Solve the Quadratic Inequality: When Does y = -2x² - 16x > 0?

To solve this problem, we'll perform the following steps:

• Step 1: Determine the roots of the equation $-2x^2 - 16x = 0$ using the quadratic formula.
• Step 2: Identify the intervals determined by these roots.
• Step 3: Test a value from each interval to determine where the quadratic function is positive.

Step 1: Finding the roots of the quadratic equation.

The quadratic equation is $-2x^2 - 16x = 0$. We can simplify this by factoring:

Factor out the common term: $-2x(x + 8) = 0$.

Setting each factor to zero, we find the roots:

• $-2x = 0 \implies x = 0$
• $x + 8 = 0 \implies x = -8$

Step 2: Use these roots to determine intervals on the number line: $(-\infty, -8)$, $(-8, 0)$, and $(0, \infty)$.

Step 3: Test each interval to see where the function is positive:

• Choose $x = -9$ in the interval $(-\infty, -8)$:
  $y = -2(-9)^2 -16(-9) = -162 + 144 = -18$. Negative, so the function is not positive here.
• Choose $x = -4$ in the interval $(-8, 0)$:
  $y = -2(-4)^2 -16(-4) = -32 + 64 = 32$. Positive, so the function is positive here.
• Choose $x = 1$ in the interval $(0, \infty)$:
  $y = -2(1)^2 -16(1) = -2 - 16 = -18$. Negative, so the function is not positive here.

Thus, the function $y = -2x^2 - 16x$ is positive for $x$ in the interval $(-8, 0)$.

Therefore, the values of $x$ for which $f(x) > 0$ are $-8 < x < 0$.