Question

The least value of n for which $(n-2)x^2+8x+n+4>0$ where $n\in N$ is.

• A. $4$
• B. $6$
• C. $5$
• D. $3$

Answer 1

The correct option is C $5$

$\Delta =b^2-4ac<0$

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

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

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

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

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

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

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

$\Rightarrow n+6<0$ or $n-4>0$

$\Rightarrow n<-6$ or $n>4$

$\Rightarrow n>4$ for $n\in N$

$\therefore n$ can take value of greater than $4$ which are natural numbers.

$\therefore n=5,6,..$

The least value of $n=5$