Find the roots of $$\frac{6}{x} - \frac{2}{x-1} - \frac{1}{x-2} = 0 \\$$​

Correct option is C

Given:

​$$\frac{6}{x} - \frac{2}{x-1} - \frac{1}{x-2} = 0 \\$$

Concept Used:

Standard form of the quadratic equation: ax² + bx + c = 0

To find the roots of the quadratic equation,

​$$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\$$

Solution:

​$$\frac{6}{x} - \frac{2}{x-1} - \frac{1}{x-2} = 0 \\\text{Take the LCM:} \\6(x-1)(x-2) = 2x(x-2) + x(x-1) \\6(x^2 - 3x + 2) = 2x^2 - 4x + x^2 - x \\6x^2 - 18x + 12 = 3x^2 - 5x \\3x^2 - 13x + 12 = 0 \\\text{Use the quadratic formula:} \\x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(3)(12)}}{2(3)} \\x = \frac{13 \pm \sqrt{169 - 144}}{6} \\x = \frac{13 \pm \sqrt{25}}{6} \\x = \frac{13 + 5}{6} \quad \text{and} \quad x = \frac{13 - 5}{6} \\x = \frac{18}{6} \quad \text{and} \quad x = \frac{8}{6} \\x = 3 \quad \text{and} \quad x = \frac{4}{3} \\\textbf{Final Answer:} \\x = 3 \quad \text{and} \quad x = \frac{4}{3}.$$​​