If a, b, c are distinct odd natural numbers, | vedclass

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(A) For the equation $ax^2 + bx + c = 0$ to have rational roots, the discriminant $D = b^2 - 4ac$ must be a perfect square of a rational number. Since

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must be $0$

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(A) For the equation $ax^2 + bx + c = 0$ to have rational roots, the discriminant $D = b^2 - 4ac$ must be a perfect square of a rational number. Since $a, b, c$ are integers, $D$ must be a perfect square of an integer.Given $a, b, c$ are odd natural numbers, $b^2$ is odd and $4ac$ is even. Thus, $D = b^2 - 4ac$ is an odd integer.Let $D = (2k + 1)^2$ for some integer $k$. Then $b^2 - 4ac = (2k + 1)^2$.Rearranging gives $4ac = b^2 - (2k + 1)^2 = (b - (2k + 1))(b + (2k + 1))$.Since $b$ is odd, let $b = 2n + 1$. Then $4ac = (2n + 1 - 2k - 1)(2n + 1 + 2k + 1) = (2n - 2k)(2n + 2k + 2) = 4(n - k)(n + k + 1)$.So, $ac = (n - k)(n + k + 1)$.Note that $(n - k)$ and $(n + k + 1)$ differ by $(n + k + 1) - (n - k) = 2k + 1$, which is odd. Thus, one of $(n - k)$ or $(n + k + 1)$ must be even, making their product $ac$ even.However, the product of two odd numbers $a$ and $c$ must be odd. This is a contradiction.Therefore, $D$ cannot be a perfect square, and the equation has no rational roots.Thus, the number of rational roots is $0$.

If $a, b, c$ are distinct odd natural numbers, then the number of rational roots of the equation $ax^2 + bx + c = 0$ is:

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