Multiple choice

The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation, $6x^ 2-11x+ \alpha = 0$ are rational numbers is

  1. $2$
  2. $5$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots are rational if the discriminant D = b^2 - 4ac = (-11)^2 - 4(6)(alpha) = 121 - 24 * alpha is a perfect square. For alpha = 1, D = 97 (no). For alpha = 2, D = 121 - 48 = 73 (no). For alpha = 3, D = 121 - 72 = 49 (yes, 7^2). For alpha = 4, D = 121 - 96 = 25 (yes, 5^2). For alpha = 5, D = 121 - 120 = 1 (yes, 1^2). For alpha > 5, D < 0. There are 3 values.

AI explanation

For the roots of the quadratic equation to be rational, its discriminant must be a perfect square. The discriminant is (-11)^2 - 4(6)(a) = 121 - 24a, so we set 121 - 24a = k^2 for some non-negative integer k. Rearranging gives 24a = 121 - k^2, meaning k^2 must be less than 121; testing values reveals that k^2 can be 1, 25, or 49. Solving for a in these three cases yields the positive integers a = 5, a = 4, and a = 3, resulting in 3 possible values.

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