Question

How do you solve and check for extraneous solutions in `abs(x + 6) = 2x`?

Answer 1

`x = 6`

Explanation:

Your absolute value equation looks like this

`|x+6| = 2x`

Right from the start, you can say that any negative value of `x` will be an extraneous solution because the absolute value of a number can only be positive.

So, you need to check two cases for your equation

• If `(x+6)>0`, you have

`|x+6| = x+6`

The equation becomes

`x+6 = 2x => x = color(green)(6)`

• If `(x+6)<0`, you have

`|x+6| = -(x+6) = -x-6`

The equation will be

`-x-6 = 2x => 3x = -6 => x = (-6)/3 = color(red)(-2)`

This solution will be extraneous because it implies that the absolute values of `4` is negative, which is false.

`|color(red)(-2) + 6| = 2 * color(red)((-2))`

`|4| = -4 <=> 4!=-4`

The first solution is valid, since you have

`|color(green)(6) + 6| = 2 * color(green)(6)`

`|12| = 12 <=> 12 = 12`