How to get rid of the absolute value in $|x + 2| > -4$?

Question

How to get rid of the absolute value in $|x + 2| > -4$? I know that $|x|> a$ is equivalent to $x > a$ or $x < -a$, but I think that this definition is not applicable here? am I right ?

Answer 1

Since the absolute value is always greater than or equal to zero, this statement is true for all values of $x$.

To further answer your question, we can write this as

$$x + 2 > -4 \mbox { or } x + 2 < 4$$

Again, since $x+2$ is either at least $-4$ or at most $4$, this is true for all $x$.

Answer 2

You can use the definition of the absolute value: $$|x| = \left\{ \begin{array}{cc} x & x\geq 0\\ -x & x < 0 \end{array}\right..$$ So for your case you have: $$|x+2| = \left\{ \begin{array}{cc} x+2 & x\geq -2\\ -(x+2) & x < -2 \end{array}\right..$$

Answer 3

The equivalence $$ |x| > a \iff x > a\ \text{or}\ x<-a $$ holds for every $a\in\mathbb{R}$. It's usually only stated for $a\ge 0$ because we have $|x|\ge 0$ for every $x\in\mathbb{R}$.

To see why it remains true, suppose $a<0$. Then the right hand side of the equivalence above becomes $$ x\in(a,\infty)\cup(-\infty,-a)=\mathbb{R}, $$ which is the same set of solutions for the left hand side.

Answer 4

Just draw a number line and think about $|x+2|>-4$ when can this happen? there can be two cases 1) when $x+2$ is positive then from number line and the inequality $x+2 $ is in the right side of $-4$ or $x+2 > -4$ or $x > -6$

2) when $x+2$ is negative then from number line nd the inequality this can happen when $x+2$ is in the left of $-4$ or $x+2 < -4$ or $x < -6$ .

Concepts about absolute values/modulus simplifies when you consider number line.

So your initial thought was correct right!.