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.
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.
\[ 30x - 9 \geq 25(x + 2)(x - 2) \]
Expand the right side:
\[ 25(x + 2)(x - 2) = 25(x^2 - 4) = 25x^2 - 100 \]
So the inequality becomes:
\[ 30x - 9 \geq 25x^2 - 100 \]
Rearranging the inequality:
\[ 0 \geq 25x^2 - 30x - 91 \]
or
\[ 25x^2 - 30x - 91 \leq 0 \]
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:
\[ x = \frac{130}{50} = 2.6 \quad \text{and} \quad x = \frac{-70}{50} = -1.4 \]
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 \( (2.6, \infty) \).
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 \( (2.6, \infty) \)): \[ 25(3)^2 - 30(3) - 91 = 225 - 90 - 91 = 44 \quad (\text{positive}) \]
Conclusion from tests: The quadratic \( 25x^2 - 30x - 91 \leq 0 \) is negative in the interval \( (-1.4, 2.6) \).
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.
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 \(2\).
