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Step 1: Identify the condition for rational roots
For a quadratic equation of the form $ax^2 + bx + c = 0$, the roots are rational if and only if the discriminant $D = b^2 - 4ac$ is a perfect square (and $a, b, c$ are integers). Here, our quadratic equation is:
$6x^2 - 11x + \alpha = 0$.
Step 2: Compute the discriminant
The discriminant $D$ is given by:
$D = (-11)^2 - 4 \times 6 \times \alpha = 121 - 24\alpha.$
Step 3: Set up the perfect square condition
For roots to be rational, $121 - 24\alpha$ must be a perfect square. Suppose
$121 - 24\alpha = k^2,$
where $k$ is an integer. Then:
$24\alpha = 121 - k^2,$
$\alpha = \frac{121 - k^2}{24}.$
Step 4: Find all positive integral values of $ \alpha$
We look for integer $k$ such that $(121 - k^2)$ is nonnegative, divisible by 24, and gives a positive integer $\alpha$. Checking values of $k$ systematically:
$k=1 \implies 121 - 1 = 120$, and $120/24 = 5 \implies \alpha = 5.$
$k=5 \implies 121 - 25 = 96$, and $96/24 = 4 \implies \alpha = 4.$
$k=7 \implies 121 - 49 = 72$, and $72/24 = 3 \implies \alpha = 3.$
These are the only integer values for $k$ that produce a positive integer $\alpha$. Hence, $\alpha$ can be $3, 4,$ or $5.$
Step 5: Count the number of valid positive integral values
Since $\alpha$ can take exactly three values (3, 4, and 5), the number of possible positive integral values of $\alpha$ is 3.
Correct Answer
3
