Solving Quadratic Inequality y = x² + 9x + 18 for f(x) > 0

To solve the inequality $x^2 + 9x + 18 > 0$, we start by finding the roots of the quadratic equation $x^2 + 9x + 18 = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Here $a = 1$, $b = 9$, and $c = 18$.

Compute the discriminant: $b^2 - 4ac = 9^2 - 4 \cdot 1 \cdot 18 = 81 - 72 = 9$.

Therefore, $x = \frac{-9 \pm \sqrt{9}}{2} = \frac{-9 \pm 3}{2}$.

The roots are $x = \frac{-9 + 3}{2} = -3$ and $x = \frac{-9 - 3}{2} = -6$ .

These roots divide the number line into the intervals: $(-\infty, -6)$, $(-6, -3)$, and $(-3, \infty)$.

We test a point from each interval in the inequality $x^2 + 9x + 18 > 0$ to determine where the function is positive:

• For $x < -6$, choose $x = -7$: $(-7)^2 + 9(-7) + 18 = 49 - 63 + 18 = 4$ (Positive).
• For $-6 < x < -3$, choose $x = -5$: $(-5)^2 + 9(-5) + 18 = 25 - 45 + 18 = -2$ (Negative).
• For $x > -3$, choose $x = 0$: $(0)^2 + 9(0) + 18 = 18$ (Positive).

The function $y = x^2 + 9x + 18$ is positive in the intervals $x < -6$ and $x > -3$.

Therefore, the solution is $x > -3$ or $x < -6$.