, -1)$ according to my textbook. Do we get it from your final interval?: – Melz Jan 13 '19 at 13:24
• That is correct, @Enzo . Your solution didn't even take into account $\;a=-4\;$ (perhaps you should have done this in the first case). That you have an inequality < or > doesn't mean you must take your cases in the same way – DonAntonio Jan 13 '19 at 13:34

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}

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 $$-x+2>-x-4$$

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

$$x:=a+1$$;

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

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 $$(-3)$$ ).

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