Solve 1-2x+x^2=1/p

$$1 - 2 x + x ^ { 2 } = \frac { 1 } { p }$$

Solve for p

$$p=\frac{1}{\left(x-1\right)^{2}}$$

$$x\neq 1$$

Steps for Solving Linear Equation

Variable p cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by p.

$$p-2xp+px^{2}=1$$

Combine all terms containing p.

$$\left(1-2x+x^{2}\right)p=1$$

The equation is in standard form.

$$\left(x^{2}-2x+1\right)p=1$$

Divide both sides by x^{2}-2x+1.

$$\frac{\left(x^{2}-2x+1\right)p}{x^{2}-2x+1}=\frac{1}{x^{2}-2x+1}$$

Dividing by x^{2}-2x+1 undoes the multiplication by x^{2}-2x+1.

$$p=\frac{1}{x^{2}-2x+1}$$

Divide 1 by x^{2}-2x+1.

$$p=\frac{1}{\left(x-1\right)^{2}}$$

Variable p cannot be equal to 0.

$$p=\frac{1}{\left(x-1\right)^{2}}\text{, }p\neq 0$$

Solve for x (complex solution)

$$x=p^{-\frac{1}{2}}+1$$

$$x=1-p^{-\frac{1}{2}}\text{, }p\neq 0$$

Solve for x

$$x=1+\frac{1}{\sqrt{p}}$$

$$x=1-\frac{1}{\sqrt{p}}\text{, }p>0$$