Explore Quadratic Inequalities: Solve y = -2x² + 32 Where f(x) < 0

Let's solve the problem by following these steps:

• Step 1: Determine the roots of $f(x) = -2x^2 + 32$.
• Step 2: Find where $f(x)$ changes sign.
• Step 3: Determine intervals where $f(x) < 0$.

Step 1: Solving the equation $-2x^2 + 32 = 0$.

Rearrange to find the roots:

$-2x^2 + 32 = 0$ implies $2x^2 = 32$, and dividing both sides by 2 gives us:

$x^2 = 16$.

Taking the square root on both sides results in:

$x = \pm 4$.

Step 2: Identify the intervals defined by the roots $x = -4$ and $x = 4$.

We have three intervals to test: $(-\infty, -4)$, $(-4, 4)$, and $(4, \infty)$.

Step 3: Analyze the sign of the function in each interval:

• For $x < -4$: Choose $x = -5$ as a test point. Substitute into the function:
  $f(-5) = -2(-5)^2 + 32 = -2(25) + 32 = -50 + 32 = -18$.
  The function is negative.
• For $-4 < x < 4$: Choose $x = 0$ as a test point. Substitute into the function:
  $f(0) = -2(0)^2 + 32 = 32$.
  The function is positive.
• For $x > 4$: Choose $x = 5$ as a test point. Substitute into the function:
  $f(5) = -2(5)^2 + 32 = -2(25) + 32 = -50 + 32 = -18$.
  The function is negative.

Therefore, the function is negative for $x > 4$ or $x < -4$.

The correct choice is: $x > 4$ or $x < -4$.