# How do you solve the inequality: 16x^2 + 24x < -9?

Aug 29, 2015

$x = \emptyset$

#### Explanation:

First, move $- 9$ to the other side of the inequality

$16 {x}^{2} + 24 x + 9 < 0$

Now, notice what happens when use $16$ as a common factor for all the terms

$16 \left({x}^{2} + \frac{24}{16} x + \frac{9}{16}\right) < 0$

You can write this as

$16 \cdot \left[{x}^{2} + 2 \cdot \left(\frac{3}{4}\right) x + {\left(\frac{3}{4}\right)}^{2}\right] < 0$

$16 \cdot {\left(x + \frac{3}{4}\right)}^{2} < 0$

The left-side of the inequality will always be positive, since ${\left(x + \frac{3}{4}\right)}^{2}$ is positive regardless of the value of $x$.

For $x = - \frac{3}{4}$, the left-hand side of the inequality will be equal to zero, the smallest possible value it can take.

This means that your original inequality will not be satisfied, regardless of the value of $x \in \mathbb{R}$ you use.