`x^2+x-12lt0`

Options

A. `-ooltx-4` and `3ltxltoo`
B. `[-4,3]`
C. `(-oo,4)uu(3,oo)`
D. `(-4,3)`

Answer

To solve the inequality \( x^2 + x - 12 < 0 \), we will follow these steps:

### Step 1: Factor the quadratic expression
We start with the quadratic equation \( x^2 + x - 12 \). We need to factor it. We look for two numbers that multiply to \(-12\) (the constant term) and add to \(1\) (the coefficient of \(x\)).

The numbers \(4\) and \(-3\) satisfy these conditions because:
- \(4 \times (-3) = -12\)
- \(4 + (-3) = 1\)

Thus, we can factor the expression as:
\[ (x + 4)(x - 3) < 0 \]

### Step 2: Find the critical points
Next, we find the critical points by setting each factor equal to zero:
1. \(x + 4 = 0 \Rightarrow x = -4\)
2. \(x - 3 = 0 \Rightarrow x = 3\)

So, the critical points are \(x = -4\) and \(x = 3\).

### Step 3: Determine the intervals
The critical points divide the number line into three intervals:
1. \( (-\infty, -4) \)
2. \( (-4, 3) \)
3. \( (3, \infty) \)

### Step 4: Test each interval
We will test a point from each interval to see where the inequality holds true.

1. **Interval \( (-\infty, -4) \)**: Choose \(x = -5\)
\[ (-5 + 4)(-5 - 3) = (-1)(-8) = 8 > 0 \quad \text{(not a solution)} \]

2. **Interval \( (-4, 3) \)**: Choose \(x = 0\)
\[ (0 + 4)(0 - 3) = (4)(-3) = -12 < 0 \quad \text{(solution)} \]

3. **Interval \( (3, \infty) \)**: Choose \(x = 4\)
\[ (4 + 4)(4 - 3) = (8)(1) = 8 > 0 \quad \text{(not a solution)} \]

### Step 5: Write the solution
The solution to the inequality \( x^2 + x - 12 < 0 \) is the interval where the inequality holds true:
\[ \boxed{(-4, 3)} \]