Question

Determine the value of x which satisfy the inequation $x^2+5x+4>0$ and $-x^2-x+42>0$

• A. (-7, -4)
• B. (-1,6)
• C. (2,3)
• D. (A) and (B) both are correct

Answer 1

The correct option is D (A) and (B) both are correct

Given,
$=>x^2+5x+4>0$
$=>x^2+4x+x+4>0$
$=>x(x+4)+1(x+4)>0$
$=>(x+4)(x+1)>0$
Case 1:
When $x+4>0;x+1>0$
$=>x>-4;x>-1$
So, $x>-1$ is the solution of case 1
Case 2:
When $x+4<0;x+1<0$
$=>x<-4;x<-1$
So, $x<-4$ is the solution of case 2
So combined solution is $x>-1$ and $x<-4$
Now, second inequation
$=>-x^2-x+42>0$
$=>x^2+x-42<0$
$=>x^2+7x-6x-42<0$
$=>x(x+7)-6(x+7)<0$
$=>(x+7)(x-6)<0$
Case 1:
When $x+7>0;x-6<0$
$=>x>-7;x<6$
So, $(-7,6)$ is the solution of case 1
Case 2:
When $x+7<0;x-6>0$
$=>x<-7;x>6$
So, all numbers except in the range $(-7,6)$ is the solution of case 2
now we have to take intersection of these answer so that x can satisfy both the equation

so take intersection and we got

$xϵ(-7,-4)\bigsqcup (-1,6)$