Find the Positive x-values in -x² - 6x - 8 > 0

To determine the values of $x$ for which the quadratic function $y = -x^2 - 6x - 8$ is greater than 0, we will first find the roots of the quadratic equation where it equals zero.

We apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute $a = -1$, $b = -6$, and $c = -8$ into the quadratic formula:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(-8)}}{2(-1)}$

Simplifying inside the square root and the rest of the expression:

$x = \frac{6 \pm \sqrt{36 - 32}}{-2}$ $x = \frac{6 \pm \sqrt{4}}{-2}$

Since $\sqrt{4} = 2$, the equation becomes:

$x = \frac{6 \pm 2}{-2}$

This gives us two potential solutions:

- $x = \frac{8}{-2} = -4$

- $x = \frac{4}{-2} = -2$

The roots divide the x-axis into three intervals: $x < -4$, $-4 < x < -2$, and $x > -2$.

To find where the function is positive, choose test points from these intervals:

• For $x < -4$ (e.g., $x = -5$): $f(-5) = -(-5)^2 - 6(-5) - 8 = -25 + 30 - 8 = -3$
• For $-4 < x < -2$ (e.g., $x = -3$): $f(-3) = -(-3)^2 - 6(-3) - 8 = -9 + 18 - 8 = 1$
• For $x > -2$ (e.g., $x = 0$): $f(0) = -(0)^2 - 6(0) - 8 = -8$

From this, the function is positive on the interval $-4 < x < -2$.

Therefore, the solution to the problem is $-4 < x < -2$.