Question

What are all the real values of x such that (x² + 2x + 5) / (3x² - x - 4) ≥ 0? Why are all your values the only values that satisfy the inequality?

Answer 1

The quadratic equation (x² + 2x + 5) / (3x² - x - 4) ≥ 0 has no real solutions, so there are no real values of x that satisfy the inequality.

Step-by-step explanation:

To find the real values of x that satisfy the inequality (x² + 2x + 5) / (3x² - x - 4) ≥ 0, we need to solve the equation (x² + 2x + 5) / (3x² - x - 4) = 0 first. We can rewrite this equation as (x² + 2x + 5) = 0 * (3x² - x - 4), which simplifies to x² + 2x + 5 = 0.

To solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a), where a = 1, b = 2, and c = 5. Substituting these values into the formula, we get:

1. x = (-2 + √(2² - 4 * 1 * 5)) / (2 * 1)
2. x = (-2 - √(2² - 4 * 1 * 5)) / (2 * 1)

Simplifying further, we have:

1. x = (-2 + √(-16)) / 2
2. x = (-2 - √(-16)) / 2

Since we cannot take the square root of a negative number and obtain real solutions, there are no real values of x that satisfy the equation. Therefore, there are no real values of x that satisfy the inequality (x² + 2x + 5) / (3x² - x - 4) ≥ 0.