1. **Problem:** Determine the condition on $\alpha$ such that the quadratic equation $6x^2 - 11x + \alpha = 0$ has rational roots. 2. **Formula:** For a quadratic equation $ax^2 + bx + c = 0$, roots are rational if the discriminant $\Delta = b^2 - 4ac$ is a perfect square of a rational number. 3. **Apply the formula:** Here, $a=6$, $b=-11$, and $c=\alpha$. The discriminant is $$\Delta = (-11)^2 - 4 \times 6 \times \alpha = 121 - 24\alpha.$$ 4. **Condition for rational roots:** $121 - 24\alpha$ must be a perfect square, say $k^2$ where $k \in \mathbb{Q}$. 5. **Express $\alpha$ in terms of $k$:** $$121 - 24\alpha = k^2 \implies 24\alpha = 121 - k^2 \implies \alpha = \frac{121 - k^2}{24}.$$ 6. **Interpretation:** For any rational $k$ such that $121 - k^2$ is rational, $\alpha$ is rational and the roots are rational. **Final answer:** The quadratic $6x^2 - 11x + \alpha = 0$ has rational roots if and only if $$\boxed{\alpha = \frac{121 - k^2}{24} \text{ for some rational } k.}$$