Question

The least value of $n$ such that $(n-2)x^2+8x+n+4>0\forall x\in R$, where $n\in N$.

Answer 1

$(n-2)x^2+8x+n+4>0$

$n-2>0$ ...(i)

$D<0$ ...(ii)

$\Rightarrow 64-4(n-2)(n+4)<0$

$\Rightarrow 16-(n^2+4n-2n-8)<0$

$\Rightarrow -n^2-2n+24<0$

$\Rightarrow n^2+2n-24>0$

$\Rightarrow n^2+6n-4n-24>0$

$\Rightarrow n(n+6)-4(n+6)>0$

$\Rightarrow (n-4)(n+6)>0$

$\therefore n\in (-\infty ,-6)\cup (4,\infty )$ ...(ii)

From (i) & (ii)

$\therefore n\in (4,\infty )$

$\therefore$ Least value of ${}^{\prime }{n}^{\prime }=5$