# In order that the equation 2ax(ax+nc) +(n^2-2)c^2=0 may have real roots , n belongs to interval

Solution:   The given equation  is $2ax(ax+nc)+(n^{2}-2)c^{2}=0$

It has real roots , so

$\\ \\ \Rightarrow \hspace{1cm} D\geq 0\Rightarrow (2anc)^{2}-4\cdot 2a^{2}(n^{2}-2)c^{2}\geq 0\\ \\ \Rightarrow \hspace{1cm}n^{2}-2(n^{2}-2)\geq0\\ \\ \Rightarrow \hspace{1cm}n^{2}-4\leq0\Rightarrow (n+2)(n-2)\leq0\\ \\ \Rightarrow \hspace{1cm}n\in [-2,2] \\ \\ \because \hspace{1cm}4a^{2}c^{2} \hspace{0.2cm}is \hspace{0.2cm}positive$

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