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

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

Solve for k (complex solution)

$$k=-\frac{x^{2}}{4x^{2}+2x+1}$$

$$x\neq \frac{-1+\sqrt{3}i}{4}\text{ and }x\neq \frac{-\sqrt{3}i-1}{4}$$

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}.

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

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

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

Use the distributive property to multiply k by 4x^{2}+4x+1.

$$x^{2}-2xk+k^{2}+4kx^{2}+4kx+k=k^{2}$$

Combine -2xk and 4kx to get 2xk.

$$x^{2}+2xk+k^{2}+4kx^{2}+k=k^{2}$$

Subtract k^{2} from both sides.

$$x^{2}+2xk+k^{2}+4kx^{2}+k-k^{2}=0$$

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

$$x^{2}+2xk+4kx^{2}+k=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$2xk+4kx^{2}+k=-x^{2}$$

Combine all terms containing k.

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

The equation is in standard form.

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

Divide both sides by 2x+4x^{2}+1.

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

Dividing by 2x+4x^{2}+1 undoes the multiplication by 2x+4x^{2}+1.

$$k=-\frac{x^{2}}{4x^{2}+2x+1}$$

Solve for k

$$k=-\frac{x^{2}}{4x^{2}+2x+1}$$

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}.

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

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

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

Use the distributive property to multiply k by 4x^{2}+4x+1.

$$x^{2}-2xk+k^{2}+4kx^{2}+4kx+k=k^{2}$$

Combine -2xk and 4kx to get 2xk.

$$x^{2}+2xk+k^{2}+4kx^{2}+k=k^{2}$$

Subtract k^{2} from both sides.

$$x^{2}+2xk+k^{2}+4kx^{2}+k-k^{2}=0$$

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

$$x^{2}+2xk+4kx^{2}+k=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$2xk+4kx^{2}+k=-x^{2}$$

Combine all terms containing k.

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

The equation is in standard form.

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

Divide both sides by 2x+4x^{2}+1.

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

Dividing by 2x+4x^{2}+1 undoes the multiplication by 2x+4x^{2}+1.

$$k=-\frac{x^{2}}{4x^{2}+2x+1}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{-k\left(3k+1\right)}-k}{4k+1}\text{; }x=-\frac{\sqrt{-k\left(3k+1\right)}+k}{4k+1}\text{, }&k\neq -\frac{1}{4}\\x=-\frac{1}{2}\text{, }&k=-\frac{1}{4}\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{-k\left(3k+1\right)}-k}{4k+1}\text{; }x=-\frac{\sqrt{-k\left(3k+1\right)}+k}{4k+1}\text{, }&k\neq -\frac{1}{4}\text{ and }k\leq 0\text{ and }k\geq -\frac{1}{3}\\x=-\frac{1}{2}\text{, }&k=-\frac{1}{4}\end{matrix}\right.$$