To solve the inequality [tex]\( f(x) = x^2 + 4x - 45 \geq 0 \)[/tex], we need to find the regions where the function [tex]\( f(x) \)[/tex] is greater than or equal to zero.
### Step-by-Step Solution:
1. Find the roots of the quadratic equation:
To determine where the function changes sign, first find the roots of the quadratic equation [tex]\( f(x) = x^2 + 4x - 45 = 0 \)[/tex]. These are the points where the function intersects the x-axis.
2. Solve the quadratic equation [tex]\( x^2 + 4x - 45 = 0 \)[/tex]:
This can be solved using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
[tex]\[
a = 1, \quad b = 4, \quad c = -45
\][/tex]
Plugging in the values, we get:
[tex]\[
x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1}
\][/tex]
This simplifies to:
[tex]\[
x = \frac{-4 \pm \sqrt{16 + 180}}{2}
\][/tex]
[tex]\[
x = \frac{-4 \pm \sqrt{196}}{2}
\][/tex]
[tex]\[
x = \frac{-4 \pm 14}{2}
\][/tex]
Thus, we have two roots:
[tex]\[
x = \frac{-4 + 14}{2} = 5
\][/tex]
[tex]\[
x = \frac{-4 - 14}{2} = -9
\][/tex]
3. Critical points and test intervals:
The roots [tex]\( x = -9 \)[/tex] and [tex]\( x = 5 \)[/tex] are critical points. We will test the intervals around these points to determine where the function is non-negative.
The intervals to test are:
- [tex]\( x \leq -9 \)[/tex]
- [tex]\( -9 \leq x \leq 5 \)[/tex]
- [tex]\( x \geq 5 \)[/tex]
4. Test the intervals:
- For [tex]\( x \leq -9 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = -10 \)[/tex]:
[tex]\[
f(-10) = (-10)^2 + 4(-10) - 45 = 100 - 40 - 45 = 15
\][/tex]
Since [tex]\( f(-10) > 0 \)[/tex], [tex]\( f(x) \)[/tex] is non-negative in the interval [tex]\( x \leq -9 \)[/tex].
- For [tex]\( -9 \leq x \leq 5 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = 0^2 + 4(0) - 45 = -45
\][/tex]
Since [tex]\( f(0) < 0 \)[/tex], [tex]\( f(x) \)[/tex] is negative in the interval [tex]\( -9 \leq x \leq 5 \)[/tex].
- For [tex]\( x \geq 5 \)[/tex]:
Choose a point in this interval, such as [tex]\( x = 6 \)[/tex]:
[tex]\[
f(6) = 6^2 + 4(6) - 45 = 36 + 24 - 45 = 15
\][/tex]
Since [tex]\( f(6) > 0 \)[/tex], [tex]\( f(x) \)[/tex] is non-negative in the interval [tex]\( x \geq 5 \)[/tex].
5. Conclusion:
The function [tex]\( f(x) = x^2 + 4x - 45 \)[/tex] is non-negative in the intervals where [tex]\( x \leq -9 \)[/tex] and [tex]\( x \geq 5 \)[/tex]. Thus, the solution to the inequality [tex]\( f(x) \geq 0 \)[/tex] is:
[tex]\[
x \in (-\infty, -9] \cup [5, \infty)
\][/tex]
Therefore, the regions where the quadratic function is greater than or equal to zero are [tex]\( x \leq -9 \)[/tex] and [tex]\( x \geq 5 \)[/tex].
