AP EAMCET · Maths · Quadratic Equation
The set of all real values of \(x\) satisfying the inequality \(\frac{7 x^2-5 x-18}{2 x^2+x-6} \lt 2\) is
- A \(\left(-\infty,-\frac{2}{3}\right] \cup[3, \infty)\)
- B \(\left(-2,-\frac{2}{3}\right) \cup\left(\frac{3}{2}, 3\right)\)
- C \((-\infty,-2) \cup\left(\frac{3}{2}, \infty\right)\)
- D \(\left[-\frac{2}{3}, \frac{3}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(-2,-\frac{2}{3}\right) \cup\left(\frac{3}{2}, 3\right)\)
Step-by-step Solution
Detailed explanation
Since, \(\frac{7 x^2-5 x-18}{2 x^2+x-6} \lt 2\) \(\Rightarrow \frac{3 x^2-7 x-6}{(x+2)(2 x-3)} \lt 0 \Rightarrow \frac{(x-3)(3 x+2)}{(x+2)(2 x-3)} \lt 0\) So, \(x \in\left(-2, \frac{-2}{3}\right) \cup\left(\frac{3}{2}, 3\right)\).
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