Solve x^2+2sqrt{2}kx+18=0

$${ x }^{ 2 } +2 \sqrt{ 2 } kx+18=0$$

Solve for k

$$k=-\frac{\sqrt{2}\left(x^{2}+18\right)}{4x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$2\sqrt{2}kx+18=-x^{2}$$

Subtract 18 from both sides.

$$2\sqrt{2}kx=-x^{2}-18$$

The equation is in standard form.

$$2\sqrt{2}xk=-x^{2}-18$$

Divide both sides by 2\sqrt{2}x.

$$\frac{2\sqrt{2}xk}{2\sqrt{2}x}=\frac{-x^{2}-18}{2\sqrt{2}x}$$

Dividing by 2\sqrt{2}x undoes the multiplication by 2\sqrt{2}x.

$$k=\frac{-x^{2}-18}{2\sqrt{2}x}$$

Divide -x^{2}-18 by 2\sqrt{2}x.

$$k=-\frac{\sqrt{2}\left(x^{2}+18\right)}{4x}$$

Solve for x (complex solution)

$$x=\sqrt{2}\left(\sqrt{k^{2}-9}-k\right)$$

$$x=\sqrt{2}\left(-\sqrt{k^{2}-9}-k\right)$$

Solve for x

$$x=\sqrt{2}\left(\sqrt{k^{2}-9}-k\right)$$

$$x=\sqrt{2}\left(-\sqrt{k^{2}-9}-k\right)\text{, }|k|\geq 3$$