Solve q_k^2-12x+4=0

$$q _ { k } ^ { 2 } - 12 x + 4 = 0$$

Solve for x

$$x=\frac{q_{k}^{2}+4}{12}$$

Steps for Solving Linear Equation

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

$$-12x+4=-q_{k}^{2}$$

Subtract 4 from both sides.

$$-12x=-q_{k}^{2}-4$$

Divide both sides by -12.

$$\frac{-12x}{-12}=\frac{-q_{k}^{2}-4}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x=\frac{-q_{k}^{2}-4}{-12}$$

Divide -q_{k}^{2}-4 by -12.

$$x=\frac{q_{k}^{2}}{12}+\frac{1}{3}$$

Solve for q_k (complex solution)

$$q_{k}=-2\sqrt{3x-1}$$

$$q_{k}=2\sqrt{3x-1}$$

Solve for q_k

$$q_{k}=2\sqrt{3x-1}$$

$$q_{k}=-2\sqrt{3x-1}\text{, }x\geq \frac{1}{3}$$