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(A) The given quadratic equation is $x^2 + x - n = 0$.For the roots to be integers,the discriminant $D = b^2 - 4ac$ must be a perfect square of an odd

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$8$

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(A) The given quadratic equation is $x^2 + x - n = 0$.For the roots to be integers,the discriminant $D = b^2 - 4ac$ must be a perfect square of an odd integer.$D = 1^2 - 4(1)(-n) = 1 + 4n$.Let $1 + 4n = k^2$,where $k$ is an odd integer. Let $k = 2\lambda + 1$ for some integer $\lambda$.Then $1 + 4n = (2\lambda + 1)^2 = 4\lambda^2 + 4\lambda + 1$.$4n = 4\lambda^2 + 4\lambda \implies n = \lambda(\lambda + 1)$.Given $n \in [5, 100]$,we have $5 \le \lambda(\lambda + 1) \le 100$.For $\lambda = 2, n = 2(3) = 6$.For $\lambda = 3, n = 3(4) = 12$.For $\lambda = 4, n = 4(5) = 20$.For $\lambda = 5, n = 5(6) = 30$.For $\lambda = 6, n = 6(7) = 42$.For $\lambda = 7, n = 7(8) = 56$.For $\lambda = 8, n = 8(9) = 72$.For $\lambda = 9, n = 9(10) = 90$.For $\lambda = 10, n = 10(11) = 110$ (which is $> 100$).Thus,there are $8$ possible values for $n$.

Consider the equation $x^2+x-n = 0$,where $n \in N$ and $n \in [5, 100]$. The total number of different values of $n$ such that the given equation has integral roots is:

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