# How do you solve the inequality 2x^2+25x+63<=0?

Feb 1, 2017

The answer is $x \in \left[- 9 , - \frac{7}{2}\right]$

#### Explanation:

Let's factorise the LHS

$2 {x}^{2} + 25 x + 63 \le 0$

$\left(2 x + 7\right) \left(x + 9\right) \le 0$

Let $f \left(x\right) = \left(2 x + 7\right) \left(x + 9\right)$

We can build the 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}$$- 9$$\textcolor{w h i t e}{a a a a}$$- \frac{7}{2}$$\textcolor{w h i t e}{a a a a}$$- \infty$

$\textcolor{w h i t e}{a a a a}$$x + 9$$\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}$$+$

$\textcolor{w h i t e}{a a a a}$$2 x + 7$$\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 a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) \le 0$, when $x \in \left[- 9 , - \frac{7}{2}\right]$