Solve v^2-2x+1=0

$$v ^ { 2 } - 2 x + 1 = 0$$

Solve for x

$$x=\frac{v^{2}+1}{2}$$

Steps for Solving Linear Equation

Subtract v^{2} from both sides. Anything subtracted from zero gives its negation.

$$-2x+1=-v^{2}$$

Subtract 1 from both sides.

$$-2x=-v^{2}-1$$

Divide both sides by -2.

$$\frac{-2x}{-2}=\frac{-v^{2}-1}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x=\frac{-v^{2}-1}{-2}$$

Divide -v^{2}-1 by -2.

$$x=\frac{v^{2}+1}{2}$$

Solve for v (complex solution)

$$v=-\sqrt{2x-1}$$

$$v=\sqrt{2x-1}$$

Solve for v

$$v=\sqrt{2x-1}$$

$$v=-\sqrt{2x-1}\text{, }x\geq \frac{1}{2}$$