# How do I use a sign chart to solve 2< -16t^2+6<5?

Jun 30, 2015

$2 < - 16 {t}^{2} + 6 < 5$ in the intervals (-0.5, -0.25) and (0.25, 0.5).

#### Explanation:

Here, we have two inequalities that must be satisfied.

(1) $2 < - 16 {t}^{2} + 6$ and
(2) $- 16 {t}^{2} + 6 < 5$

(1) The first inequality

We write the inequality in standard form by putting all non-zero terms on the left side.

$16 {t}^{2} - 6 + 2 < 0$
$16 {t}^{2} - 4 < 0$

We start by finding the critical numbers.

Set ${f}_{1} \left(t\right) = 16 {t}^{2} - 4 = 0$ and solve for $t$.

$4 {t}^{2} - 1 = 0$

$\left(2 t + 1\right) \left(2 t - 1\right) = 0$

$2 t + 1 = 0$ or $2 t - 1 = 0$

$t = - 0.5$ or $t = + 0.5$

The critical numbers are $- 0.5$ and $+ 0.5$.

We have three intervals to consider: (-∞, -0.5), (-0.5, 0.5), and (0.5, ∞).

We pick a test number and evaluate the function and its sign at that number. (2) The second inequality

The inequality in standard form is

$- 16 {t}^{2} + 6 - 5 < 0$
$- 16 {t}^{2} + 1 < 0$ or $1 - 16 {t}^{2} < 0$

Set ${f}_{2} \left(t\right) = 1 - 16 {t}^{2} = 0$ and solve for $t$.

$\left(1 + 4 t\right) \left(1 - 4 t\right) = 0$

$1 + 4 t = 0$ or $1 - 4 t = 0$

$t = - 0.25$ or $t = + 0.25$

The critical numbers are $- 0.25$ and $+ 0.25$.

The three intervals to consider are: (-∞, -0.25), (-0.25, 0.25), and (0.25, ∞). Now we create a sign chart for the two functions. The only intervals for which the two signs are both negative are (-0.5, -0.25) and (0.25, 0.5).