\[b_{1} = x - 4;\ \]
\[b_{2} = \sqrt{6x};\ \]
\[b_{3} = x + 12.\]
\[\frac{b_{2}}{b_{1}} = \frac{b_{3}}{b_{2}} = q;\ \]
\[\frac{\sqrt{6x}}{x - 4} = \frac{x + 12}{\sqrt{6x}};\ \ x > 0;\ \ x \neq 4;\]
\[6x = (x - 4)(x + 12)\]
\[6x = x^{2} - 4x + 12x - 48\]
\[x^{2} + 2x - 48 = 0\]
\[D_{1} = 1 + 48 = 49\]
\[x_{1} = - 1 + 7 = 6;\ \]
\[x_{2} = - 1 - 7 = - 8 < 0.\]
\[Ответ:при\ x = 6.\]