Solve R^2x^2-8x+4kx+4=0

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

Solve for k

$$k=\frac{8x-\left(Rx\right)^{2}-4}{4x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

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

Add 8x to both sides.

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

Subtract 4 from both sides.

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

The equation is in standard form.

$$4xk=-R^{2}x^{2}+8x-4$$

Divide both sides by 4x.

$$\frac{4xk}{4x}=\frac{-R^{2}x^{2}+8x-4}{4x}$$

Dividing by 4x undoes the multiplication by 4x.

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

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

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

Solve for R (complex solution)

$$R=-\frac{2i\sqrt{kx-2x+1}}{x}$$

$$R=\frac{2i\sqrt{kx-2x+1}}{x}\text{, }x\neq 0$$

Solve for R

$$R=\frac{2\sqrt{-kx+2x-1}}{|x|}$$

$$R=-\frac{2\sqrt{-kx+2x-1}}{|x|}\text{, }\left(x=-\frac{1}{k-2}\text{ and }k\neq 2\right)\text{ or }\left(x\geq -\frac{1}{k-2}\text{ and }k<2\text{ and }x\neq 0\right)\text{ or }\left(k>2\text{ and }x\leq -\frac{1}{k-2}\text{ and }x\neq 0\right)$$