แก้โจทย์ -2x_{2}=8(x-1)(x-4)/(k-1)(k+1)

$$- 2 x _ { 2 } = \frac { 8 ( x - 1 ) ( x - 4 ) } { ( k - 1 ) ( k + 1 ) }$$

หาค่า x (complex solution)

$$x=\frac{\sqrt{9+x_{2}-x_{2}k^{2}}+5}{2}$$

$$x=\frac{-\sqrt{9+x_{2}-x_{2}k^{2}}+5}{2}\text{, }k\neq 1\text{ and }k\neq -1$$

หาค่า k

$$\left\{\begin{matrix}k=\sqrt{-\frac{4x^{2}-20x-x_{2}+16}{x_{2}}}\text{; }k=-\sqrt{-\frac{4x^{2}-20x-x_{2}+16}{x_{2}}}\text{, }&x\neq 4\text{ and }x\neq 1\text{ and }\left(x_{2}>0\text{ or }x_{2}\leq 4x^{2}-20x+16\right)\text{ and }\left(x_{2}<0\text{ or }x_{2}\geq 4x^{2}-20x+16\right)\text{ and }\left(x_{2}=4x^{2}-20x+16\text{ or }x_{2}\neq 0\right)\\k\in \mathrm{R}\setminus 1,-1\text{, }&\left(x_{2}=0\text{ and }x=1\right)\text{ or }\left(x_{2}=0\text{ and }x=4\right)\end{matrix}\right.$$

หาค่า x

$$x=\frac{\sqrt{9+x_{2}-x_{2}k^{2}}+5}{2}$$

$$x=\frac{-\sqrt{9+x_{2}-x_{2}k^{2}}+5}{2}\text{, }\left(x_{2}\geq -\frac{9}{1-k^{2}}\text{ and }k>-1\text{ and }|k|<1\right)\text{ or }\left(x_{2}=-\frac{9}{1-k^{2}}\text{ and }|k|\neq 1\right)\text{ or }\left(|k|>1\text{ and }x_{2}\leq -\frac{9}{1-k^{2}}\right)$$