Solve Quadratic Inequality: When is x²+9x+18 Less Than Zero?

Look at the following function:

$y=x^2+9x+18$

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

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

To find for which values of $x$ the function $f(x) = x^2 + 9x + 18$ is less than 0, we first find the roots of the quadratic equation:

Step 1: Calculate the discriminant $b^2 - 4ac$ from the quadratic formula:
For $f(x) = x^2 + 9x + 18$, we have $a = 1$, $b = 9$, and $c = 18$.
The discriminant is $9^2 - 4 \times 1 \times 18 = 81 - 72 = 9$.

Step 2: Find the roots using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{9}}{2 \times 1} = \frac{-9 \pm 3}{2}$

Thus, the roots are:
$x_1 = \frac{-9 + 3}{2} = -3$
$x_2 = \frac{-9 - 3}{2} = -6$

Step 3: Analyze the sign of $f(x)$ around these roots:

The parabola opens upwards (since the coefficient of $x^2$ is positive), so it will be below the x-axis between the roots. This means $f(x) < 0$ for $-6 < x < -3$.

Therefore, the solution is:

$-6 < x < -3$.