Question

Find the real solution(s) of `sqrt(3-x)-sqrt(x+1) > 1/3` ?

Answer 1

`x in [-1, 1 - sqrt(71)/18)`

Explanation:

First find values of `x` where:

`sqrt(3-x)-sqrt(x+1) = 1/3`

Squaring both sides (which may introduce spurious solutions), we get:

`(3-x)-2sqrt((3-x)(x+1))+(x+1) = 1/9`

which simplifies to:

`4-2 sqrt(3+2x-x^2) = 1/9`

Multiply both sides by `9` to get:

`36-18 sqrt(3+2x-x^2) = 1`

Hence:

`35 = 18 sqrt(3+2x-x^2)`

Square both sides to get:

`1225 = 324(3+2x-x^2) = 972+648x-324x^2`

Rearrange to get:

`324x^2-648x+253 = 0`

Divide through by `324` to get:

`0 = x^2-2x+253/324 = (x-1)^2-71/324`

Hence `x = 1+-sqrt(71)/18`

Note that if `x = 1+sqrt(71)/18` then `3-x < 2`, `x+1 > 2` and:

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

If `x = 1-sqrt(71)/18` then:

`sqrt(3-x)-sqrt(x+1) = sqrt(2+sqrt(71)/18)-sqrt(2-sqrt(71)/18) > 0`

and we find:

`(sqrt(3-x)-sqrt(x+1))^2 = (sqrt(2+sqrt(71)/18)-sqrt(2-sqrt(71)/18))^2`

`color(white)((sqrt(3-x)-sqrt(x+1))^2) = (2+sqrt(71)/18)-2sqrt((2+sqrt(71)/18)(2-sqrt(71)/18))+(2-sqrt(71)/18)`

`color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2sqrt(4-71/324)`

`color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2sqrt(1225/324)`

`color(white)((sqrt(3-x)-sqrt(x+1))^2) = 4-2*35/18`

`color(white)((sqrt(3-x)-sqrt(x+1))^2) = 36/9-35/9`

`color(white)((sqrt(3-x)-sqrt(x+1))^2) = 1/9`

So this is a valid solution.

Now if `-1 <= x < 1-sqrt(71)/18` then both square roots are defined and `sqrt(3-x)-sqrt(x+1) > 1/3`