# How do you solve |x+2|+|x-2|=4?

Jun 28, 2018

Every number between $- 2$ and $2$ (included) is a solution for this equation:

$- 2 \setminus \le x \setminus \le 2$

#### Explanation:

The absolute value is a compact notation to write two possibilities: $| x |$ equals $x$ if $x$ is positive, $- x$ if $x$ is negative.

This means that, to solve the absolute values in this equation, we have to consider all possible cases, depending on the sign of $x + 2$ and $x - 2$. Let's consider them all:

Case 1: $x < - 2$
In this case, both $x + 2$ and $x - 2$ are negative. This means that the absolute value flips the sign of both expressions: $| x + 2 | = - x - 2$ and $| x - 2 | = - x + 2$. The equation becomes

$- x - 2 - x + 2 = 4 \setminus \iff - 2 x = 4 \setminus \iff x = - 2$

But we are supposing $x < - 2$, so we reject this solution (for now)

Case 2: $- 2 \setminus \le x < 2$
In this case, $x + 2$ is positive and $x - 2$ is still negative. This means that the absolute value flips only the sign of the second expression: $| x + 2 | = x + 2$ and $| x - 2 | = - x + 2$. The equation becomes

$x + 2 - x + 2 = 4 \setminus \iff 4 = 4$

This means that every numer between $- 2$ (included) and $2$ (excluded, for now) is a solution for this equation.

Case 3: $x \ge 2$
In this case, both $x + 2$ and $x - 2$ are positive. This means that the absolute value has no effect anymore: $| x + 2 | = x + 2$ and $| x - 2 | = x - 2$. The equation becomes

$x + 2 + x - 2 = 4 \setminus \iff 2 x = 4 \setminus \iff x = 2$

This means that $2$ is actually a solution as well.

Here you can see that this function equals $4$ on the whole $\left[- 2 , 2\right]$ interval.