Question

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

Answer 1

Given, $(n-2)x^2+8x+n+4>0\forall x$ where $n\in N$

$\therefore$ The above given quadratic equation has real roots.

$\Rightarrow$ Discriminant$=b^2-4ac>0$

$\Rightarrow (8{)}^2-4(n-2\left)\right(n+4)>0$

$\Rightarrow 64-4\{n^2+4n-2n-8\}>0$

$\Rightarrow 64-4\{n^2+2n-8\}>0$

$\Rightarrow 64-4n^2-8n+32>0$

$\Rightarrow -4n^2-8n+96>0$

$\Rightarrow 4n^2+8n-96<0$

$\Rightarrow 4\{n^2+2n-24\}<0$

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

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

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

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

$\therefore n-4<0$ or $n+6<0$

$\Rightarrow n<4$ or $n<-6$ which is neglected since $n\in N$

Here $n<4$, which can be written as $n\le 3$

$\therefore$ The least value of n is ${}^{\prime }{3}^{\prime }$.