Solve 4x^{2}-2(a^2+b^2)x+(a^4-b^4)=0

$$4 x ^ { 2 } - 2 ( a ^ { 2 } + b ^ { 2 } ) x + ( a ^ { 4 } - b ^ { 4 } ) = 0$$

Solve for a (complex solution)

$$a=\sqrt{-\sqrt{\left(b^{2}-x\right)\left(3x+b^{2}\right)}+x}$$

$$a=-\sqrt{-\sqrt{\left(b^{2}-x\right)\left(3x+b^{2}\right)}+x}$$

$$a=-\sqrt{\sqrt{\left(b^{2}-x\right)\left(3x+b^{2}\right)}+x}$$

$$a=\sqrt{\sqrt{\left(b^{2}-x\right)\left(3x+b^{2}\right)}+x}$$

Solve for b (complex solution)

$$b=\sqrt{\sqrt{5x^{2}-2xa^{2}+a^{4}}-x}$$

$$b=-\sqrt{\sqrt{5x^{2}-2xa^{2}+a^{4}}-x}$$

$$b=-i\sqrt{\sqrt{5x^{2}-2xa^{2}+a^{4}}+x}$$

$$b=i\sqrt{\sqrt{5x^{2}-2xa^{2}+a^{4}}+x}$$

Solve for b

$$\left\{\begin{matrix}\\b=-\sqrt{\sqrt{5x^{2}-2xa^{2}+a^{4}}-x}\text{; }b=\sqrt{\sqrt{5x^{2}-2xa^{2}+a^{4}}-x}\text{, }&\text{unconditionally}\\b=0\text{, }&x=0\text{ and }a=0\end{matrix}\right.$$