Solve kf(x)=(x-p)^2-q

$$k f ( x ) = ( x - p ) ^ { 2 } - q$$

Solve for f (complex solution)

$$\left\{\begin{matrix}f=-\frac{q-p^{2}+2px-x^{2}}{kx}\text{, }&x\neq 0\text{ and }k\neq 0\\f\in \mathrm{C}\text{, }&\left(q=p^{2}\text{ and }x=0\right)\text{ or }\left(q=\left(x-p\right)^{2}\text{ and }k=0\text{ and }x\neq 0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$kfx=x^{2}-2xp+p^{2}-q$$

The equation is in standard form.

$$kxf=x^{2}-2px+p^{2}-q$$

Divide both sides by kx.

$$\frac{kxf}{kx}=\frac{\left(x-p\right)^{2}-q}{kx}$$

Dividing by kx undoes the multiplication by kx.

$$f=\frac{\left(x-p\right)^{2}-q}{kx}$$

Divide -q+\left(x-p\right)^{2} by kx.

$$f=\frac{x^{2}-2px+p^{2}-q}{kx}$$

Solve for k (complex solution)

$$\left\{\begin{matrix}k=-\frac{q-p^{2}+2px-x^{2}}{fx}\text{, }&x\neq 0\text{ and }f\neq 0\\k\in \mathrm{C}\text{, }&\left(q=p^{2}\text{ and }x=0\right)\text{ or }\left(q=\left(x-p\right)^{2}\text{ and }f=0\text{ and }x\neq 0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$kfx=x^{2}-2xp+p^{2}-q$$

The equation is in standard form.

$$fxk=x^{2}-2px+p^{2}-q$$

Divide both sides by fx.

$$\frac{fxk}{fx}=\frac{\left(x-p\right)^{2}-q}{fx}$$

Dividing by fx undoes the multiplication by fx.

$$k=\frac{\left(x-p\right)^{2}-q}{fx}$$

Divide -q+\left(x-p\right)^{2} by fx.

$$k=\frac{x^{2}-2px+p^{2}-q}{fx}$$

Solve for f

$$\left\{\begin{matrix}f=-\frac{q-p^{2}+2px-x^{2}}{kx}\text{, }&x\neq 0\text{ and }k\neq 0\\f\in \mathrm{R}\text{, }&\left(q=p^{2}\text{ and }x=0\right)\text{ or }\left(q=\left(x-p\right)^{2}\text{ and }k=0\text{ and }x\neq 0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$kfx=x^{2}-2xp+p^{2}-q$$

The equation is in standard form.

$$kxf=x^{2}-2px+p^{2}-q$$

Divide both sides by kx.

$$\frac{kxf}{kx}=\frac{\left(x-p\right)^{2}-q}{kx}$$

Dividing by kx undoes the multiplication by kx.

$$f=\frac{\left(x-p\right)^{2}-q}{kx}$$

Divide -q+\left(x-p\right)^{2} by kx.

$$f=\frac{x^{2}-2px+p^{2}-q}{kx}$$

Solve for k

$$\left\{\begin{matrix}k=-\frac{q-p^{2}+2px-x^{2}}{fx}\text{, }&x\neq 0\text{ and }f\neq 0\\k\in \mathrm{R}\text{, }&\left(q=p^{2}\text{ and }x=0\right)\text{ or }\left(q=\left(x-p\right)^{2}\text{ and }f=0\text{ and }x\neq 0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$kfx=x^{2}-2xp+p^{2}-q$$

The equation is in standard form.

$$fxk=x^{2}-2px+p^{2}-q$$

Divide both sides by fx.

$$\frac{fxk}{fx}=\frac{\left(x-p\right)^{2}-q}{fx}$$

Dividing by fx undoes the multiplication by fx.

$$k=\frac{\left(x-p\right)^{2}-q}{fx}$$

Divide -q+\left(x-p\right)^{2} by fx.

$$k=\frac{x^{2}-2px+p^{2}-q}{fx}$$