# How do you solve 3(2w^2-5)<w using a sign chart?

Jan 31, 2017

The answer is w in]-3/2, 5/3 [

#### Explanation:

Let's rewrite the inequality

$6 {w}^{2} - w - 6 < 0$

Let's factorise

$\left(3 w - 5\right) \left(2 w + 3\right) < 0$

Let $f \left(w\right) = \left(3 w - 5\right) \left(2 w + 3\right)$

Now, we can build the sign chart

$\textcolor{w h i t e}{a a a a}$$w$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \frac{3}{2}$$\textcolor{w h i t e}{a a a a}$$\frac{5}{3}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$2 w + 3$$\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}$$3 w - 5$$\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}$$f \left(w\right)$$\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,

$f \left(w\right) < 0$ when w in]-3/2, 5/3 [