To solve the quadratic inequality [tex]\(x^2 + 3x - 28 < 0\)[/tex], follow these steps:
1. Factor the Quadratic Expression:
First, we need to factor the quadratic expression [tex]\( x^2 + 3x - 28 \)[/tex]. To do this, we look for two numbers that multiply to [tex]\(-28\)[/tex] (the constant term) and add up to [tex]\(3\)[/tex] (the coefficient of the [tex]\(x\)[/tex] term). These numbers are [tex]\(7\)[/tex] and [tex]\(-4\)[/tex]. Thus, we can write:
[tex]\[
x^2 + 3x - 28 = (x + 7)(x - 4)
\][/tex]
So, the inequality becomes:
[tex]\[
(x + 7)(x - 4) < 0
\][/tex]
2. Determine the Critical Points:
The critical points occur when the expression equals zero:
[tex]\[
(x + 7)(x - 4) = 0
\][/tex]
Solving for [tex]\(x\)[/tex], we get:
[tex]\[
x = -7 \quad \text{or} \quad x = 4
\][/tex]
These points divide the number line into three intervals: [tex]\((-\infty, -7)\)[/tex], [tex]\((-7, 4)\)[/tex], and [tex]\((4, \infty)\)[/tex].
3. Test the Intervals:
We need to determine where the product [tex]\((x + 7)(x - 4)\)[/tex] is less than zero. We do this by testing a point from each interval:
- For the interval [tex]\((-\infty, -7)\)[/tex], pick [tex]\(x = -8\)[/tex]:
[tex]\[
(x + 7)(x - 4) \quad \text{becomes} \quad (-8 + 7)(-8 - 4) = (-1)(-12) = 12 \quad (\text{positive})
\][/tex]
- For the interval [tex]\((-7, 4)\)[/tex], pick [tex]\(x = 0\)[/tex]:
[tex]\[
(x + 7)(x - 4) \quad \text{becomes} \quad (0 + 7)(0 - 4) = 7 \times (-4) = -28 \quad (\text{negative})
\][/tex]
- For the interval [tex]\((4, \infty)\)[/tex], pick [tex]\(x = 5\)[/tex]:
[tex]\[
(x + 7)(x - 4) \quad \text{becomes} \quad (5 + 7)(5 - 4) = 12 \times 1 = 12 \quad (\text{positive})
\][/tex]
4. Identify the Solution:
We are looking for the values of [tex]\(x\)[/tex] for which [tex]\((x + 7)(x - 4) < 0\)[/tex]. From the interval testing, we see that this occurs in the interval where the product is negative, which is [tex]\((-7, 4)\)[/tex].
Therefore, the solution to the inequality [tex]\(x^2 + 3x - 28 < 0\)[/tex] is:
[tex]\[
-7 < x < 4
\][/tex]
