# How do solve x^2+x<12 and write the answer as a inequality and interval notation?

Jan 16, 2017

The answer is x in ] -4,3 [
$- 4 < x < 3$

#### Explanation:

The inequation is

${x}^{2} + x - 12 < 0$

Let's factorise the expression

${x}^{2} + x - 12 = \left(x - 3\right) \left(x + 4\right)$

Let $f \left(x\right) = {x}^{2} + x - 12$

Now, we can do the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$- \infty$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 3$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) < 0$, when x in ] -4,3 [

or $- 4 < x < 3$