x ∈ ℝ : 6 + x - x 2 2 x + 5 ≥ 6 + x - x 2 x + 4 =

It is given that, $\frac{\sqrt{6+x-x^2}}{2x+5}\ge \frac{\sqrt{6+x-x^2}}{x+4}$
Rewrite the above equation.

$\Rightarrow \sqrt{6+x-x^2}\left(\frac{1}{2x+5}-\frac{1}{x+4}\right)\ge 0$

$\Rightarrow \sqrt{6+x-x^2}\left(\frac{x+4-2x-5}{(2x+5)(x+4)}\right)\ge 0$

$\Rightarrow \sqrt{6+x-x^2}\left(\frac{-(x+1)}{(2x+5)(x+4)}\right)\ge 0$

$\Rightarrow (x+1)(2x+5)(x+4)\le 0$

Now, $x\in (-\infty ,-4)\cup (-\frac{5}{2},-1] ...\left(\mathrm{i}\right)$ 

For the expression to exist

$6+x-x^2\ge 0$

$\Rightarrow x^2-x-6\le 0$

$\Rightarrow (x-3)(x+2)\le 0$

$x\in [-2,3] ...\left(\mathrm{ii}\right)$  

From Equations $\left(\mathrm{i}\right)$ and

$x\in [-2,-1]\cup \left\{3\right\}$