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Q.
The smallest natural number 'n' so that $(2-n) x^2-8 x-4-n<0, \forall x \in R$ is equal to
Complex Numbers and Quadratic Equations
Solution:
$\text { We have }( n -2) x ^2+8 x +( n +4)>0 . \forall x \in R $
$\Rightarrow n -2>0 \text { and } D <0 $
$ \Rightarrow 64-4( n -2)( n +4)<0$
$\Rightarrow n ^2+2 n -24>0$
$ \Rightarrow( n +6)( n -4)>0$
$\Rightarrow n >4 \text { as } n \in N$
$\therefore n _{\text {smallest }}=5$