3x ^ 2 - 12 >= 0
To solve the inequality \( 3x^2 - 12 \geq 0 \), follow these steps:
Rearrange the inequality: \[ 3x^2 - 12 \geq 0 \]
Factor the quadratic: First, we can factor out a 3 from the inequality: \[ 3(x^2 - 4) \geq 0 \] Now, we can factor \( x^2 - 4 \) further using the difference of squares: \[ 3(x - 2)(x + 2) \geq 0 \]
Identify critical points: Set the factors equal to zero: \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
Determine the intervals: The critical points divide the number line into three intervals:
- \( (-\infty, -2) \)
- \( (-2, 2) \)
- \( (2, \infty) \)
Test each interval: Choose a test point from each interval to determine where the inequality holds:
- For the interval \( (-\infty, -2) \), choose \( x = -3 \): \[ 3(-3 - 2)(-3 + 2) = 3(-5)(-1) = 15 \quad (\text{positive}) \]
- For the interval \( (-2, 2) \), choose \( x = 0 \): \[ 3(0 - 2)(0 + 2) = 3(-2)(2) = -12 \quad (\text{negative}) \]
- For the interval \( (2, \infty) \), choose \( x = 3 \): \[ 3(3 - 2)(3 + 2) = 3(1)(5) = 15 \quad (\text{positive}) \]
Combine the results: Based on the test points, we find:
- The inequality is satisfied for \( x < -2 \) and \( x > 2 \).
- The points where the product is zero (i.e., \( x = -2 \) and \( x = 2 \)) should also be included because we have a "greater than or equal to" inequality.
Write the final solution: \[ x \in (-\infty, -2] \cup [2, \infty) \]
Thus, the solution to the inequality \( 3x^2 - 12 \geq 0 \) is: \[ \boxed{(-\infty, -2] \cup [2, \infty)} \]
