다음을 풀어보세요: (2ax^2+a^2x)neq4

$$( 2 a x ^ { 2 } + a ^ { 2 } x ) \neq 4$$

계산 (complex solution)

$$ax\left(2x+a\right)\neq 4$$

x에 대한 해

$$\left\{\begin{matrix}x\in \mathrm{R}\setminus -\frac{\sqrt{a^{4}+32a}}{4a}-\frac{a}{4},\frac{\sqrt{a^{4}+32a}}{4a}-\frac{a}{4}\text{, }&a>0\text{ or }a\leq -2\sqrt[3]{4}\\x\in \mathrm{R}\text{, }&a>-2\sqrt[3]{4}\text{ and }a\leq 0\end{matrix}\right.$$

a에 대한 해

$$\left\{\begin{matrix}a\in \mathrm{R}\setminus -\frac{\sqrt{x^{4}+4x}}{x}-x,\frac{\sqrt{x^{4}+4x}}{x}-x\text{, }&x>0\text{ or }x\leq -\sqrt[3]{4}\\a\in \mathrm{R}\text{, }&x>-\sqrt[3]{4}\text{ and }x\leq 0\end{matrix}\right.$$