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

Sep 4, 2016

$x \in \left[- 1 , 1 - \frac{\sqrt{71}}{18}\right)$

#### Explanation:

First find values of $x$ where:

$\sqrt{3 - x} - \sqrt{x + 1} = \frac{1}{3}$

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

$\left(3 - x\right) - 2 \sqrt{\left(3 - x\right) \left(x + 1\right)} + \left(x + 1\right) = \frac{1}{9}$

which simplifies to:

$4 - 2 \sqrt{3 + 2 x - {x}^{2}} = \frac{1}{9}$

Multiply both sides by $9$ to get:

$36 - 18 \sqrt{3 + 2 x - {x}^{2}} = 1$

Hence:

$35 = 18 \sqrt{3 + 2 x - {x}^{2}}$

Square both sides to get:

$1225 = 324 \left(3 + 2 x - {x}^{2}\right) = 972 + 648 x - 324 {x}^{2}$

Rearrange to get:

$324 {x}^{2} - 648 x + 253 = 0$

Divide through by $324$ to get:

$0 = {x}^{2} - 2 x + \frac{253}{324} = {\left(x - 1\right)}^{2} - \frac{71}{324}$

Hence $x = 1 \pm \frac{\sqrt{71}}{18}$

Note that if $x = 1 + \frac{\sqrt{71}}{18}$ then $3 - x < 2$, $x + 1 > 2$ and:

$\sqrt{3 - x} - \sqrt{x + 1} < 0$

If $x = 1 - \frac{\sqrt{71}}{18}$ then:

$\sqrt{3 - x} - \sqrt{x + 1} = \sqrt{2 + \frac{\sqrt{71}}{18}} - \sqrt{2 - \frac{\sqrt{71}}{18}} > 0$

and we find:

${\left(\sqrt{3 - x} - \sqrt{x + 1}\right)}^{2} = {\left(\sqrt{2 + \frac{\sqrt{71}}{18}} - \sqrt{2 - \frac{\sqrt{71}}{18}}\right)}^{2}$

$\textcolor{w h i t e}{{\left(\sqrt{3 - x} - \sqrt{x + 1}\right)}^{2}} = \left(2 + \frac{\sqrt{71}}{18}\right) - 2 \sqrt{\left(2 + \frac{\sqrt{71}}{18}\right) \left(2 - \frac{\sqrt{71}}{18}\right)} + \left(2 - \frac{\sqrt{71}}{18}\right)$

$\textcolor{w h i t e}{{\left(\sqrt{3 - x} - \sqrt{x + 1}\right)}^{2}} = 4 - 2 \sqrt{4 - \frac{71}{324}}$

$\textcolor{w h i t e}{{\left(\sqrt{3 - x} - \sqrt{x + 1}\right)}^{2}} = 4 - 2 \sqrt{\frac{1225}{324}}$

$\textcolor{w h i t e}{{\left(\sqrt{3 - x} - \sqrt{x + 1}\right)}^{2}} = 4 - 2 \cdot \frac{35}{18}$

$\textcolor{w h i t e}{{\left(\sqrt{3 - x} - \sqrt{x + 1}\right)}^{2}} = \frac{36}{9} - \frac{35}{9}$

$\textcolor{w h i t e}{{\left(\sqrt{3 - x} - \sqrt{x + 1}\right)}^{2}} = \frac{1}{9}$

So this is a valid solution.

Now if $- 1 \le x < 1 - \frac{\sqrt{71}}{18}$ then both square roots are defined and $\sqrt{3 - x} - \sqrt{x + 1} > \frac{1}{3}$