6.1.2 Exam: Semester 2 Exam Question 40 of 40 What is the solution to \( \sqrt{7 x-4}=2 \sqrt{x} \) ? A. \( -\frac{4}{3} \) B. \( -\frac{4}{5} \) C. \( \frac{4}{5} \) D. \( \frac{4}{3} \)

Solve the equation \( \sqrt{7x-4}=2\sqrt{x} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{7x-4}=2\sqrt{x}\) - step1: Find the domain: \(\sqrt{7x-4}=2\sqrt{x},x\geq \frac{4}{7}\) - step2: Swap the sides: \(2\sqrt{x}=\sqrt{7x-4}\) - step3: Rewrite the expression: \(\sqrt{x}=\frac{\sqrt{7x-4}}{2}\) - step4: Raise both sides to the \(2\)-th power\(:\) \(\left(\sqrt{x}\right)^{2}=\left(\frac{\sqrt{7x-4}}{2}\right)^{2}\) - step5: Evaluate the power: \(x=\frac{7x-4}{4}\) - step6: Cross multiply: \(x\times 4=7x-4\) - step7: Simplify the equation: \(4x=7x-4\) - step8: Move the variable to the left side: \(4x-7x=-4\) - step9: Subtract the terms: \(-3x=-4\) - step10: Change the signs: \(3x=4\) - step11: Divide both sides: \(\frac{3x}{3}=\frac{4}{3}\) - step12: Divide the numbers: \(x=\frac{4}{3}\) - step13: Check if the solution is in the defined range: \(x=\frac{4}{3},x\geq \frac{4}{7}\) - step14: Find the intersection: \(x=\frac{4}{3}\) - step15: Check the solution: \(x=\frac{4}{3}\) The solution to the equation \( \sqrt{7x-4}=2\sqrt{x} \) is \( x = \frac{4}{3} \). Therefore, the correct answer is D. \( \frac{4}{3} \).