Solve the Quadratic Inequality for y=-2x²+32: When is f(x) Greater Than Zero?

Look at the following function:

$y=-2x^2+32$

Determine for which values of $x$ the following is true:

$f\left(x\right) > 0$

To solve this problem, we'll follow these steps:

• Step 1: Solve for the roots of the equation $f(x) = -2x^2 + 32 = 0$.
• Step 2: Use the roots to determine intervals and test where the function is positive.

Let's work through each step:

Step 1: Solve the equation for roots:
The equation given is $-2x^2 + 32 = 0$. We can find the roots by isolating $x^2$:

$-2x^2 + 32 = 0$
$-2x^2 = -32$
$x^2 = 16$

Taking the square root of both sides gives $x = \pm 4$. So, the roots are $x = 4$ and $x = -4$.

Step 2: Determine the intervals and test for positivity:

The roots split the real number line into the intervals $(-\infty, -4)$, $(-4, 4)$, and $(4, \infty)$. We test the sign of $f(x)$ within these intervals:

• For $x \in (-\infty, -4)$, choose $x = -5$: $f(-5) = -2(-5)^2 + 32 = -50 + 32 = -18$. So, $f(x) < 0$.
• For $x \in (-4, 4)$, choose $x = 0$: $f(0) = -2(0)^2 + 32 = 32$. So, $f(x) > 0$.
• For $x \in (4, \infty)$, choose $x = 5$: $f(5) = -2(5)^2 + 32 = -50 + 32 = -18$. So, $f(x) < 0$.

Therefore, the function $f(x) = -2x^2 + 32$ is positive only between the roots, i.e., in the interval $-4 < x < 4$.

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