Key Concepts and Formulas

• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Discriminant: The discriminant of the quadratic equation $ax^2 + bx + c = 0$ is $D = b^2 - 4ac$.
• Rational Roots: A quadratic equation with rational coefficients has rational roots if and only if its discriminant is a perfect square.

Step-by-Step Solution

Step 1: Identify the coefficients We are given the quadratic equation $6x^2 - 11x + \alpha = 0$. Comparing this to the general form $ax^2 + bx + c = 0$, we identify the coefficients: $a = 6$, $b = -11$, and $c = \alpha$. We note that $\alpha$ must be a positive integer. This is crucial for finding the possible values of $\alpha$.

Step 2: Calculate the discriminant For the roots to be rational, the discriminant $D$ must be a perfect square. We calculate the discriminant: $D = b^2 - 4ac = (-11)^2 - 4(6)(\alpha) = 121 - 24\alpha$. Thus, we require $121 - 24\alpha$ to be a perfect square.

Step 3: Set the discriminant equal to a perfect square Let $k^2 = 121 - 24\alpha$, where $k$ is a non-negative integer. This means $24\alpha = 121 - k^2$. Since $\alpha$ is a positive integer, $24\alpha$ must be a positive integer multiple of 24. Therefore, $121 - k^2$ must be a positive integer multiple of 24. Also, since $\alpha$ is positive, $121 - k^2 > 0$, so $k^2 < 121$, meaning $0 \le k \le 10$ since $k$ is a non-negative integer.

Step 4: Find possible values for $\alpha$ by testing integer values of $k$ We need to find integer values of $k$ between 0 and 10 such that $121 - k^2$ is a positive multiple of 24. Rearranging the equation $24\alpha = 121 - k^2$, we get $\alpha = \frac{121 - k^2}{24}$. We test integer values of $k$ from 0 to 10:

• $k = 0$: $\alpha = \frac{121 - 0}{24} = \frac{121}{24}$. Not an integer.
• $k = 1$: $\alpha = \frac{121 - 1}{24} = \frac{120}{24} = 5$. Integer, valid.
• $k = 2$: $\alpha = \frac{121 - 4}{24} = \frac{117}{24}$. Not an integer.
• $k = 3$: $\alpha = \frac{121 - 9}{24} = \frac{112}{24}$. Not an integer.
• $k = 4$: $\alpha = \frac{121 - 16}{24} = \frac{105}{24}$. Not an integer.
• $k = 5$: $\alpha = \frac{121 - 25}{24} = \frac{96}{24} = 4$. Integer, valid.
• $k = 6$: $\alpha = \frac{121 - 36}{24} = \frac{85}{24}$. Not an integer.
• $k = 7$: $\alpha = \frac{121 - 49}{24} = \frac{72}{24} = 3$. Integer, valid.
• $k = 8$: $\alpha = \frac{121 - 64}{24} = \frac{57}{24}$. Not an integer.
• $k = 9$: $\alpha = \frac{121 - 81}{24} = \frac{40}{24}$. Not an integer.
• $k = 10$: $\alpha = \frac{121 - 100}{24} = \frac{21}{24}$. Not an integer.

Therefore, the possible positive integral values of $\alpha$ are 3, 4, and 5.

Step 5: Count the number of possible values of $\alpha$ The possible values of $\alpha$ are 3, 4, and 5. There are 3 such values.

Common Mistakes & Tips

• Forgetting the positive integral constraint: If $\alpha$ could be zero or negative, more values might be possible.
• Not systematically checking all possibilities: It's crucial to check all integer values of $k$ within the determined range ($0 \le k \le 10$) and verify if $\frac{121 - k^2}{24}$ results in a positive integer $\alpha$.
• Arithmetic errors: Small calculation mistakes when checking for perfect squares or divisibility can lead to incorrect results.

Summary For the quadratic equation $6x^2 - 11x + \alpha = 0$ to have rational roots, its discriminant $D = 121 - 24\alpha$ must be a perfect square. By setting the discriminant equal to a perfect square ($k^2$) and applying the constraint that $\alpha$ must be a positive integer, we found that there are exactly 3 possible values for $\alpha$: 3, 4, and 5.

Final Answer The final answer is \boxed{3}, which corresponds to option (A).