Solve x^2+k=kx-3

$$x ^ { 2 } + k = k x - 3$$

Solve for k

$$k=-\frac{x^{2}+3}{1-x}$$

$$x\neq 1$$

Steps for Solving Linear Equation

Subtract kx from both sides.

$$x^{2}+k-kx=-3$$

Subtract x^{2} from both sides.

$$k-kx=-3-x^{2}$$

Combine all terms containing k.

$$\left(1-x\right)k=-3-x^{2}$$

The equation is in standard form.

$$\left(1-x\right)k=-x^{2}-3$$

Divide both sides by 1-x.

$$\frac{\left(1-x\right)k}{1-x}=\frac{-x^{2}-3}{1-x}$$

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

$$k=\frac{-x^{2}-3}{1-x}$$

Divide -3-x^{2} by 1-x.

$$k=-\frac{x^{2}+3}{1-x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(k-6\right)\left(k+2\right)}+k}{2}$$

$$x=\frac{-\sqrt{\left(k-6\right)\left(k+2\right)}+k}{2}$$

Solve for x

$$x=\frac{\sqrt{\left(k-6\right)\left(k+2\right)}+k}{2}$$

$$x=\frac{-\sqrt{\left(k-6\right)\left(k+2\right)}+k}{2}\text{, }k\leq -2\text{ or }k\geq 6$$