Solve Az(x^2-2)^2=x^2

$$Az ( x ^ { 2 } - 2 ) ^ { 2 } = x ^ { 2 }$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=\frac{\left(\frac{x}{x^{2}-2}\right)^{2}}{z}\text{, }&x\neq -\sqrt{2}\text{ and }x\neq \sqrt{2}\text{ and }z\neq 0\\A\in \mathrm{C}\text{, }&x=0\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-2\right)^{2}.

$$Az\left(\left(x^{2}\right)^{2}-4x^{2}+4\right)=x^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$Az\left(x^{4}-4x^{2}+4\right)=x^{2}$$

Use the distributive property to multiply Az by x^{4}-4x^{2}+4.

$$Azx^{4}-4Azx^{2}+4Az=x^{2}$$

Combine all terms containing A.

$$\left(zx^{4}-4zx^{2}+4z\right)A=x^{2}$$

Divide both sides by zx^{4}+4z-4zx^{2}.

$$\frac{\left(zx^{4}-4zx^{2}+4z\right)A}{zx^{4}-4zx^{2}+4z}=\frac{x^{2}}{zx^{4}-4zx^{2}+4z}$$

Dividing by zx^{4}+4z-4zx^{2} undoes the multiplication by zx^{4}+4z-4zx^{2}.

$$A=\frac{x^{2}}{zx^{4}-4zx^{2}+4z}$$

Divide x^{2} by zx^{4}+4z-4zx^{2}.

$$A=\frac{x^{2}}{z\left(x^{2}-2\right)^{2}}$$

Solve for A

$$\left\{\begin{matrix}A=\frac{\left(\frac{x}{x^{2}-2}\right)^{2}}{z}\text{, }&z\neq 0\text{ and }|x|\neq \sqrt{2}\\A\in \mathrm{R}\text{, }&x=0\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-2\right)^{2}.

$$Az\left(\left(x^{2}\right)^{2}-4x^{2}+4\right)=x^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$Az\left(x^{4}-4x^{2}+4\right)=x^{2}$$

Use the distributive property to multiply Az by x^{4}-4x^{2}+4.

$$Azx^{4}-4Azx^{2}+4Az=x^{2}$$

Combine all terms containing A.

$$\left(zx^{4}-4zx^{2}+4z\right)A=x^{2}$$

Divide both sides by zx^{4}+4z-4zx^{2}.

$$\frac{\left(zx^{4}-4zx^{2}+4z\right)A}{zx^{4}-4zx^{2}+4z}=\frac{x^{2}}{zx^{4}-4zx^{2}+4z}$$

Dividing by zx^{4}+4z-4zx^{2} undoes the multiplication by zx^{4}+4z-4zx^{2}.

$$A=\frac{x^{2}}{zx^{4}-4zx^{2}+4z}$$

Divide x^{2} by zx^{4}+4z-4zx^{2}.

$$A=\frac{x^{2}}{z\left(x^{2}-2\right)^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{A^{-\frac{1}{2}}z^{-\frac{1}{2}}\sqrt{2\left(4Az+\sqrt{8Az+1}+1\right)}}{2}\text{; }x=-\frac{A^{-\frac{1}{2}}z^{-\frac{1}{2}}\sqrt{2\left(4Az+\sqrt{8Az+1}+1\right)}}{2}\text{; }x=-\frac{iA^{-\frac{1}{2}}z^{-\frac{1}{2}}\sqrt{2\left(-4Az+\sqrt{8Az+1}-1\right)}}{2}\text{; }x=\frac{iA^{-\frac{1}{2}}z^{-\frac{1}{2}}\sqrt{2\left(-4Az+\sqrt{8Az+1}-1\right)}}{2}\text{, }&z\neq 0\text{ and }A\neq 0\\x=-2\left(4Az+1\right)^{-\frac{1}{2}}\sqrt{A}\sqrt{z}\text{; }x=2\left(4Az+1\right)^{-\frac{1}{2}}\sqrt{A}\sqrt{z}\text{, }&z=0\text{ or }A=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{\frac{2\left(4Az+\sqrt{8Az+1}+1\right)}{Az}}}{2}\text{; }x=\frac{\sqrt{\frac{2\left(4Az+\sqrt{8Az+1}+1\right)}{Az}}}{2}\text{; }x=\frac{\sqrt{\frac{2\left(4Az-\sqrt{8Az+1}+1\right)}{Az}}}{2}\text{; }x=-\frac{\sqrt{\frac{2\left(4Az-\sqrt{8Az+1}+1\right)}{Az}}}{2}\text{, }&\left(A>-\frac{1}{4z}\text{ and }z>0\text{ and }A>0\text{ and }A>-\frac{1}{8z}\right)\text{ or }\left(A<-\frac{1}{4z}\text{ and }A<0\text{ and }z>0\text{ and }A>-\frac{1}{8z}\right)\\x=-\frac{\sqrt{\frac{2\left(4Az+\sqrt{8Az+1}+1\right)}{Az}}}{2}\text{; }x=\frac{\sqrt{\frac{2\left(4Az+\sqrt{8Az+1}+1\right)}{Az}}}{2}\text{; }x=\frac{\sqrt{\frac{2\left(4Az-\sqrt{8Az+1}+1\right)}{Az}}}{2}\text{; }x=-\frac{\sqrt{\frac{2\left(4Az-\sqrt{8Az+1}+1\right)}{Az}}}{2}\text{, }&\left(A>-\frac{1}{4z}\text{ and }z<0\text{ and }A>0\text{ and }A<-\frac{1}{8z}\right)\text{ or }\left(A<-\frac{1}{4z}\text{ and }z<0\text{ and }A<0\text{ and }A<-\frac{1}{8z}\right)\\x=2\sqrt{\frac{Az}{4Az+1}}\text{; }x=-2\sqrt{\frac{Az}{4Az+1}}\text{, }&z=0\text{ or }A=0\end{matrix}\right.$$