Answer
The solutions are $\frac{-1±\sqrt{19}}{6}$, and the solution set is {$\frac{-1+\sqrt{19}}{6}, \frac{-1-\sqrt{19}}{6}$}.
Work Step by Step
$\frac{x^2}{3}+\frac{x}{9}-\frac{1}{6}=0$
The least common multiple of $3$ and $9$ is $9$. Multiply the equation by the LCM.
$\frac{x^2}{3}·9+\frac{x}{9}·9-\frac{1}{6}·9=0·9$
$3x^2+x-\frac{3}{2}=0$
Divide both sides by $3$. Thus, it becomes:
$x^2+\frac{x}{3}-\frac{1}{2}=0$
Add $\frac{1}{2}$ to both sides.
$x^2+\frac{x}{3}-\frac{1}{2}+\frac{1}{2}=0+\frac{1}{2}$
$x^2+\frac{x}{3}=\frac{1}{2}$
The coefficient of the x-term is $\frac{1}{3}$. Half of $\frac{1}{3}$ is $\frac{1}{6}$, and $(\frac{1}{6})^2$ is $\frac{1}{36}$.
Add $\frac{1}{36}$ to both sides of the equation to complete the square.
$x^2+\frac{x}{3}+\frac{1}{36}=\frac{1}{2}+\frac{1}{36}$
$(x+\frac{1}{6})^2 = \frac{19}{36}$
$x+\frac{1}{6}=\sqrt\frac{19}{36}$ or $x+\frac{1}{6}=-\sqrt\frac{19}{36}$
$x=-\frac{1}{6}+\sqrt\frac{19}{36}$ or $x=-\frac{1}{6}-\sqrt\frac{19}{36}$
$x=-\frac{1}{6}+\frac{\sqrt19}{6}$ or $x=x=-\frac{1}{6}-\frac{\sqrt19}{6}$
The solutions are $\frac{-1±\sqrt{19}}{6}$, and the solution set is {$\frac{-1+\sqrt{19}}{6}, \frac{-1-\sqrt{19}}{6}$}.
