Question

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

Answer 1

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

`-2\le x \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 \iff -2x=4 \iff x=-2`

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

Case 2: `-2 \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 \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>=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 \iff 2x=4 \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 `[-2,2]` interval.