Question

If $x^2+6x-27>0$ and $x^2-3x-4<0$ then

• A. $x>3$
• B. $x<4$
• C. $3<x<4$
• D. $x=3\frac{1}{2}$

Answer 1

Answer: (C)

$x^2+6x-27\ge 0$ and $x^2-3x+4<0$
$\Rightarrow$ $(x+9)(x-3)>0$ and $(x-4)(x+1)<0$.
$\Rightarrow$ $x>-9,x>3x-4>0$
or $x<-9,x<3x+1<0$
$\Rightarrow$ $x>3$ or $x<-9$ or $x-4<0,x+1>0$
$i.e.,x>4,x<-1$ or $x<4,x>-1$
$\therefore -1<x<4$, other case is not possible.
Thus $x>3$ or $x<-9$ and $-1<x<4$.
Thus $3<x<4$.