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

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

Solve for k (complex solution)

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

Steps for Solving Linear Equation

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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

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

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

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

Add 2x to both sides.

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

Combine all terms containing k.

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for k

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

Steps for Solving Linear Equation

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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

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

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

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

Add 2x to both sides.

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

Combine all terms containing k.

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for x

$$\left\{\begin{matrix}\\x=1\text{, }&\text{unconditionally}\\x=\frac{k+2}{k}\text{, }&k\neq 0\end{matrix}\right.$$