Consider the quadratic equation nx^2 + 7sqrt{ | vedclass

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(B) The given equation is $nx^2 + 7\sqrt{n}x + n = 0$.The discriminant $D$ is given by $D = (7\sqrt{n})^2 - 4(n)(n) = 49n - 4n^2 = n(49 - 4n)$.For the

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$I$ and $III$

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(B) The given equation is $nx^2 + 7\sqrt{n}x + n = 0$.The discriminant $D$ is given by $D = (7\sqrt{n})^2 - 4(n)(n) = 49n - 4n^2 = n(49 - 4n)$.For the roots to be distinct,$D 
eq 0$. Since $n$ is a positive integer,$49 - 4n = 0$ implies $n = 12.25$,which is not an integer. Thus,$D 
eq 0$ for all $n \in \mathbb{Z}^+$. Hence,statement $I$ is correct.For the roots to be real,$D \geq 0$,which implies $n(49 - 4n) \geq 0$. Since $n > 0$,we have $49 - 4n \geq 0$,so $n \leq 12.25$. The possible values for $n$ are $\{1, 2, 3, \dots, 12\}$. This is a finite set,so statement $II$ is incorrect.The product of the roots of a quadratic equation $ax^2 + bx + c = 0$ is given by $\frac{c}{a}$. Here,the product is $\frac{n}{n} = 1$,which is an integer. Hence,statement $III$ is correct.Therefore,statements $I$ and $III$ are correct.

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