# How do you solve the inequality 16x^2+8x+1>0 and write your answer in interval notation?

Mar 8, 2017

$\left(- \infty , - \frac{1}{4}\right) U \left(- \frac{1}{4} , \infty\right)$

#### Explanation:

Factor the parabola: $y = \left(4 x + 1\right) \left(4 x + 1\right) > 0$

Find out when the function $= 0$: $\left(4 x + 1\right) = 0$
$4 x = - 1$ ; $x = - \frac{1}{4}$

Since both factors are the same: $y = {\left(4 x + 1\right)}^{2}$, $x = - \frac{1}{4}$ is a touch point on the $x$-axis.

This means we have two choices for answers:
$\left(- \infty , - \frac{1}{4}\right) U \left(- \frac{1}{4} , \infty\right)$ or no solution

To find out for sure, you can either graph the equation to see where the $y$-values are positive or you can select test points in the interval:
Let $x = - 1$, $y = 9$ so $\left(- \infty , - \frac{1}{4}\right) > 0$

Let $x = 1$, $y = 25$ so $\left(- \frac{1}{4} , \infty\right) > 0$

Therefore the solution set in interval notation is: $\left(- \infty , - \frac{1}{4}\right) U \left(- \frac{1}{4} , \infty\right)$