# How do you solve 2v^2+7v<4 using a sign chart?

Dec 31, 2016

The answer is v in ] -4,1/2 [

#### Explanation:

Let's rewrite the inequality as

$2 {v}^{2} + 7 v - 4 < 0$

Let $f \left(v\right) = 2 {v}^{2} + 7 v - 4$

The domain of $f \left(v\right)$ is ${D}_{f} \left(v\right) = \mathbb{R}$

Let's factorise the expression

$2 {v}^{2} + 7 v - 4 = \left(2 v - 1\right) \left(v + 4\right)$

Now we establish the sign chart

$\textcolor{w h i t e}{a a a a}$$v$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a}$$\frac{1}{2}$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

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

$\textcolor{w h i t e}{a a a a}$$2 v - 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$

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

Therefore,

$f \left(v\right) < 0$ when v in ] -4,1/2 [