Question

Find the values of $m$ for which $(m-2)x^2+8x+m+4>0$ for all real $x$

• A. $[2,3]$
• B. $(2,3)$
• C. $(4,\infty )$
• D. $(-\infty ,-6)$

Answer 1

Answer: (D)

$(-\infty ,-6)$

Equation is positive when m-2>0 and discriminant is less than zero

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

$\Rightarrow 64-4m^2-8m+32<0$

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

$\Rightarrow 4(m^2+2m-24)>0$

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

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

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

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

But in first condition $m<2$

taking intersection of $m<2$ and $m<-6$ and $m>4$

$\therefore m\in (-\infty ,-6)$