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The number of integers $n$ such that the quadratic equation $nx^2+(n+1)x+(n+2)=0$ has rational roots is
$(A)0\hspace{1cm}(B)1\hspace{1cm}(C)2\hspace{1cm}(D)3$

The condition for the rational roots is that the discriminant should be greater than or equal to zero and discriminant should be a perfect square.
Discriminant$=(n+1)^2-4n(n+2)=n^2+2n+1-4n^2-8n=-3n^2-6n+1$
$-3n^2-6n+1\geq 0.......................(1)$
I dont know how to apply the perfect square condition here.With the equation $(1)$,i get only one integer $n$.But the correct answer given in my book is $2$.Two integer $n$ are possible.$n=-1,n=-2$.

Please help me.Thanks

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Complete the square: $$-3n^2-6n+1 = -3n^2-6n-3+3+1 = -3(n+1)^2+4.$$ For this to be nonnegative, we must have $(n+1)^2 \le \frac{4}{3}$. So $n=0, -1, -2$ are the only solutions, and $n=0$ gives a linear, not a quadratic, equation (though this special case would probably stop me from using this problem).

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