Solve (2+3k_{1})x^2+6k_{1}k_{2}x+3k_2^2-6=0

$$( 2 + 3 k _ { 1 } ) x ^ { 2 } + 6 k _ { 1 } k _ { 2 } x + 3 k _ { 2 } ^ { 2 } - 6 = 0$$

Solve for k_1 (complex solution)

$$\left\{\begin{matrix}k_{1}=-\frac{2x^{2}+3k_{2}^{2}-6}{3x\left(x+2k_{2}\right)}\text{, }&x\neq -2k_{2}\text{ and }x\neq 0\\k_{1}\in \mathrm{C}\text{, }&\left(x=0\text{ and }k_{2}=\sqrt{2}\right)\text{ or }\left(x=0\text{ and }k_{2}=-\sqrt{2}\right)\text{ or }\left(x=\frac{2\sqrt{66}}{11}\text{ and }k_{2}=-\frac{\sqrt{66}}{11}\right)\text{ or }\left(x=-\frac{2\sqrt{66}}{11}\text{ and }k_{2}=\frac{\sqrt{66}}{11}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2+3k_{1} by x^{2}.

$$2x^{2}+3k_{1}x^{2}+6k_{1}k_{2}x+3k_{2}^{2}-6=0$$

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

$$3k_{1}x^{2}+6k_{1}k_{2}x+3k_{2}^{2}-6=-2x^{2}$$

Subtract 3k_{2}^{2} from both sides.

$$3k_{1}x^{2}+6k_{1}k_{2}x-6=-2x^{2}-3k_{2}^{2}$$

Add 6 to both sides.

$$3k_{1}x^{2}+6k_{1}k_{2}x=-2x^{2}-3k_{2}^{2}+6$$

Combine all terms containing k_{1}.

$$\left(3x^{2}+6k_{2}x\right)k_{1}=-2x^{2}-3k_{2}^{2}+6$$

The equation is in standard form.

$$\left(3x^{2}+6k_{2}x\right)k_{1}=6-3k_{2}^{2}-2x^{2}$$

Divide both sides by 3x^{2}+6k_{2}x.

$$\frac{\left(3x^{2}+6k_{2}x\right)k_{1}}{3x^{2}+6k_{2}x}=\frac{6-3k_{2}^{2}-2x^{2}}{3x^{2}+6k_{2}x}$$

Dividing by 3x^{2}+6k_{2}x undoes the multiplication by 3x^{2}+6k_{2}x.

$$k_{1}=\frac{6-3k_{2}^{2}-2x^{2}}{3x^{2}+6k_{2}x}$$

Divide -2x^{2}-3k_{2}^{2}+6 by 3x^{2}+6k_{2}x.

$$k_{1}=\frac{6-3k_{2}^{2}-2x^{2}}{3x\left(x+2k_{2}\right)}$$

Solve for k_1

$$\left\{\begin{matrix}k_{1}=-\frac{2x^{2}+3k_{2}^{2}-6}{3x\left(x+2k_{2}\right)}\text{, }&x\neq -2k_{2}\text{ and }x\neq 0\\k_{1}\in \mathrm{R}\text{, }&\left(k_{2}=\frac{\sqrt{66}}{11}\text{ and }x=-\frac{2\sqrt{66}}{11}\right)\text{ or }\left(k_{2}=-\frac{\sqrt{66}}{11}\text{ and }x=\frac{2\sqrt{66}}{11}\right)\text{ or }\left(x=0\text{ and }|k_{2}|=\sqrt{2}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2+3k_{1} by x^{2}.

$$2x^{2}+3k_{1}x^{2}+6k_{1}k_{2}x+3k_{2}^{2}-6=0$$

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

$$3k_{1}x^{2}+6k_{1}k_{2}x+3k_{2}^{2}-6=-2x^{2}$$

Subtract 3k_{2}^{2} from both sides.

$$3k_{1}x^{2}+6k_{1}k_{2}x-6=-2x^{2}-3k_{2}^{2}$$

Add 6 to both sides.

$$3k_{1}x^{2}+6k_{1}k_{2}x=-2x^{2}-3k_{2}^{2}+6$$

Combine all terms containing k_{1}.

$$\left(3x^{2}+6k_{2}x\right)k_{1}=-2x^{2}-3k_{2}^{2}+6$$

The equation is in standard form.

$$\left(3x^{2}+6k_{2}x\right)k_{1}=6-3k_{2}^{2}-2x^{2}$$

Divide both sides by 3x^{2}+6k_{2}x.

$$\frac{\left(3x^{2}+6k_{2}x\right)k_{1}}{3x^{2}+6k_{2}x}=\frac{6-3k_{2}^{2}-2x^{2}}{3x^{2}+6k_{2}x}$$

Dividing by 3x^{2}+6k_{2}x undoes the multiplication by 3x^{2}+6k_{2}x.

$$k_{1}=\frac{6-3k_{2}^{2}-2x^{2}}{3x^{2}+6k_{2}x}$$

Divide -2x^{2}-3k_{2}^{2}+6 by 3x^{2}+6k_{2}x.

$$k_{1}=\frac{6-3k_{2}^{2}-2x^{2}}{3x\left(x+2k_{2}\right)}$$

Solve for k_2 (complex solution)

$$k_{2}=\frac{\sqrt{18+9\left(k_{1}x\right)^{2}-6x^{2}-9k_{1}x^{2}}}{3}-k_{1}x$$

$$k_{2}=-\frac{\sqrt{18+9\left(k_{1}x\right)^{2}-6x^{2}-9k_{1}x^{2}}}{3}-k_{1}x$$

Solve for k_2

$$k_{2}=\frac{\sqrt{18+9\left(k_{1}x\right)^{2}-6x^{2}-9k_{1}x^{2}}}{3}-k_{1}x$$

$$k_{2}=-\frac{\sqrt{18+9\left(k_{1}x\right)^{2}-6x^{2}-9k_{1}x^{2}}}{3}-k_{1}x\text{, }|x|\leq \frac{2\sqrt{66}}{11}\text{ or }k_{1}\geq \frac{\sqrt{297x^{4}-648x^{2}}}{18x^{2}}+\frac{1}{2}\text{ or }k_{1}\leq -\frac{\sqrt{297x^{4}-648x^{2}}}{18x^{2}}+\frac{1}{2}$$