Chapter 3.5, Problem 13AYU

$x^2\text{ + }x\text{ > 12}$

To determine

To find: The solution of the inequality $x^2+x>12$.

Answer to Problem 13AYU

Solution:

$x<-4$ or $x>3$

Explanation of Solution

Given:

The inequality is

$x^2+x>12$

Calculation:

$x^2+x-12>0$

$x^2+4x-3x-12>0$

$x\left(x+4\right)-3\left(x+4\right)>0$

$\left(x+4\right)\left(x-3\right)>0$

$\left(x+4\right)<0$ or $\left(x-3\right)>0$

$x<-4$ or $x>3$

Therefore the solution to the given inequality is $x<-4$ or $x>3$.