Solve kx^2-4x+8=0

$$k x ^ { 2 } - 4 x + 8 = 0$$

Solve for k

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

$$x\neq 0$$

Steps for Solving Linear Equation

Add 4x to both sides. Anything plus zero gives itself.

$$kx^{2}+8=4x$$

Subtract 8 from both sides.

$$kx^{2}=4x-8$$

The equation is in standard form.

$$x^{2}k=4x-8$$

Divide both sides by x^{2}.

$$\frac{x^{2}k}{x^{2}}=\frac{4x-8}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$k=\frac{4x-8}{x^{2}}$$

Divide -8+4x by x^{2}.

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

Solve for x (complex solution)

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

Solve for x

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