Question

If the expression $(n-2)x^2+8x+(n+4)$ is negative $\forall x\in R$, then $n$ lies in

• A. $(-\infty ,-6)\cup (4,\infty )$
• B. $(4,\infty )$
• C. $(-\infty ,2)$
• D. $(-\infty ,-6)$

Answer 1

The correct option is D $(-\infty ,-6)$

$(n-2)x^2+8x+(n+4)<0\forall x\in R$ is possible when
$n-2<0\Rightarrow n<2\cdots \left(1\right)$
and $\Delta <0$
$\Rightarrow 64-4(n-2)(n+4)<0$
$\Rightarrow n^2+2n-24>0$
$\Rightarrow (n+6)(n-4)>0$
$\Rightarrow n\in (-\infty ,-6)\cup (4,\infty )\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$n\in (-\infty ,-6)$