# How do you solve and write the following in interval notation: x^2 + 7x + 10 >0?

May 23, 2017

The solution is $x \in \left(- \infty , - 5\right) \cup \left(- 2 , + \infty\right)$

#### Explanation:

Let's factorise the inequality

${x}^{2} + 7 x + 10 = \left(x + 2\right) \left(x + 5\right)$

So,

$\left(x + 2\right) \left(x + 5\right) > 0$

Let $f \left(x\right) = \left(x + 2\right) \left(x + 5\right)$

We build a 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}$$- 5$$\textcolor{w h i t e}{a a a a}$$- 2$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 5$$\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}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 2$$\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}$$+$

$\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 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 $x \in \left(- \infty , - 5\right) \cup \left(- 2 , + \infty\right)$