Solve f(x-2)^2=1-(x+2)^2

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

Solve for f

$$f=-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-2\right)^{2}}$$

$$x\neq 2$$

Steps for Solving Linear Equation

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

$$f\left(x^{2}-4x+4\right)=1-\left(x+2\right)^{2}$$

Use the distributive property to multiply f by x^{2}-4x+4.

$$fx^{2}-4fx+4f=1-\left(x+2\right)^{2}$$

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

$$fx^{2}-4fx+4f=1-\left(x^{2}+4x+4\right)$$

To find the opposite of x^{2}+4x+4, find the opposite of each term.

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

Subtract 4 from 1 to get -3.

$$fx^{2}-4fx+4f=-3-x^{2}-4x$$

Combine all terms containing f.

$$\left(x^{2}-4x+4\right)f=-3-x^{2}-4x$$

The equation is in standard form.

$$\left(x^{2}-4x+4\right)f=-x^{2}-4x-3$$

Divide both sides by x^{2}-4x+4.

$$\frac{\left(x^{2}-4x+4\right)f}{x^{2}-4x+4}=-\frac{\left(x+1\right)\left(x+3\right)}{x^{2}-4x+4}$$

Dividing by x^{2}-4x+4 undoes the multiplication by x^{2}-4x+4.

$$f=-\frac{\left(x+1\right)\left(x+3\right)}{x^{2}-4x+4}$$

Divide -\left(1+x\right)\left(3+x\right) by x^{2}-4x+4.

$$f=-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-2\right)^{2}}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{2f+\sqrt{1-15f}-2}{f+1}\text{; }x=\frac{2f-\sqrt{1-15f}-2}{f+1}\text{, }&f\neq -1\text{ and }f\leq \frac{1}{15}\\x=\frac{1}{8}\text{, }&f=-1\end{matrix}\right.$$