حل x^{2}-2(a^{2}+b^2)x+(a^2-b^2)^2=0

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

حل مسائل a (complex solution)

$$a=\sqrt{-2\sqrt{xb^{2}}+b^{2}+x}$$

$$a=-\sqrt{-2\sqrt{xb^{2}}+b^{2}+x}$$

$$a=-\sqrt{2\sqrt{xb^{2}}+b^{2}+x}$$

$$a=\sqrt{2\sqrt{xb^{2}}+b^{2}+x}$$

حل مسائل b (complex solution)

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

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

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

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

حل مسائل a

$$\left\{\begin{matrix}a=-b-\sqrt{x}\text{; }a=b+\sqrt{x}\text{; }a=-b+\sqrt{x}\text{; }a=b-\sqrt{x}\text{, }&x\geq 0\\a=\sqrt{x}\text{; }a=-\sqrt{x}\text{, }&x\geq 0\text{ and }b=0\end{matrix}\right.$$

حل مسائل b

$$\left\{\begin{matrix}b=-a-\sqrt{x}\text{; }b=a+\sqrt{x}\text{; }b=-a+\sqrt{x}\text{; }b=a-\sqrt{x}\text{, }&x\geq 0\\b=\sqrt{x}\text{; }b=-\sqrt{x}\text{, }&x\geq 0\text{ and }a=0\end{matrix}\right.$$