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

Solve the equation by following steps: - step0: Solve for \(x\): \(x\sqrt{1-6x^{-1}}-4=0\) - step1: Find the domain: \(x\sqrt{1-6x^{-1}}-4=0,x \in \left(-\infty,0\right)\cup \left[6,+\infty\right)\) - step2: Rewrite the expression: \(x\sqrt{\frac{x-6}{x}}-4=0\) - step3: Move the constant to the right-hand side: \(x\sqrt{\frac{x-6}{x}}=4\) - step4: Raise both sides to the \(2\)-th power\(:\) \(\left(x\sqrt{\frac{x-6}{x}}\right)^{2}=4^{2}\) - step5: Evaluate the power: \(x^{2}-6x=16\) - step6: Move the expression to the left side: \(x^{2}-6x-16=0\) - step7: Factor the expression: \(\left(x-8\right)\left(x+2\right)=0\) - step8: Separate into possible cases: \(\begin{align}&x-8=0\\&x+2=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=8\\&x=-2\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=8\\&x=-2\end{align},x \in \left(-\infty,0\right)\cup \left[6,+\infty\right)\) - step11: Find the intersection: \(\begin{align}&x=8\\&x=-2\end{align}\) - step12: Check the solution: \(x=8\) To solve the equation \( x \sqrt{1 - 6 x^{-1}} - 4 = 0 \), we can follow these steps: 1. **Isolate the square root**: \[ x \sqrt{1 - 6 x^{-1}} = 4 \] 2. **Divide both sides by \( x \)** (assuming \( x \neq 0 \)): \[ \sqrt{1 - 6 x^{-1}} = \frac{4}{x} \] 3. **Square both sides** to eliminate the square root: \[ 1 - 6 x^{-1} = \left(\frac{4}{x}\right)^2 \] 4. **Simplify the right side**: \[ 1 - 6 x^{-1} = \frac{16}{x^2} \] 5. **Multiply through by \( x^2 \)** to eliminate the fractions: \[ x^2 - 6x = 16 \] 6. **Rearrange the equation**: \[ x^2 - 6x - 16 = 0 \] 7. **Now, we can solve this quadratic equation** using the quadratic formula or factoring. In this case, we find that: \[ x = 8 \] Thus, the solution to the equation is \( x = 8 \).