Question

Find the range of the values of $x$ that satisfies the following inequation:
$x^2+8x>-12;x\in R$

• A. $x<6$ or $x>-2$
• B. $x<-6$ or $x>-2$
• C. $x<6$ or $x>2$
• D. $x<-6$ or $x>2$

Answer 1

The correct option is B $x<-6$ or $x>-2$

Given: $x^2+8x>-12$

Rule: If a term of an inequation is transferred from one side to the other side of the inequation, the sign of the term gets changed. Let's apply this rule in the above inequation.

Transfer -12 to the left side.​

$\Rightarrow x^2+8x+12>0$

$\Rightarrow x^2+6x+2x+12>0$

$\Rightarrow x(x+6)+2(x+6)>0$​

$(x+2)(x+6)>0$​

Case 1:​

$x+2>0\text{and}x+6>0$​

Rule: If a term of an inequation is transferred from one side to the other side of the inequation, the sign of the term gets changed. Let's apply this rule in the above inequation.

Transfer 2 and 6 to the right side. ​

$x>-2\text{and}x>-6$

So, $x>-2$

Case 2:​

$x+2<0\text{and}x+6<0$​

Transfer 2 and 6 to the right side.​

$x<-2\text{and}x<-6$​

So, $x<-6$​

Hence, the final solution is $x<-6$ or $x>-2$.