Multiple choice

The integral values of $m$ for which the roots of the equation $mx^2+(2m-1)x+(m-2)=0$ are rational are given by the expression [where $m$ is integer]

  1. $n^2$
  2. $n(n+2)$
  3. $n(n+1)$

  4. none of these
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C Correct answer
Explanation

For the roots of a quadratic equation to be rational, its discriminant (D = b^2 - 4ac) must be a perfect square. Here, D = (2m-1)^2 - 4m(m-2) = 4m^2 - 4m + 1 - 4m^2 + 8m = 4m + 1. For 4m + 1 to be a perfect square, say k^2, then k must be odd, so k = 2n + 1. Then 4m + 1 = (2n + 1)^2 = 4n^2 + 4n + 1, which simplifies to m = n^2 + n = n(n+1).

AI explanation

For the roots of mx^2 + (2m - 1)x + (m - 2) = 0 to be rational, its discriminant must be a perfect square. The discriminant is (2m - 1)^2 - 4(m)(m - 2) = 4m^2 - 4m + 1 - 4m^2 + 8m = 4m + 1. Let the perfect square be k^2, so 4m + 1 = k^2, which rearranges to 4m = k^2 - 1. Factoring the right side gives 4m = (k - 1)(k + 1), and since k - 1 and k + 1 are consecutive even numbers when k is odd, dividing by 4 gives m = n(n + 1) where n = (k - 1)/2. Therefore, the integral values of m are given by the expression n(n + 1).

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