Certainly! To determine the solution for the quadratic inequality [tex]\(x^2 - 3x - 28 \geq 0\)[/tex], we follow these steps:
1. Identify the quadratic equation: We start with the quadratic expression given in the inequality:
[tex]\[
x^2 - 3x - 28 \geq 0.
\][/tex]
2. Find the roots of the quadratic equation: The next step is to find the roots of the quadratic equation by setting it equal to zero:
[tex]\[
x^2 - 3x - 28 = 0.
\][/tex]
To factorize this equation, we look for two numbers that multiply to [tex]\(-28\)[/tex] and add up to [tex]\(-3\)[/tex]. The numbers [tex]\(-7\)[/tex] and [tex]\(4\)[/tex] satisfy these conditions:
[tex]\[
(x - 7)(x + 4) = 0.
\][/tex]
So, the roots are:
[tex]\[
x = 7 \quad \text{and} \quad x = -4.
\][/tex]
3. Analyze the intervals determined by the roots: The roots [tex]\(x = 7\)[/tex] and [tex]\(x = -4\)[/tex] divide the real number line into three intervals:
- [tex]\( (-\infty, -4) \)[/tex]
- [tex]\( [-4, 7] \)[/tex]
- [tex]\( (7, \infty) \)[/tex]
We will test the sign of [tex]\(x^2 - 3x - 28\)[/tex] in each of these intervals to see where the quadratic expression is greater than or equal to zero.
4. Test intervals:
- For the interval [tex]\( (-\infty, -4) \)[/tex]:
- Choose a test point [tex]\(x = -5\)[/tex]:
[tex]\[
(-5)^2 - 3(-5) - 28 = 25 + 15 - 28 = 12 \quad (\geq 0).
\][/tex]
- For the interval [tex]\( (-4, 7) \)[/tex]:
- Choose a test point [tex]\(x = 0\)[/tex]:
[tex]\[
0^2 - 3(0) - 28 = -28 \quad (< 0).
\][/tex]
- For the interval [tex]\( (7, \infty) \)[/tex]:
- Choose a test point [tex]\(x = 8\)[/tex]:
[tex]\[
8^2 - 3(8) - 28 = 64 - 24 - 28 = 12 \quad (\geq 0).
\][/tex]
It's also important to verify the values at [tex]\(x = -4\)[/tex] and [tex]\(x = 7\)[/tex]:
- At [tex]\(x = -4\)[/tex]:
[tex]\[
(-4)^2 - 3(-4) - 28 = 16 + 12 - 28 = 0 \quad (\geq 0).
\][/tex]
- At [tex]\(x = 7\)[/tex]:
[tex]\[
7^2 - 3(7) - 28 = 49 - 21 - 28 = 0 \quad (\geq 0).
\][/tex]
5. Combine the results: From the test points, we observe that the quadratic expression [tex]\(x^2 - 3x - 28\)[/tex] is greater than or equal to zero in the intervals [tex]\((-\infty, -4]\)[/tex] and [tex]\([7, \infty)\)[/tex].
The solution to the inequality [tex]\(x^2 - 3x - 28 \geq 0\)[/tex] is:
[tex]\[
(-\infty, -4] \cup [7, \infty).
\][/tex]
Therefore, the correct answer is:
[tex]\[
(-\infty, -4] \text{ or } [7, \infty).
\][/tex]
