Solve 8x^2-2kx+k=0

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

Solve for k

$$k=-\frac{8x^{2}}{1-2x}$$

$$x\neq \frac{1}{2}$$

Steps for Solving Linear Equation

Subtract 8x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-2kx+k=-8x^{2}$$

Combine all terms containing k.

$$\left(-2x+1\right)k=-8x^{2}$$

The equation is in standard form.

$$\left(1-2x\right)k=-8x^{2}$$

Divide both sides by -2x+1.

$$\frac{\left(1-2x\right)k}{1-2x}=-\frac{8x^{2}}{1-2x}$$

Dividing by -2x+1 undoes the multiplication by -2x+1.

$$k=-\frac{8x^{2}}{1-2x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{k\left(k-8\right)}+k}{8}$$

$$x=\frac{-\sqrt{k\left(k-8\right)}+k}{8}$$

Solve for x

$$x=\frac{\sqrt{k\left(k-8\right)}+k}{8}$$

$$x=\frac{-\sqrt{k\left(k-8\right)}+k}{8}\text{, }k\geq 8\text{ or }k\leq 0$$