Solve the Quadratic Inequality: When is -x² + 9 Greater Than Zero?

We begin with the function $y = -x^2 + 9$ and want to find for which values of $x$, $y > 0$.

Step 1: Solve for the roots of the related equation $-x^2 + 9 = 0$.
Rearranging gives: $x^2 = 9$.

Step 2: Solve for $x$.
Taking the square root of both sides gives: $x = \pm 3$.

Step 3: Determine intervals for the inequality $-x^2 + 9 > 0$.
Consider test points in the intervals determined by $x = -3$ and $x = 3$.

• For $x$ in the interval: $(-\infty, -3)$, select $x = -4$.
  Calculate: $y = -(-4)^2 + 9 = -16 + 9 = -7$. Thus, $y < 0$.
• For $x$ in the interval: $(-3, 3)$, select $x = 0$.
  Calculate: $y = -(0)^2 + 9 = 9$. Thus, $y > 0$.
• For $x$ in the interval: $(3, \infty)$, select $x = 4$.
  Calculate: $y = -(4)^2 + 9 = -16 + 9 = -7$. Thus, $y < 0$.

Therefore, $y = -x^2 + 9$ is positive for $x$ values within the interval $-3 < x < 3$.

Therefore, the solution is $-3 < x < 3$.