Answer
$x = \frac{1}{3}$, or $x = -\frac{2}{3}$
Work Step by Step
$$9 + \frac{3}{x} = \frac{2}{x^{2}}$$
To solve the exercise, we can re-write the equation as follows:
$$x^{2}(9 + \frac{3}{x}) = x^{2}(\frac{2}{x^{2}})$$
$$9x^{2} + 3x = 2$$
$$9x^{2} + 3x - 2= 0$$
where $x \ne 0$. Using the quadratic formula $\frac{-b\frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$:
$$x = \frac{-(3)\frac{+}{} \sqrt {(3)^{2} - 4(9)(-2)}}{2(9)}$$
$$x = \frac{-3\frac{+}{} \sqrt {9 + 72}}{18}$$
$$x = \frac{-3\frac{+}{} \sqrt {81}}{18}$$
$$x = \frac{-3\frac{+}{} 9}{18} = -\frac{12}{18} or \frac{6}{18} = -\frac{2}{3} or \frac{1}{3}$$
Finally, we substitute the values in the original equation to verify:
$$9 + \frac{3}{x} = \frac{2}{x^{2}}$$
$$9 + \frac{3}{-\frac{2}{3}} = \frac{2}{(-\frac{2}{3})^{2}}$$
$$9 - \frac{9}{2} = 2\times\frac{9}{4}$$
$$\frac{9}{2} = \frac{9}{2}$$
AND
$$9 + \frac{3}{x} = \frac{2}{x^{2}}$$
$$9 + \frac{3}{\frac{1}{3}} = \frac{2}{(\frac{1}{3})^{2}}$$
$$9 + 9 = \frac{2}{\frac{1}{9}}$$
$$18 = 2 \times 9 = 18$$
where both solutions hold.