Multiple choice

The number of integral values of $m$ less than $50$ such that the roots of the quadratic equation $mx^{2}+\left ( 2m-1 \right )x+\left ( m-2 \right )=0$ are rational are

  1. $6$
  2. $7$
  3. $13$
  4. $14$
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A Correct answer
Explanation

For roots to be rational, the discriminant D = (2m-1)^2 - 4(m)(m-2) must be a perfect square. D = 4m^2 - 4m + 1 - 4m^2 + 8m = 4m + 1. For 4m+1 to be a perfect square, let 4m+1 = k^2. Since 4m+1 is odd, k must be odd. k^2 = 1, 9, 25, 49, 81, 121, 169, 225, 289. 4m+1 = k^2 => m = (k^2-1)/4. For m < 50, k^2 < 201. Possible odd k^2 values: 1, 9, 25, 49, 81, 121, 169. This gives m = 0, 2, 6, 12, 20, 30, 42. There are 7 values, but m cannot be 0 for a quadratic equation. So 6 values.

AI explanation

For the roots to be rational, the discriminant D = (2m - 1)^2 - 4m(m - 2) must be a perfect square. Expanding gives D = 4m^2 - 4m + 1 - 4m^2 + 8m = 4m + 1. For m < 50, the maximum perfect square is achieved at 4m + 1 = 225, giving m = 56, which is too large. The largest perfect square less than 50 occurs at 4m + 1 = 169, so m = 42. The perfect squares are 1, 9, 25, 49, 81, 121, 169, yielding m = 0, 2, 6, 12, 20, 30, 42. Because the quadratic requires m to not be zero, there are exactly 6 valid positive integral values of m.

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