Given a quadratic $ax^2 + bx + c,$
$ - \frac{b}{a} = \frac{11}{12}$
$\frac{c}{a} = \frac{1}{8}$
$a = \text{LCM}(12, 8) = 24$
$- \frac{b}{24} = \frac{11}{12}$
$-b = 22$
$b = -22$
$\frac{c}{24} = \frac{1}{8}$
$c = 3$
$24x^2 - 22x + 3 = 0$
$x = \frac{-(-22) \pm \sqrt{(-22)^2 - 4 \cdot 24 \cdot 3}}{24 \cdot 2} = \frac{22 \pm \sqrt{484 - 288}}{48} = \frac{22 \pm \sqrt{196}}{48} = \frac{22 \pm 14}{48} = \frac{11 \pm 7}{24}.$
$x = \frac{18}{24} = \frac{3}{4}$
$x = \frac{4}{24} = \frac{1}{6}$
$\boxed{\frac{1}{6}}$ is the answer
