# How do you solve the inequality x^2 - 9x> -18?

Apr 23, 2018

The solutions are $x \in \left(- \infty , 3\right) \cup \left(6 , + \infty\right)$

#### Explanation:

The inequality is

${x}^{2} - 9 x > - 18$

${x}^{2} - 9 x + 18 > 0$

$\left(x - 3\right) \left(x - 6\right) > 0$

Let $f \left(x\right) = \left(x - 3\right) \left(x - 6\right)$

Let's 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}$$- \infty$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a a}$$6$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - 3$$\textcolor{w h i t e}{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}$$x - 6$$\textcolor{w h i t e}{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}$$+$$\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 $x \in \left(- \infty , 3\right) \cup \left(6 , + \infty\right)$

graph{x^2-9x+18 [-4.29, 15.71, -3.96, 6.04]}

Apr 23, 2018

(-inf., 3) and (6, +inf.)

#### Explanation:

x^2 - 9x > - 18\$ f(x) = x^2 - 9x + 18 > 0#
Find 2 real roots, that have same sign(ac > 0), knowing the sum (- b = 9) and the product (c = 18). They are 6 and 3.
The graph of f(x) is an upward parabola (a > 0).
Inside the interval (3, 6) --> f(x) < 0 as the graph is below the x-axis.
Outside the interval (3, 6) --> f (x )> 0. Therefor, the solution set is the 2 open intervals:
(- inf., 3) and (6, +inf.)