Do we need to build an inequality for $|x+4| -4 =x $?

Question

$$|x+4| -4 =x $$

I've two questions about this equation.

• Why do we need to build an inequality?

• If we build an inequality, in what cases do we need to analyse?

Also I'm trying to find the negative values that $x$ can take.

Answer 1

Reasoning for your new question is the same.

There are two cases for killing of the absolute value:

1. $x\geq-4$, which gives $x+4-4=x,$ which says that the equation has infenitely many solutions in this case:

Any number from the set $[-4,+\infty)$ is valid;

2. $x<-4$, which gives $-x-4-4=x$ or $x=-4$, which is not valid in this case.

The answer is: $[-4,+\infty).$

Answer 2

Building an inequality for cases $x < -4$ and $x \ge -4$ is simply in order to make things easier. Because,

Case 1 $(x < -4)$: In this case, $|x+4| = -x-4$ so we have $-x-4-4 = x \implies x = -4$ but $-4 \not\lt -4$ so this solution is not valid for this case.

Case 2 $(x \ge -4)$: In this case, $|x+4| = x+4$ so we have $x+4-4 = x \implies 0 = 0$ so for all $x \ge-4$, this equation holds (Notice that it also includes the solution from case 1, which was not valid). So the answer is equation holds for all $x \ge -4$.

Answer 3

First of all, note that this isn’t an inequality, but an equation. There are two methods to solve this:

• Case by case analysis: Note that the equation can be written as: $$|x+4|= x +4 $$ When $x\leq -4$, we know that $(x+4)\leq 0$, hence, $|x+4|=-4-x$. Thus, it boils down to solving: $$-4-x=x+4 \implies x = -4$$ which is valid in the range. When $x > -4$, we know that $(x+4)>0$, hence, $|x+4|=x+4$. Thus, it boils down to solving $$x+4=x+4 \implies \text{ Holds } \forall x>-4$$

• By squaring: Note that when we square the absolute value, we can get rid of the anonymity surrounding it, thus: $$|x+4|=x+4 \implies |x+4|^2=(x+4)^2 \implies x^2+8x+16 = x^2+8x + 16 \implies \text{ Holds } \forall x\geq -4$$

Note that the second method is easier than the first.