Réitigh A=sqrt{4-x}+sqrt{x+2}

$$A = \sqrt { 4 - x } + \sqrt { x + 2 }$$

Réitigh do A. (complex solution)

$$A=\sqrt{x+2}+\sqrt{4-x}$$

Réitigh do x. (complex solution)

$$x=-\frac{\sqrt{12A^{2}-A^{4}}}{2}+1$$

$$x=\frac{\sqrt{12A^{2}-A^{4}}}{2}+1\text{, }\left(|-arg(-A)+arg(-\sqrt{x+2}-\sqrt{4-x})|<\pi \text{ and }|arg(\sqrt{-\left(-\frac{\sqrt{12A^{2}-A^{4}}}{2}\right)^{2}+9})-arg(\frac{A^{2}}{2}-3)|<\pi \text{ and }A\neq -\sqrt{6}\text{ and }A\neq \sqrt{6}\right)\text{ or }\left(|arg(-\sqrt{x+2}-\sqrt{4-x})-\pi |<\pi \text{ and }A=\sqrt{6}\right)\text{ or }\left(arg(-\sqrt{x+2}-\sqrt{4-x})<\pi \text{ and }A=-\sqrt{6}\right)$$

Réitigh do A.

$$A=\sqrt{x+2}+\sqrt{4-x}$$

$$x\geq -2\text{ and }x\leq 4$$

Réitigh do x.

$$x=\frac{A\sqrt{12-A^{2}}+2}{2}$$

$$x=\frac{-A\sqrt{12-A^{2}}+2}{2}\text{, }\left(A\leq -\sqrt{6}\text{ and }A\geq -2\sqrt{3}\text{ and }-\frac{\sqrt{2}\left(\sqrt{A\sqrt{12-A^{2}}+6}+\sqrt{-A\sqrt{12-A^{2}}+6}\right)}{2}\geq 0\text{ and }-A\sqrt{12-A^{2}}\leq 6\right)\text{ or }A=\sqrt{6}\text{ or }\left(A\leq 2\sqrt{3}\text{ and }A\geq \sqrt{6}\text{ and }A\sqrt{12-A^{2}}\leq 6\right)$$