Solve (f(x)=[B(x)-C(x)]^2

$$( f ( x ) = [ B ( x ) - C ( x ) ] ^ { 2 }$$

Solve for B (complex solution)

$$\left\{\begin{matrix}B\in \mathrm{C}\text{, }&x=0\\B=C-x^{-\frac{1}{2}}\sqrt{f}\text{; }B=C+x^{-\frac{1}{2}}\sqrt{f}\text{, }&x\neq 0\end{matrix}\right.$$

Solve for C (complex solution)

$$\left\{\begin{matrix}C\in \mathrm{C}\text{, }&x=0\\C=B+x^{-\frac{1}{2}}\sqrt{f}\text{; }C=B-x^{-\frac{1}{2}}\sqrt{f}\text{, }&x\neq 0\end{matrix}\right.$$

Solve for B

$$\left\{\begin{matrix}B\in \mathrm{R}\text{, }&x=0\\B=C-\frac{\sqrt{fx}}{x}\text{; }B=C+\frac{\sqrt{fx}}{x}\text{, }&f\leq 0\text{ and }x<0\\B=C-\sqrt{\frac{f}{x}}\text{; }B=C+\sqrt{\frac{f}{x}}\text{, }&f\geq 0\text{ and }x>0\end{matrix}\right.$$

Solve for C

$$\left\{\begin{matrix}C\in \mathrm{R}\text{, }&x=0\\C=B+\frac{\sqrt{fx}}{x}\text{; }C=B-\frac{\sqrt{fx}}{x}\text{, }&f\leq 0\text{ and }x<0\\C=B+\sqrt{\frac{f}{x}}\text{; }C=B-\sqrt{\frac{f}{x}}\text{, }&f\geq 0\text{ and }x>0\end{matrix}\right.$$