30x-9/x-2>=25(x+2)

To solve the inequality \( \frac{30x - 9}{x - 2} \geq 25(x + 2) \), we can start by simplifying and rearranging it. 1. **Multiply both sides by \( x - 2 \)**: We need to be careful here; if \( x - 2 < 0 \), the inequality will flip. We'll handle that case later. For now, we'll assume \( x - 2 \) is positive. \[ 30x - 9 \geq 25(x + 2)(x - 2) \] 2. **Expand the right side**: \[ 25(x + 2)(x - 2) = 25(x^2 - 4) = 25x^2 - 100 \] So the inequality becomes: \[ 30x - 9 \geq 25x^2 - 100 \] 3. **Rearranging the inequality**: \[ 0 \geq 25x^2 - 30x - 91 \] or \[ 25x^2 - 30x - 91 \leq 0 \] 4. **Finding the roots of the quadratic** using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 25 \cdot (-91)}}{2 \cdot 25} \] \[ = \frac{30 \pm \sqrt{900 + 9100}}{50} \] \[ = \frac{30 \pm \sqrt{10000}}{50} \] \[ = \frac{30 \pm 100}{50} \] This gives us: \[ x = \frac{130}{50} = 2.6 \quad \text{and} \quad x = \frac{-70}{50} = -1.4 \] 5. **Identifying the intervals** based on the roots \( x = -1.4 \) and \( x = 2.6 \). - We will test three intervals: \( (-\infty, -1.4) \), \( (-1.4, 2.6) \), and \( (2.6, \infty) \). 6. **Testing intervals**: - For \( x = -2 \) (in \( (-\infty, -1.4) \)): \[ 25(-2)^2 - 30(-2) - 91 = 100 + 60 - 91 = 69 \quad (\text{positive}) \] - For \( x = 0 \) (in \( (-1.4, 2.6) \)): \[ 25(0)^2 - 30(0) - 91 = -91 \quad (\text{negative}) \] - For \( x = 3 \) (in \( (2.6, \infty) \)): \[ 25(3)^2 - 30(3) - 91 = 225 - 90 - 91 = 44 \quad (\text{positive}) \] 7. **Conclusion from tests**: The quadratic \( 25x^2 - 30x - 91 \leq 0 \) is negative in the interval \( (-1.4, 2.6) \). 8. **Considering the point where \( x - 2 = 0 \)**: \( x = 2 \) is where the original expression is undefined. Thus, we need to exclude it from our solution set. 9. **Final Result**: The solution to the inequality is: \[ x \in [-1.4, 2) \cup (2, 2.6] \] This means \( x \) can take any value from \(-1.4\) to \(2.6\) but cannot include \(2\).