Multiple choice

If the roots of the equation $mx^{2}+(2m-1)x+m-2=0$ are rational, then if $m\in I$ it will be

  1. odd integer

  2. even integer

  3. zero only

  4. none of these
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A Correct answer
AI explanation

For the quadratic equation mx^2 + (2m - 1)x + m - 2 = 0 to have rational roots, its discriminant D = (2m - 1)^2 - 4m(m - 2) must be a perfect square. This simplifies to D = 4m + 1, meaning 4m + 1 must be a perfect square. Let 4m + 1 = n^2, and since n^2 - 1 = 4m, n must be an odd integer for m to be an integer. Therefore, m will always be an odd integer.

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