Question

How do you solve the inequality `abs(x-3)-abs(2x+1)<0` and write your answer in interval notation?

Answer 1

Assuming x is real, use the alternate form:

`sqrt((x-3)^2)-sqrt((2x+1)^2)< 0`

This implies that:

`(x-3)^2 < (2x+1)^2`

Expand the squares and solve the resulting quadratic inequality.

Explanation:

Starting with the above:

`(x-3)^2 < (2x+1)^2`

Expand the squares

`x^2 - 2x +9 < 4x^2 + 4x + 1`

Combine like terms:

`0 < 3x^2+ 6x - 8`

The quadratic will be greater than 0 to the left and to the right of the roots, therefore, we should find the roots:

`0 = 3x^2+ 6x - 8`

Factor:

`0 = (3x - 2)(x+4)`

`x = 2/3` and `x = -4`

The intervals to the left and to the right of these roots are:

`(-oo, -4)` and `(2/3, oo)`

Here is a graph of `y = |x - 3|- |2x+1|`

Please observe that it drops below the `y = 0` line at `x = -4` and `x = 2/3`