Question

Solution of the inequation $x^2+20x+96>0$.
A. $x<-12$ B. $x<-16$
C. $x>-10$ D. $x>-8$

• A. A or C
• B. A or D
• C. B or C
• D. B or D

Answer 1

The correct option is B A or D

Given, $x^2+20x+96>0$
$=>(x+12)(x+8)>0$
Case 1: When both $:(x+12)>0;(x+8)>0$
$=>x>-12;x>-8$
So, the solution for this case is $=>x>-8$ --- (1)
Case 2: When both $:(x+12)<0;(x+8)<0$
$=>x<-12;x<-8$
So, the solution for this case is $=>x<-12$--- (2)
Both A and D are the solutions.