\( \left. \begin{array} { l | l } { } & { } \\ \hline 1.4 & { x \sqrt { 1 - 6 x ^ { - 1 } } - 4 = 0 } \end{array} \right. \)

To solve the equation \( x \sqrt{1 - 6x^{-1}} - 4 = 0 \), we can start by isolating the radical. Rearranging gives us \( x \sqrt{1 - 6x^{-1}} = 4 \). Next, divide each side by \( x \) (assuming \( x \neq 0 \)) to get \( \sqrt{1 - 6/x} = \frac{4}{x} \). Now, squaring both sides eliminates the square root: \[ 1 - \frac{6}{x} = \frac{16}{x^2}. \] Rearranging leads to: \[ 16 = x^2 - 6x, \] which can be rearranged to form a quadratic equation: \[ x^2 - 6x - 16 = 0. \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -6 \), and \( c = -16 \), we find: \[ x = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm 10}{2}. \] This yields two potential solutions: \( x = 8 \) and \( x = -2 \). However, checking back in the original context, we find \( x \) must be positive. Thus, the only valid solution is \( x = 8 \). Happy solving!