Solve x^2-2(k+1)x+4k=0

$$x ^ { 2 } - 2 ( k + 1 ) x + 4 k = 0$$

Solve for k (complex solution)

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

Steps for Solving Linear Equation

Multiply -1 and 2 to get -2.

$$x^{2}-2\left(k+1\right)x+4k=0$$

Use the distributive property to multiply -2 by k+1.

$$x^{2}+\left(-2k-2\right)x+4k=0$$

Use the distributive property to multiply -2k-2 by x.

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

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

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

Add 2x to both sides.

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

Combine all terms containing k.

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

The equation is in standard form.

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

Divide both sides by -2x+4.

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

Dividing by -2x+4 undoes the multiplication by -2x+4.

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

Divide x\left(2-x\right) by -2x+4.

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

Solve for k

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

Steps for Solving Linear Equation

Multiply -1 and 2 to get -2.

$$x^{2}-2\left(k+1\right)x+4k=0$$

Use the distributive property to multiply -2 by k+1.

$$x^{2}+\left(-2k-2\right)x+4k=0$$

Use the distributive property to multiply -2k-2 by x.

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

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

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

Add 2x to both sides.

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

Combine all terms containing k.

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

The equation is in standard form.

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

Divide both sides by -2x+4.

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

Dividing by -2x+4 undoes the multiplication by -2x+4.

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

Divide x\left(2-x\right) by -2x+4.

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

Solve for x

$$x=2k$$

$$x=2$$