Understanding Rational Roots of a Quadratic Equation

The question asks for the number of possible positive integral values for the parameter $\alpha$ such that the quadratic equation $10x^2 - 27x + \alpha = 0$ has rational roots.

For a quadratic equation of the form $ax^2 + bx + c = 0$, the roots are given by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ The roots of this equation are rational if and only if the discriminant, $D = b^2 - 4ac$, is a perfect square of a rational number. Since the coefficients $a$, $b$, and $c$ are integers (or can be considered integers after potential scaling), the discriminant $D$ must be a perfect square of an integer.

Conditions for Rational Roots

In the given equation, $10x^2 - 27x + \alpha = 0$, we have the coefficients:

• $a = 10$
• $b = -27$
• $c = \alpha$

The discriminant is $D = b^2 - 4ac$. Substituting the coefficients:

For the roots to be rational, $D$ must be a perfect square. Let $D = k^2$, where $k$ is a non-negative integer.

Finding Possible Values for $\alpha$

We are looking for positive integral values of $\alpha$. This imposes constraints on $k$.

1. Condition 1: $\alpha$ must be a positive integer ($\alpha \ge 1$). From the equation $k^2 = 729 - 40\alpha$, we can express $\alpha$ in terms of $k$: $$40\alpha = 729 - k^2$$ $$\alpha = \frac{729 - k^2}{40}$$ Since $\alpha \ge 1$, we must have $\frac{729 - k^2}{40} \ge 1$. $$729 - k^2 \ge 40$$ $$729 - 40 \ge k^2$$ $$689 \ge k^2$$ Also, since $\alpha$ must be positive, $729 - k^2 > 0$, which means $k^2 < 729$. Taking the square root, $k < \sqrt{729} = 27$. So, $k$ must be an integer such that $0 \le k < 27$.
2. Condition 2: $\alpha$ must be an integer. For $\alpha$ to be an integer, $729 - k^2$ must be perfectly divisible by 40. In terms of modular arithmetic: $$729 - k^2 \equiv 0 \pmod{40}$$ Let's find the remainder of 729 when divided by 40: $729 = 18 \times 40 + 9$. So, $729 \equiv 9 \pmod{40}$. The condition becomes: $$9 - k^2 \equiv 0 \pmod{40}$$ $$k^2 \equiv 9 \pmod{40}$$

Now, we need to find integer values of $k$ such that $0 \le k < 27$ and $k^2 \equiv 9 \pmod{40}$. Let's test values of $k$ in the range $[0, 26]$:

• If $k=3$, $k^2 = 9$. $9 \equiv 9 \pmod{40}$. This value of $k$ is valid.
• If $k=7$, $k^2 = 49$. $49 = 1 \times 40 + 9$, so $49 \equiv 9 \pmod{40}$. This value of $k$ is valid.
• If $k=13$, $k^2 = 169$. $169 = 4 \times 40 + 9$, so $169 \equiv 9 \pmod{40}$. This value of $k$ is valid.
• If $k=17$, $k^2 = 289$. $289 = 7 \times 40 + 9$, so $289 \equiv 9 \pmod{40}$. This value of $k$ is valid.
• If $k=23$, $k^2 = 529$. $529 = 13 \times 40 + 9$, so $529 \equiv 9 \pmod{40}$. This value of $k$ is valid.

Let's check if there are other possibilities. Consider squares modulo 40. We are looking for $k^2 \equiv 9 \pmod{40}$. The possible values of $k \pmod{40}$ are $3, 7, 13, 17, 23, 27, 33, 37$. We need $k$ values in the range $[0, 26]$. The values are $k = 3, 7, 13, 17, 23$. Note that $k=27$ gives $k^2=729$, making $\alpha=0$, which is not a positive integer.

Calculating the Corresponding $\alpha$ Values

For each valid value of $k$, we calculate the corresponding value of $\alpha$ using $\alpha = \frac{729 - k^2}{40}$:

• For $k=3$: $\alpha = \frac{729 - 3^2}{40} = \frac{729 - 9}{40} = \frac{720}{40} = 18$.
• For $k=7$: $\alpha = \frac{729 - 7^2}{40} = \frac{729 - 49}{40} = \frac{680}{40} = 17$.
• For $k=13$: $\alpha = \frac{729 - 13^2}{40} = \frac{729 - 169}{40} = \frac{560}{40} = 14$.
• For $k=17$: $\alpha = \frac{729 - 17^2}{40} = \frac{729 - 289}{40} = \frac{440}{40} = 11$.
• For $k=23$: $\alpha = \frac{729 - 23^2}{40} = \frac{729 - 529}{40} = \frac{200}{40} = 5$.

All these calculated values of $\alpha$ (18, 17, 14, 11, 5) are positive integers.

Conclusion on Number of Values

We have found 5 distinct positive integral values for $\alpha$ that satisfy the condition for rational roots:

The set of possible values for $\alpha$ is $\{5, 11, 14, 17, 18\}$.

Therefore, the total number of possible positive integral values of $\alpha$ is 5.