Solve 2M=(V^2+1)^2

$$2 M = ( V ^ { 2 } + 1 ) ^ { 2 }$$

Solve for M

$$M=\frac{\left(V^{2}+1\right)^{2}}{2}$$

Steps for Solving Linear Equation

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

$$2M=\left(V^{2}\right)^{2}+2V^{2}+1$$

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

$$2M=V^{4}+2V^{2}+1$$

Divide both sides by 2.

$$\frac{2M}{2}=\frac{\left(V^{2}+1\right)^{2}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$M=\frac{\left(V^{2}+1\right)^{2}}{2}$$

Solve for V (complex solution)

$$V=\sqrt{\sqrt{2M}-1}$$

$$V=-\sqrt{\sqrt{2M}-1}$$

$$V=-i\sqrt{\sqrt{2M}+1}$$

$$V=i\sqrt{\sqrt{2M}+1}$$

Solve for V

$$V=-\sqrt{\sqrt{2M}-1}$$

$$V=\sqrt{\sqrt{2M}-1}\text{, }M\geq \frac{1}{2}$$