# How do you solve 2u^2+5u>=12 using a sign chart?

##### 1 Answer
Nov 28, 2017

The solution is $u \in \left(- \infty , - 4\right] \cup \left[\frac{3}{2} , + \infty\right)$

#### Explanation:

Lets 's factorise the function

$f \left(u\right) = 2 {u}^{2} + 5 u - 12 = \left(2 u - 3\right) \left(u + 4\right)$

The equation we have to solve is

$f \left(u\right) = \left(2 u - 3\right) \left(u + 4\right) \ge 0$

Let's build the sign chart

$\textcolor{w h i t e}{a a a a}$$u$$\textcolor{w h i t e}{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 a a a a a a}$$\frac{3}{2}$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$u + 4$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$2 u - 3$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(u\right)$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$

Therefore,

$f \left(u\right) \ge 0$ when $u \in \left(- \infty , - 4\right] \cup \left[\frac{3}{2} , + \infty\right)$

graph{2x^2+5x-12 [-29.27, 28.45, -13.66, 15.22]}