Solve fx=(2x+1)^2

$$f x = ( 2 x + 1 ) ^ { 2 }$$

Solve for f

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

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$fx=4x^{2}+4x+1$$

The equation is in standard form.

$$xf=4x^{2}+4x+1$$

Divide both sides by x.

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

Dividing by x undoes the multiplication by x.

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{f\left(f-8\right)}+f-4}{8}$$

$$x=\frac{\sqrt{f\left(f-8\right)}+f-4}{8}$$

Solve for x

$$x=\frac{-\sqrt{f\left(f-8\right)}+f-4}{8}$$

$$x=\frac{\sqrt{f\left(f-8\right)}+f-4}{8}\text{, }f\geq 8\text{ or }f\leq 0$$