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

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

Solve for k (complex solution)

$$\left\{\begin{matrix}\\k=\frac{x-1}{2}\text{, }&\text{unconditionally}\\k\in \mathrm{C}\text{, }&x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}-2xk+k^{2}=k^{2}+2k+1$$

Subtract k^{2} from both sides.

$$x^{2}-2xk+k^{2}-k^{2}=2k+1$$

Combine k^{2} and -k^{2} to get 0.

$$x^{2}-2xk=2k+1$$

Subtract 2k from both sides.

$$x^{2}-2xk-2k=1$$

Subtract x^{2} from both sides.

$$-2xk-2k=1-x^{2}$$

Combine all terms containing k.

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

Divide both sides by -2x-2.

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

Dividing by -2x-2 undoes the multiplication by -2x-2.

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

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

$$k=\frac{x-1}{2}$$

Solve for k

$$\left\{\begin{matrix}\\k=\frac{x-1}{2}\text{, }&\text{unconditionally}\\k\in \mathrm{R}\text{, }&x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}-2xk+k^{2}=k^{2}+2k+1$$

Subtract k^{2} from both sides.

$$x^{2}-2xk+k^{2}-k^{2}=2k+1$$

Combine k^{2} and -k^{2} to get 0.

$$x^{2}-2xk=2k+1$$

Subtract 2k from both sides.

$$x^{2}-2xk-2k=1$$

Subtract x^{2} from both sides.

$$-2xk-2k=1-x^{2}$$

Combine all terms containing k.

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

Divide both sides by -2x-2.

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

Dividing by -2x-2 undoes the multiplication by -2x-2.

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

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

$$k=\frac{x-1}{2}$$

Solve for x

$$x=2k+1$$

$$x=-1$$