Solve f*y=(sqrt{alpha}-1)^2-(alpha^2+1)^4

$$f \cdot y = ( \sqrt { \alpha } - 1 ) ^ { 2 } - ( \alpha ^ { 2 } + 1 ) ^ { 4 }$$

Solve for f (complex solution)

$$\left\{\begin{matrix}f=\frac{\sqrt{\alpha }\left(\alpha ^{\frac{7}{2}}+2\alpha ^{\frac{3}{2}}+1\right)\left(-\alpha ^{4}-2\alpha ^{2}+\sqrt{\alpha }-2\right)}{y}\text{, }&y\neq 0\\f\in \mathrm{C}\text{, }&\left(-\left(\alpha ^{2}+1\right)^{2}+\sqrt{\alpha }-1\right)\left(\left(\alpha ^{2}+1\right)^{2}+\sqrt{\alpha }-1\right)=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{\alpha }-1\right)^{2}.

$$fy=\left(\sqrt{\alpha }\right)^{2}-2\sqrt{\alpha }+1-\left(\alpha ^{2}+1\right)^{4}$$

Calculate \sqrt{\alpha } to the power of 2 and get \alpha .

$$fy=\alpha -2\sqrt{\alpha }+1-\left(\alpha ^{2}+1\right)^{4}$$

Reorder the terms.

$$fy=-\left(\alpha ^{2}+1\right)^{4}+\alpha -2\sqrt{\alpha }+1$$

The equation is in standard form.

$$yf=-\left(\alpha ^{2}+1\right)^{4}+\alpha -2\sqrt{\alpha }+1$$

Divide both sides by y.

$$\frac{yf}{y}=\frac{-\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2}}{y}$$

Dividing by y undoes the multiplication by y.

$$f=\frac{-\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2}}{y}$$

Divide -\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2} by y.

$$f=\frac{\sqrt{\alpha }\left(\alpha ^{\frac{7}{2}}+2\alpha ^{\frac{3}{2}}+1\right)\left(-\alpha ^{4}-2\alpha ^{2}+\sqrt{\alpha }-2\right)}{y}$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=\frac{\sqrt{\alpha }\left(\alpha ^{\frac{7}{2}}+2\alpha ^{\frac{3}{2}}+1\right)\left(-\alpha ^{4}-2\alpha ^{2}+\sqrt{\alpha }-2\right)}{f}\text{, }&f\neq 0\\y\in \mathrm{C}\text{, }&\left(-\left(\alpha ^{2}+1\right)^{2}+\sqrt{\alpha }-1\right)\left(\left(\alpha ^{2}+1\right)^{2}+\sqrt{\alpha }-1\right)=0\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{\alpha }-1\right)^{2}.

$$fy=\left(\sqrt{\alpha }\right)^{2}-2\sqrt{\alpha }+1-\left(\alpha ^{2}+1\right)^{4}$$

Calculate \sqrt{\alpha } to the power of 2 and get \alpha .

$$fy=\alpha -2\sqrt{\alpha }+1-\left(\alpha ^{2}+1\right)^{4}$$

Reorder the terms.

$$fy=-\left(\alpha ^{2}+1\right)^{4}+\alpha -2\sqrt{\alpha }+1$$

Divide both sides by f.

$$\frac{fy}{f}=\frac{-\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2}}{f}$$

Dividing by f undoes the multiplication by f.

$$y=\frac{-\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2}}{f}$$

Divide -\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2} by f.

$$y=\frac{\sqrt{\alpha }\left(\alpha ^{\frac{7}{2}}+2\alpha ^{\frac{3}{2}}+1\right)\left(-\alpha ^{4}-2\alpha ^{2}+\sqrt{\alpha }-2\right)}{f}$$

Solve for f

$$\left\{\begin{matrix}f=\frac{\sqrt{\alpha }\left(\alpha ^{\frac{7}{2}}+2\alpha ^{\frac{3}{2}}+1\right)\left(-\alpha ^{4}-2\alpha ^{2}+\sqrt{\alpha }-2\right)}{y}\text{, }&y\neq 0\text{ and }\alpha \geq 0\\f\in \mathrm{R}\text{, }&0=-\left(\alpha ^{2}+1\right)^{4}+\alpha -2\sqrt{\alpha }+1\text{ and }\alpha \geq 0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{\alpha }-1\right)^{2}.

$$fy=\left(\sqrt{\alpha }\right)^{2}-2\sqrt{\alpha }+1-\left(\alpha ^{2}+1\right)^{4}$$

Calculate \sqrt{\alpha } to the power of 2 and get \alpha .

$$fy=\alpha -2\sqrt{\alpha }+1-\left(\alpha ^{2}+1\right)^{4}$$

Reorder the terms.

$$fy=-\left(\alpha ^{2}+1\right)^{4}+\alpha -2\sqrt{\alpha }+1$$

The equation is in standard form.

$$yf=-\left(\alpha ^{2}+1\right)^{4}+\alpha -2\sqrt{\alpha }+1$$

Divide both sides by y.

$$\frac{yf}{y}=\frac{-\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2}}{y}$$

Dividing by y undoes the multiplication by y.

$$f=\frac{-\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2}}{y}$$

Divide -\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2} by y.

$$f=\frac{\sqrt{\alpha }\left(\alpha ^{\frac{7}{2}}+2\alpha ^{\frac{3}{2}}+1\right)\left(-\alpha ^{4}-2\alpha ^{2}+\sqrt{\alpha }-2\right)}{y}$$

Solve for y

$$\left\{\begin{matrix}y=\frac{\sqrt{\alpha }\left(\alpha ^{\frac{7}{2}}+2\alpha ^{\frac{3}{2}}+1\right)\left(-\alpha ^{4}-2\alpha ^{2}+\sqrt{\alpha }-2\right)}{f}\text{, }&f\neq 0\text{ and }\alpha \geq 0\\y\in \mathrm{R}\text{, }&0=-\left(\alpha ^{2}+1\right)^{4}+\alpha -2\sqrt{\alpha }+1\text{ and }\alpha \geq 0\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{\alpha }-1\right)^{2}.

$$fy=\left(\sqrt{\alpha }\right)^{2}-2\sqrt{\alpha }+1-\left(\alpha ^{2}+1\right)^{4}$$

Calculate \sqrt{\alpha } to the power of 2 and get \alpha .

$$fy=\alpha -2\sqrt{\alpha }+1-\left(\alpha ^{2}+1\right)^{4}$$

Reorder the terms.

$$fy=-\left(\alpha ^{2}+1\right)^{4}+\alpha -2\sqrt{\alpha }+1$$

Divide both sides by f.

$$\frac{fy}{f}=\frac{-\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2}}{f}$$

Dividing by f undoes the multiplication by f.

$$y=\frac{-\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2}}{f}$$

Divide -\left(\alpha ^{2}+1\right)^{4}+\left(\sqrt{\alpha }-1\right)^{2} by f.

$$y=\frac{\sqrt{\alpha }\left(\alpha ^{\frac{7}{2}}+2\alpha ^{\frac{3}{2}}+1\right)\left(-\alpha ^{4}-2\alpha ^{2}+\sqrt{\alpha }-2\right)}{f}$$