Solve the following inequality. \( 2 x^{2}

Ask by Schmidt Hampton. in Qatar

Dec 07,2024

To solve the inequality \(2x^2 < x + 15\), we first rearrange it to the standard form: \[2x^2 - x - 15 < 0.\] Next, we can factor the quadratic: \[(2x + 5)(x - 3) < 0.\] To solve the inequality, we find the critical points by setting each factor to zero: 1. \(2x + 5 = 0 \rightarrow x = -\frac{5}{2}\) 2. \(x - 3 = 0 \rightarrow x = 3\) Now we test the intervals created by these points: \((-\infty, -\frac{5}{2})\), \((-frac{5}{2}, 3)\), and \((3, \infty)\). Choosing test points from each interval gives: 1. For \(x = -3\) in \((-\infty, -\frac{5}{2})\): \((2(-3) + 5)(-3 - 3) < 0 \) → Positive 2. For \(x = 0\) in \((- \frac{5}{2}, 3)\): \( (2(0) + 5)(0 - 3) < 0 \) → Negative 3. For \(x = 4\) in \((3, \infty)\): \( (2(4) + 5)(4 - 3) < 0 \) → Positive The solution is where the product is negative, which occurs in the interval: \((- \frac{5}{2}, 3)\). Thus, the solution set is: A. The solution set is \( (-\frac{5}{2}, 3) \).